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SECTION 16.1 (PAGE 858)
CHAPTER 16. VECTOR CALCULUS Section 16.1 (page 858) 1.
2.
F = xi + yj ∂ ∂ ∂ (x) + (y) + (0) = 1 + 1 = 2 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = =0 ∂ x ∂ y ∂z x y 0 F = yi + xj ∂ ∂ ∂ (y) + (x) + (0) = 0 + 0 = 0 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = (1 − 1)k = 0 ∂ x ∂ y ∂z y x 0 F = yi + zj + xk ∂ ∂ ∂ div F = (y) + (z) + (x) = 0 ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = −i − j − k ∂ x ∂ y ∂z y z x
4.
F = yzi + x zj + x yk ∂ ∂ ∂ (yz) + (x z) + (x y) = 0 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = ∂ x ∂ y ∂z yz x z x y = (x − x)i + (y − y)j + (z − z)k = 0
5.
F = xi + xk ∂ ∂ ∂ (x) + (0) + (x) = 1 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = −j ∂ x ∂ y ∂z x 0 x F = x y 2 i − yz 2j + zx 2 k ∂ 2 ∂ ∂ 2 div F = xy + −yz 2 + zx ∂x ∂y ∂z = y2 − z2 + x 2 j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 2 2 x y −yz zx = 2yzi − 2x zj − 2x yk
600
7.
F = f (x)i + g(y)j + h(z)k ∂ ∂ ∂ f (x) + g(y) + h(z) div F = ∂x ∂y ∂z = f (x) + g (y) + h (z) i j k ∂ ∂ ∂ curl F = =0 ∂x ∂y ∂z f (x) g(y) h(z)
8.
F = f (z)i − f (z)j ∂ ∂ f (z) + div F = − f (z) = 0 ∂x ∂y i j k ∂ ∂ ∂ curl F = = f (z)(i + j) ∂x ∂y ∂z f (z) − f (z) 0
Gradient, Divergence, and Curl
3.
6.
R. A. ADAMS: CALCULUS
9. Since x = r cos θ , and y = r sin θ , we have r 2 = x 2 + y 2 , and so
x ∂r = = cos θ ∂x r y ∂r = = sin θ ∂y r ∂ ∂ y −x y cos θ sin θ sin θ = = 3 =− ∂x ∂x r r r ∂ y 1 y2 ∂ sin θ = = − 3 ∂y ∂y r r r 2 2 x cos θ = 3 = r r ∂ x 1 x2 ∂ cos θ = = − 3 ∂x ∂x r r r y2 sin2 θ = 3 = r r ∂ ∂ x −x y cos θ sin θ cos θ = = 3 =− . ∂y ∂y r r r (The last two derivatives are not needed for this exercise, but will be useful for the next two exercises.) For F = r i + sin θ j, we have ∂ cos2 θ ∂r + sin θ = cos θ + ∂x ∂y r i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z r sin θ 0 sin θ cos θ − sin θ k. = − r
div F =
INSTRUCTOR’S SOLUTIONS MANUAL
10.
F = rˆ = cos θ i + sin θ j sin2 θ
11.
It follows that
cos2 θ
1 1 + = = r r r x 2 + y2 i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z cos θ sin θ 0 cos θ sin θ cos θ sin θ =− − k=0 r r
div F =
SECTION 16.1 (PAGE 858)
F = θˆ = − sin θ i + cos θ j cos θ sin θ cos θ sin θ − =0 div F = r r i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z − sin θ cos θ 0 2 sin θ cos2 θ 1 1 = + k= k= k r r r x 2 + y2
1 ˆ d S = div F(0, 0, 0). F•N a,b,c→0+ 8abc Ba,b,c lim
13. This proof just mimics that of Theorem 1. F can be expanded in Maclaurin series F = F 0 + F1 x + F2 y + · · · , where F0 = F(0, 0)
∂ ∂ F1 F1 = = F(x, y) i+ ∂x ∂x (0,0) ∂ ∂ F1 F(x, y) i+ F2 = = ∂y ∂y
12. We use the Maclaurin expansion of F, as presented in the proof of Theorem 1: F = F 0 + F1 x + F2 y + F3 z + · · · , where F0 = F(0, 0, 0)
∂ F2 ∂ F3 ∂ ∂ F1 F(x, y, z) i+ j+ k = ∂x ∂x ∂x ∂x (0,0,0) (0,0,0) ∂ F ∂ F ∂ ∂ F 1 2 3 F(x, y, z) i+ j+ k F2 = = ∂y ∂y ∂y ∂y (0,0,0) (0,0,0) ∂ F ∂ F ∂ ∂ F 1 2 3 F(x, y, z) i+ j+ k F3 = = ∂z ∂z ∂z ∂z (0,0,0) (0,0,0) F1 =
and where · · · represents terms of degree 2 and higher in x, y, and z. ˆ = k. On the top of the box Ba,b,c , we have z = c and N ˆ = −k. On the bottom of the box, we have z = −c and N On both surfaces d S = d x d y. Thus ˆ dS + F•N
top a
bottom b
d y cF3 • k − cF3 • (−k) + · · · −a −b ∂ = 8abcF3 • k + · · · = 8abc F3 (x, y, z) + ···, ∂z (0,0,0)
=
dx
where · · · represents terms of degree 4 and higher in a, b, and c. Similar formulas obtain for the two other pairs of faces, and the three formulas combine into ˆ d S = 8abcdiv F(0, 0, 0) + · · · . F•N Ba,b,c
(0,0)
∂ F2 j ∂x (0,0) ∂ F2 j ∂y (0,0)
and where · · · represents terms of degree 2 and higher in x and y. On the curve C of radius centred at (0, 0), we have ˆ = 1 (xi + yj). Therefore, N ˆ = 1 F0 • ix + F0 • jy + F1 • ix 2 F•N + F1 • jx y + F2 • ix y + F2 • jy 2 + · · · where · · · represents terms of degree 3 or higher in x and y. Since C C
x ds = x 2 ds =
C
y ds =
C
C y 2 ds =
x y ds = 0
0
2π
2 cos2 θ dθ = π 3 ,
we have 3 1 ˆ ds = 1 π (F1 • i + F2 • j) + · · · F•N 2 2 π C π = div F(0, 0) + · · · where · · · represents terms of degree 1 or higher in . Therefore, taking the limit as → 0 we obtain lim
→0
1 ˆ ds = div F(0, 0). F•N π 2 C
601
SECTION 16.1 (PAGE 858)
R. A. ADAMS: CALCULUS
14. We use the same Maclaurin expansion for F as in Exercises 12 and 13. On C we have r = cos θ i + sin θ j, (0 ≤ θ ≤ 2π ) dr = − sin θ i + cos θ j F • dr = − sin θ F0 • i + cos θ F0 • j − 2 sin θ cos θ F1 • i + 2 cos2 θ F1 • j
− 2 sin2 θ F2 • i + 2 sin θ cos θ F2 • j + · · · ds,
where · · · represents terms of degree 3 or higher in . Since 0
0
2π
2π
sin θ dθ = cos2 θ dθ =
2π
0
0
cos θ dθ =
2π
2π 0
sin θ cos θ dθ = 0
sin2 θ dθ = π,
we have 1 F • dr = F1 • j − F2 • i + · · · , π 2 C where · · · represents terms of degree at least 1 in . Hence 1 F • dr = F1 • j − F2 • i lim →0+ π 2 C ∂ F2 ∂ F1 = − ∂x ∂y ˆ = curl F • k = curl F • N.
Section 16.2 Some Identities Involving Grad, Div, and Curl (page 864) 1. Theorem 3(a): ∂ ∂ ∂ (φψ) + (φψ) + (φψ) ∇(φψ) = ∂x ∂y ∂z ∂ψ ∂φ ∂ψ ∂φ = φ + ψ i + ··· + φ + ψ k ∂x ∂x ∂z ∂z = φ∇ψ + ψ∇φ.
2. Theorem 3(b): ∂ ∂ ∂ (φ F1 ) + (φ F2 ) + (φ F3 ) ∂x ∂y ∂z ∂φ ∂φ ∂ F1 ∂ F3 F1 + φ + ··· + F3 + φ +··· = ∂x ∂x ∂z ∂z = ∇φ • F + φ∇ • F.
∇ • (φF) =
602
3. Theorem 3(d): ∂ (F2 G 3 − F3 G 2 ) + · · · ∂x ∂ F3 ∂G 3 ∂G 2 ∂ F2 G 3 + F2 − G 2 − F3 + ··· = ∂x ∂x ∂x ∂x = (∇ × F) • G − F • (∇ × G).
∇ • (F × G) =
4. Theorem 3(f). The first component of ∇(F • G) is ∂ F1 ∂G 1 ∂ F2 ∂G 2 ∂ F3 ∂G 3 G 1 + F1 + G 2 + F2 + G 3 + F3 . ∂x ∂x ∂x ∂x ∂x ∂x We calculate the first components of the four terms on the right side of the identity to be proved. The first component of F × (∇ × G) is ∂G 1 ∂G 3 ∂G 2 ∂G 1 F2 − − F3 − . ∂x ∂y ∂z ∂x The first component of G × (∇ × F) is ∂ F1 ∂ F3 ∂ F2 ∂ F1 − − G3 − . G2 ∂x ∂y ∂z ∂x The first component of (F • ∇)G is F1
∂G 1 ∂G 1 ∂G 1 + F2 + F3 . ∂x ∂y ∂z
The first component of (G • ∇)F is G1
∂ F1 ∂ F1 ∂ F1 + G2 + G3 . ∂x ∂y ∂z
When we add these four first components, eight of the fourteen terms cancel out and the six remaining terms are the six terms of the first component of ∇(F • G), as calculated above. Similar calculations show that the second and third components of both sides of the identity agree. Thus ∇(F•G) = F×(∇ ×G)+G×(∇ ×F)+(F•∇)G+(G•∇)F.
5. Theorem 3(h). By equality of mixed partials, i j k ∂ ∂ ∂ ∇ × ∇φ = ∂ x ∂ y ∂z ∂φ ∂φ ∂φ ∂ x ∂ y ∂z ∂ ∂φ ∂ ∂φ = − i + · · · = 0. ∂ y ∂z ∂z ∂ y
6. Theorem 3(i). We examine the first components of the terms on both sides of the identity ∇ × (∇ × F) = ∇(∇ • F) − ∇ 2 F.
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.2 (PAGE 864)
The first component of ∇ × (∇ × F) is ∂ F2 ∂ ∂ F1 ∂ F1 ∂ F3 − − − ∂x ∂y ∂z ∂z ∂x ∂ 2 F2 ∂ 2 F1 ∂ 2 F1 ∂ 2 F3 = − . − + ∂ y∂ x ∂ y2 ∂z 2 ∂z∂ x
∂ ∂y
The first component of ∇(∇ • F) is
−
f (r )r
= 0, so that
du + 3u = 0 dr du 3 dr =− u r ln |u| = −3 ln |r | + ln |C|
r
Thus f (r ) = Cr −3 , for some constant C.
10. Given that div F = 0 and curl F = 0, Theorem 3(i)
The first component of −∇ 2 F is −∇ 2 F1 =
u = Cr −3 .
∂ 2 F2 ∂ 2 F1 ∂ 2 F3 ∂ + ∇•F= + . 2 ∂x ∂x ∂ x∂ y ∂ x∂z
∂ 2 F1 − 2 ∂x
If f (r )r is solenoidal then ∇ • u = f (r ) satisfies
∂ 2 F1 ∂ y2
−
implies that ∇ 2 F = 0 too. Hence the components of F are harmonic functions. If F = ∇φ, then
∂ 2 F1 . ∂z 2
Evidently the first components of both sides of the given identity agree. By symmetry, so do the other components.
7. If the field lines of F(x, y, z) are parallel straight lines, in
∇ 2 φ = ∇ • ∇φ = ∇ • F = 0, so φ is also harmonic.
11. By Theorem 3(e) and 3(f),
the direction of the constant nonzero vector a say, then F(x, y, z) = φ(x, y, z)a for some scalar field φ, which we assume to be smooth. By Theorem 3(b) and (c) we have
div F = div (φa) = ∇φ • a curl F = curl (φa) = ∇φ × a.
8. If r = xi + yj + zk and r = |r|, then ∇ × r = 0,
∇r =
r . r
If c is a constant vector, then its divergence and curl are both zero. By Theorem 3(d), (e), and (f) we have ∇ • (c × r) = (∇ × c) • r − c • (∇ × r) = 0 ∇ × (c × r) = (∇ • r)c + (r • ∇)c − (∇ • c)r − (c • ∇)r = 3c + 0 − 0 − c = 2c ∇(c • r) = c × (∇ × r) + r × (∇ × c) + (c • ∇)r + (r • ∇)c = 0 + 0 + c + 0 = c.
9.
∇ • f (r )r = ∇ f (r ) • r + f (r )(∇ • r) r•r + 3 f (r ) = f (r ) r = r f (r ) + 3 f (r ).
If r = xi + yj + zk, then ∇ • r = 3 and ∇ × r = 0. Also, (F • ∇)r = F1
Since ∇φ is an arbitrary gradient, div F can have any value, but curl F is perpendicular to a, and thereofore to F.
∇ • r = 3,
∇ × (F × r) = (∇ • r)F + (r • ∇)F − (∇ • F)r − (F • ∇)r ∇(F • r) = F × (∇ × r) + r × (∇ × F) + (F • ∇)r + (r • ∇)F.
∂r ∂r ∂r + F2 + F3 = F. ∂x ∂y ∂z
Combining all these results, we obtain ∇ × (F × r) − ∇(F • r) = 3F − 2(F • ∇)r − (∇ • F)r − r × (∇ × F) = F − (∇ • F)r − r × (∇ × F). In particular, if ∇ • F = 0 and ∇ × F = 0, then ∇ × (F × r) − ∇(F • r) = F.
12. If ∇ 2 φ = 0 and ∇ 2 ψ = 0, then ∇ • (φ∇ψ − ψ∇φ) = ∇φ • ∇ψ + φ∇ 2 ψ − ∇ψ • ∇φ − ψ∇ 2 φ = 0, so φ∇ψ − ψ∇φ is solenoidal.
13. By Theorem 3(c) and (h), ∇ × (φ∇ψ) = ∇φ × ∇ψ + φ∇ × ∇ψ = ∇φ × ∇ψ −∇ × (ψ∇φ) = −∇ψ × ∇φ − ψ∇ × ∇φ = ∇φ × ∇ψ.
603
SECTION 16.2 (PAGE 864)
14. By Theorem 3(b), (d), and (h), we have
∇ • f (∇g × ∇h) = ∇ f • (∇g × ∇h) + f ∇ • (∇g × ∇h) = ∇ f • (∇g × ∇h) + f (∇ × ∇g) • ∇h − ∇g • (∇ × ∇h) = ∇ f • (∇g × ∇h) + 0 − 0 = ∇ f • (∇g × ∇h).
R. A. ADAMS: CALCULUS
Look for a solution with G 2 = 0. We have G3 =
Try M(x, z) = 0. Then G 3 = x ye2z , and ∂G 1 ∂G 3 = ye2z + = 2ye2z . ∂z ∂x
15. If F = ∇φ and G = ∇ψ, then ∇ × F = 0 and ∇ × G = 0 by Theorem 3(h). Therefore, by Theorem 3(d) we have ∇ • (F × G) = (∇ × F) • G + F • (∇ × G) = 0.
Thus
Thus F × G is solenoidal. By Exercise 13, ∇ × (φ∇ψ) = ∇φ × ∇ψ = F × G, so φ∇ψ is a vector potential for F × G. (So is −ψ∇φ.)
16. If ∇ × G = F = −yi + xj, then
xe2z d y = x ye2z + M(x, z).
G1 =
2ye2z dz = ye2z + N (x, y).
Since −e2z = −
∂G 1 ∂N = −e2z − , ∂y ∂y
∂G 3 ∂G 2 − = −y ∂y ∂z ∂G 3 ∂G 1 − =x ∂z ∂x ∂G 1 ∂G 2 − = 0. ∂x ∂y
we can take N (x, y) = 0. Thus G = ye2z i + x ye2z k is a vector potential for F.
As in Example 1, we try to find a solution with G2 = 0. Then y2 G 3 = − y d y = − + M(x, z). 2
18. For (x, y, z) in D let v = xi + yj + zk. The line segment r(t) = tv, (0 ≤ t ≤ 1), lies in D, so div F = 0 on the
y2 Again we try M(x, z) = 0, so G 3 = − . Thus 2 ∂G 3 = 0 and ∂x G 1 = x dz = x z + N (x, y). ∂G 1 = 0 we may take N (x, y) = 0. ∂y 1 G = x zi − y 2 k is a vector potential for F. (Of course, 2 this answer is not unique.) Since
17. If F = xe 2z i + ye2z j − e2z k, then div F = e 2z + e2z − 2e2z = 0, so F is solenoidal. If F = ∇ × G, then ∂G 2 ∂G 3 − = xe2z ∂y ∂z ∂G 3 ∂G 1 − = ye2z ∂z ∂x ∂G 1 ∂G 2 − = −e2z . ∂x ∂y
604
path. We have
G(x, y, z) = =
1
tF r(t) × v dt
1
tF ξ(t), η(t), ζ (t) × v dt
0
0
where ξ = t x, η = t y, ζ = t z. The first component of curl G is (curl G)1 1 = t curl (F × v) dt 1 0 1 ∂ ∂ (F × v)3 − (F × v)2 dt t = ∂y ∂z 0 1 ∂ ∂ = (F1 y − F2 x) − (F3 x − F1 z) dt t ∂y ∂z 0 1 ∂ F ∂ F2 ∂ F3 1 = − t2x − t2x t F1 + t 2 y ∂η ∂η ∂ζ 0 ∂ F1 dt + t F1 + t 2 z ∂ζ 1 ∂ F1 ∂ F1 ∂ F1 = + t2 y + t 2z dt. 2t F1 + t 2 x ∂ξ ∂η ∂ζ 0
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.3 (PAGE 868)
To get the last line we used the fact that di vF = 0 to ∂ F2 ∂ F3 ∂ F1 replace −t 2 x − t2x with t 2 x . Continuing the ∂η ∂ζ ∂ξ calculation, we have
(d) LHS := Del . (F &x G):
RHS := (Del &x F) . G - F . (Del &x G): simplify(LHS - RHS); 0
d2 t F1 (ξ, η, ζ ) dt 0 dt 1 = t 2 F1 (t x, t y, t z) = F1 (x, y, z).
(curl G)1 =
1
(e) LHS := Del &x (F &x G):
RHS1 := (Del . G)*F: RHS2 := G[1]*diff(F,x) +G[2]*diff(F,y)+G[3]*diff(F,z): RHS3 := (Del . F)*G: RHS4 := F[1]*diff(G,x) +F[2]*diff(G,y)+F[3]*diff(G,z): RHS := RHS1 + RHS2 - RHS3 - RHS4: simplify(LHS - RHS);
0
Similarly, (curl G)2 = F2 and (curl G)3 = F3 . Thus curl G = F, as required.
19. In the following we suppress output (which for some calculations can be quite lengthy) except for the final check on each inequality. You may wish to use semicolons instead of colons to see what the output actually looks like. >
0 e¯ x (f) LHS := Del(F . G):
RHS1 := F &x (Del &x G): RHS2 := G &x (Del &x F): RHS3 := F[1]*diff(G,x) +F[2]*diff(G,y)+F[3]*diff(G,z): RHS4 := G[1]*diff(F,x) +G[2]*diff(F,y)+G[3]*diff(F,z): RHS := RHS1 + RHS2 + RHS3 + RHS4: simplify(LHS - RHS);
with(VectorCalculus):
>
SetCoordinates(’cartesian’[x,y,z]): > F := VectorField ():
0 e¯ x
> G := VectorField ():
All these zero outputs indicate that the inequalities (a)–(f) of the theorem are valid.
Section 16.3 (page 868)
(a) LHS := Del(phi(x,y,z)*psi(x,y,z)):
RHS := phi(x,y,z)*Del(psi(x,y,z)) + psi(x,y,z)*Del(phi(x,y,z)): simplify(LHS - RHS); 0 e¯ x (b) LHS := Del . (F*phi(x,y,z)):
RHS := (Del(phi(x,y,z))).F + phi(x,y,z)*(Del.F): simplify(LHS - RHS); 0
1.
Green’s Theorem in the Plane
2
(sin x + 3y 2 ) d x + (2x − e−y ) d y C
∂ ∂ 2 = (2x − e−y ) − (sin x + 3y 2 ) d A ∂x ∂y R (2 − 6y) d A = πR a dθ (2 − 6r sin θ )r dr = 0 0 a π sin θ dθ r 2 dr = π a2 − 6 0
0
= π a 2 − 4a 3 . y
(c) LHS := Del &x (phi(x,y,z)*F):
RHS := RHS := (Del(phi(x,y,z))) &x F + phi(x,y,z)*(Del &x F): simplify(LHS - RHS); 0 e¯ x
C
R a
−a
x
Fig. 16.3.1
605
SECTION 16.3 (PAGE 868)
2.
R. A. ADAMS: CALCULUS
(x 2 − x y) d x + (x y − y 2 ) d y C
∂ 2 ∂ =− (x y − y 2 ) − (x − x y) d A ∂x ∂y T (y + x) d A =− T 1 4 = −( y¯ + x) ¯ × (area of T ) = − +1 ×1= − . 3 3
5. By Example 1,
1 x dy − y dx 2 C 2π 1 a cos3 t 3b sin2 t cos t = 2 0
Area =
− b sin3 t (−3a cos2 t sin t) dt
3ab 2π 2 sin t cos2 t dt 2 0 3π ab 3ab 2π sin2 (2t) dt = . = 2 0 4 8
=
y (1,1)
T
C
2
6. Let R, C, and F be as in the statement of Green’s The-
x
orem. As noted in the proof of Theorem 7, the unit ˆ satisfy tangent Tˆ to C and the unit exterior normal N ˆ = Tˆ × k. Let N
Fig. 16.3.2
G = F2 (x, y)i − F1 (x, y)j.
3.
(x sin y 2 − y 2 ) d x + (x 2 y cos y 2 + 3x) d y C = 2x y cos y 2 + 3 − (2x y cos y 2 − 2y) d A T = (3 + 2y) d A = 3 d A + 0 = 3 × 3 = 9. T
ˆ Applying the 2-dimensional DiverThen F • Tˆ = G • N. gence Theorem to G, we obtain ˆ ds F1 d x + F2 d y = F • Tˆ ds = G•N C C C =
T
div G d A
R
=
y 2
R
∂ F1 ∂ F2 − ∂x ∂y
dA
as required
7. r = sin ti + sin 2tj,
(1,1)
T
(0 ≤ t ≤ 2π ) y
C
C x
R1
R2 x
(1,−1)
−2
Fig. 16.3.3 Fig. 16.3.7 2
4. Let D be the region x 2 + y 2 ≤ 9, y ≥ 0. Since C is the clockwise boundary of D, C
x 2 y d x − x y2 d y
∂ 2 ∂ =− (−x y 2 ) − (x y) d x d y ∂y D ∂x π 3 81π 2 2 = . (y + x ) d A = dθ r 3 dr = 4 D 0 0
F = ye x i + x 3 e y j i j k ∂ ∂ ∂ 2 = (3x 2 e y − e x )k. curl F = ∂ x ∂y ∂z x2 ye x 3e y 0 Observe that C bounds two congruent regions, R1 and R2 , one counterclockwise and the other clockwise. ˆ = k; for R2 , N ˆ = −k. Since R1 and R2 For R1 , N are mirror images of each other in the y-axis, and since curl F is an even function of x, we have curl F • Nˆ d S = − curl F • Nˆ d S. R1
606
R2
INSTRUCTOR’S SOLUTIONS MANUAL
Thus C
8.
F • dr =
R1
+
R2
SECTION 16.4 (PAGE 873)
curl F • Nˆ d S = 0.
a) F = x 2 j F • dr = x2 dy = 2x d A = 2 Ax¯ . C C R b) F = x yi F • dr = xy dx = − x d A = − Ax. ¯ C C R c) F = y 2 i + 3x yj F • dr = y 2 d x + 3x y d y C C = (3y − 2y) d A = A y¯ .
ˆ dS = F•N 0 d V = 0. S B
3. If F = (x 2 + y 2 )i + (y 2 − z 2 )j + zk, then div F = 2x + 2y + 1, and 4 ˆ F•N d S = (2x+2y+1) d V = 1 d V = π a3 . 3 S B B
4. If F = x 3 i + 3yz 2 j + (3y 2 z + x 2 )k, then div F = 3x 2 + 3z 2 + 3y 2 , and
R
9. The circle Cr of radius r and centre at r0 has parametrization
r = r0 + r cos ti + r sin tj,
2. If F = ye z i + x 2 e z j + x yk, then div F = 0, and
(0 ≤ t ≤ 2π ).
ˆ dS = 3 F•N (x 2 + y 2 + z 2 ) d V S B π a 2π dθ sin φ dφ ρ 4 dρ =3 0
ˆ the unit normal to Note that dr/dt = cos ti + sin tj = N, Cr exterior to the disk Dr of which Cr is the boundary. The average value of u(x, y) on Cr is u¯ r =
1 2π
2π
0
u(x0 + r cos t, y0 + r sin t) dt,
12 5 πa . = 5
Therefore the flux of F out of any solid region R is
2π ∂u d u¯ r 1 ∂u = cos t + sin t dr 2π 0 ∂x ∂y 1 ˆ ds ∇u • N = 2πr Cr
Flux =
R
dt
since ds = r dt. By the (2-dimensional) divergence theorem, and since u is harmonic, 1 d u¯ r = ∇ • ∇u d x d y dr 2πr Dr 2 1 ∂ u ∂2u = d x d y = 0. + 2πr ∂x2 ∂ y2 Dr Thus u¯ r = limr→0 u¯ r = u(x0 , y0 ).
Section 16.4 The Divergence Theorem in 3-Space (page 873) 1. In this exercise, the sphere S bounds the ball B of radius
a centred at the origin. If F = xi − 2yj + 4zk, then div F = 1 − 2 + 4 = 3. Thus ˆ dS = F•N 3 d V = 4π a 3 . S B
0
5. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z).
and so
0
=2
div F d V R
(x + y + z) d V = 2(x¯ + y¯ + z¯ )V
where (x, ¯ y¯ , z¯ ) is the centroid of R and V is the volume of R. If R is the ball (x − 2)2 + y 2 + (z − 3)2 ≤ 9, then x¯ = 2, y¯ = 0, z¯ = 3, and V = (4π/3)33 = 36π . The flux of F out of R is 2(2 + 0 + 3)(36π ) = 360π .
6. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is Flux = =2
R
div F d V R
(x + y + z) d V = 2(x¯ + y¯ + z¯ )V
where (x, ¯ y¯ , z¯ ) is the centroid of R and V is the volume of R. If R is the ellipsoid x 2 + y 2 + 4(z − 1)2 ≤ 4, then x¯ = 0, y¯ = 0, z¯ = 1, and V = (4π/3)(2)(2)(1) = 16π/3. The flux of F out of R is 2(0 + 0 + 1)(16π/3) = 32π/3.
607
SECTION 16.4 (PAGE 873)
R. A. ADAMS: CALCULUS
7. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z).
can be calculated directly by the methods of Section 6.6. We will do it here by using the Divergence Theorem instead. S is one face of a tetrahedral domain D whose other faces are in the coordinate planes, as shown in the figure. Since φ = x y + z 2 , we have
Therefore the flux of F out of any solid region R is Flux = div F d V R =2 (x + y + z) d V = 2(x¯ + y¯ + z¯ )V
∇ • ∇φ = ∇ 2 φ = 2.
∇φ = yi + xj + 2zk,
R
where (x¯ , y¯ , z¯ ) is the centroid of R and V is the volume of R.
Thus
If R is the tetrahedron with vertices (3, 0, 0), (0, 3, 0), (0, 0, 3), and (0, 0, 0), then x¯ = y¯ = z¯ = 3/4, and V = (1/6)(3)(3)(3) = 9/2. The flux of F out of R is 2((3/4) + (3/4) + (3/4))(9/2) = 81/4.
D
∇ • ∇φ d V = 2 ×
the volume of the tetrahedron D being abc/6 cubic units. z c
8. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is Flux = div F d V R =2 (x + y + z) d V = 2(x¯ + y¯ + z¯ )V
back side
S D
R
x
ˆ = −k and z = 0, so ∇φ • N ˆ = 0, and On the bottom, N the flux out of the bottom face is 0.
region having volume V , then div F d V = 3V .
ˆ = −j, so ∇φ • N ˆ = −x. The On the side, y = 0 and N flux out of the side face is
C
The region C described in the statement of the problem is the part of a solid cone with vertex at the origin that lies inside a ball of radius R with centre at the origin. The surface S of C consists of two parts, the conical wall S1 , and the region D on the spherical boundary of the ball. At any point P on S1 , the outward normal ˆ is perpendicular to the line O P, that is, to F, so field N ˆ = 0. At any point P on D, N ˆ is parallel to F, in F•N ˆ = F/|F| = F/R. Thus fact N ˆ dS = ˆ dS + ˆ dS F•N F•N F•N S S1 D R2 F•F =0+ dS = d S = AR R R D D
side
x d x dz = −
a2 c ac a × =− . 2 3 6
(We used the fact that Mx=0 = area × x¯ and x¯ = a/3 for that face.) ˆ = −i, so the flux out of On the back face, x = 0 and N that face is back
ˆ dS = − ∇φ • N
back
y d y dz = −
b2 c bc b × =− . 2 3 6
Therefore, by the Divergence Theorem I−
a 2 c b2 c abc − +0 = , 6 6 3
so
10. The required surface integral,
608
ˆ dS = − ∇φ • N
side
where A is the area of D. By the Divergence Theorem, 3V = AR, so V = AR/3.
S
bottom
The flux of ∇φ out of D is the sum of its fluxes out of the four faces of the tetrahedron.
9. If F = xi + yj + zk, then div F = 3. If C is any solid
ˆ d S, ∇φ • N
a
Fig. 16.4.10
If R is the cylinder x 2 + y 2 ≤ 2y (or, equivalently, x 2 + (y − 1)2 ≤ 1), 0 ≤ z ≤ 4, then x¯ = 0, y¯ = 1, z¯ = 2, and V = (π 12 )(4) = 4π . The flux of F out of R is 2(0 + 1 + 2)(4π ) = 24π .
y
b
where (x¯ , y¯ , z¯ ) is the centroid of R and V is the volume of R.
I =
abc abc = , 6 3
11.
2 2 ˆ d S = I = abc + c(a + b ) . ∇φ • N 3 6 S
F = (x + y 2 )i + (3x 2 y + y 3 − x 3 )j + (z + 1)k
div F = 1 + 3(x 2 + y 2 ) + 1 = 2 + 3(x 2 + y 2 ).
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.4 (PAGE 873)
z
z
a b
back
ˆ N
side
S
S
D
D B
y
a
a
x
bottom
−k Fig. 16.4.11
Fig. 16.4.12 ˆ = −j, F • N ˆ = 0, so On the side, y = 0, N ˆ d S = 0. F•N
Let D be the conical domain, S its conical surface, and B its base disk, as shown in the figure. We have D
div F d V =
2π
0
dθ a
a 0
r dr
b(1−(r/a)) 0
side
2
(2 + 3r ) dz
ˆ = −i, F • N ˆ = −y, so On the back, x = 0, N π/2 a ˆ dS = − F•N dθ r cos θ r dr
r
dr r (2 + 3r 2 ) 1 − a a 2r 2 3r 4 dr − 2r + 3r 3 − = 2π b a a 0 2π a 2 b 3π a 4 b = + . 3 10 = 2π b
0
B
By the Divergence Theorem 3 3 3 ˆ d S + 0 − a + 2a = π a . F•N 3 3 6 S
ˆ d S = −area of B = −π a2 . F•N
S
ˆ dS + F•N
B
ˆ dS = F•N
D
Hence the flux of F upward through S is 3 3 ˆ dS = πa − a . F•N 6 3 S
div F d V ,
so the flux of F upward through the conical surface S is
3π a 4 b 2π a 2 b + + π a2 . = 3 10 S
0
π/2 a3 a3 = − sin θ × =− . 3 3 0
ˆ = −k, F • N ˆ = 2x, so On the bottom, z = 0, N π/2 a 2a 3 ˆ dS = 2 . F•N dθ r cos θ r dr = 3 bottom 0 0
By the Divergence Theorem,
0
back
ˆ = −k, F • N ˆ = −1, so On B we have z = 0, N
y
a
a
x
13.
F = (x + yz)i + (y − x z)j + (z − e x sin y)k div F = 1 + 1 + 1 = 3. z
ˆ N
12. F = (y + x z)i + (y + yz)j − (2x + z 2 )k div F = z + (1 + z) − 2z = 1. Thus
π a3 , div F d V = volume of D = 6 D
where D is the region in the first octant bounded by the sphere and the coordinate planes. The boundary of D consists of the spherical part S and the four planar parts, called the bottom, side, and back in the figure.
2a a x
ˆ N
S1
S2
D y
Fig. 16.4.13
609
SECTION 16.4 (PAGE 873)
R. A. ADAMS: CALCULUS
z
a) The flux of F out of D through S = S1 ∪ S2 is
back
S front
x
a
4a 2
= 6π
3a 2
1
u 0
S1
Fig. 16.4.14
√ du = 12 3π a 3 .
ˆ = −j, F • N ˆ = x, so On the side, y = 0, N 1 1 1 ˆ dS = F•N x dx dz = . 2 0 0 side ˆ = −k, F • N ˆ = y, so On the bottom, z = 0, N 1 1 1 ˆ dS = F•N y dy dx = . 2 bottom 0 0
ˆ = −i, F • N ˆ = 0, so On the back, x = 0, N ˆ d S = 0. F•N
− 3a
0
back
ˆ = i, F • N ˆ = 3z 2 , so On the front, x = 1, N π/2 1 3π ˆ dS = 3 . F•N dθ r 2 cos2 θ r dr = 16 front 0 0
c) The flux of F out of D through the spherical part S2 is
ˆ dS = F • N ˆ dS − ˆ dS F•N F•N S2 S S1 √ √ √ = 12 3π a 3 + 4 3π a 3 = 16 3π a 3 .
Hence, ˆ d S = 3π − 1 − 1 −0− 3π = −1. (3x z 2 i− xj− yk)• N 16 2 2 16 S
15. 14. Let D be the domain bounded by S, the coordinate planes, and the plane x = 1. If
F = 3x z 2 i − xj − yk, then div F = 3z 2 , so the total flux of F out of D is
bdry of D
ˆ dS = F•N
=3
=3×
D 1
π/2 0
dθ
3π 1 π × = . 4 4 16
0
1
r 2 cos2 θ r dr
The boundary of D consists of the cylindrical surface S and four planar surfaces, the side, bottom, back, and front.
610
F = (x 2 − x − 2y)i + (2y 2 + 3y − z)j − (z 2 − 4z + x y)k div F = 2x − 1 + 4y + 3 − 2z + 4 = 2x + 4y − 2z + 6. The flux of F out of R through its surface S is ˆ dS = F•N (2x + 4y − 2z + 6) d V . S R Now x d V = M x=0 = V x, ¯ where R has volume R
3z 2 d V dx
0
y
bottom
−x 2 − x yz − y 2 + x yz a dθ dz a S1 √ 2π 3a √ dθ √ dz = −4 3π a 3 . = −a 2
ˆ dS = F•N
1
− r2
ˆ = − xi + yj , d S = a dθ dz. The flux of F b) On S1 , N a out of D through S1 is
D
Let u = du = −2r dr 1/2
1
side
ˆ dS = F•N div F d V S D 2π 2a √4a 2 −r 2 =3 dθ r dr 2 dz 0 0 a 2a r 4a 2 − r 2 dr = 12π
V and centroid (x, ¯ y¯ z¯ ). Similar formulas obtain for the other variables, so the required flux is ˆ d S = 2V x¯ + 4V y¯ − 2V z¯ + 6V . F•N S
16. F = xi + yj + zk implies that div F = 3. The total flux of F out of D is
bdry of D
ˆ dS = 3 F•N
D
d V = 12,
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.4 (PAGE 873)
since the volume of D is half that of a cube of side 2, that is, 4 square units. D has three triangular faces, three pentagonal faces, and a hexagonal face. By symmetry, the flux of F out of each triangular face is equal to that out of the triangular face ˆ = k • k = 1 on that T in the plane z = 1. Since F • N face, these fluxes are d x d y = area of T =
T
18. φ = x 2 − y 2 + z 2 , G = 13 (−y 3 i + x 3 j + z 3 k). F = ∇φ + µcurl G. Let R be the region of 3-space occupied by the sandpile. Then R is bounded by the upper surface S of the sandpile and by the disk D: x 2 + y 2 ≤ 1 in the plane z = 0. The outward (from R) normal on D is −k. The flux of F out of R is given by
1 . 2
Similarly, the flux of F out of each pentagonal face is equal to the flux out of the pentagonal face P in the ˆ = −k • (−k) = 1; that plane z = −1, where F • N flux is 7 1 d x d y = area of P = 4 − = . 2 2 P
S
R
div F d V =
y
19.
above z = 0 and the disk D: x 2 + y 2 = 3a 2 in the x yplane form the boundary of a region R in 3-space. The outward normal from R on D is −k. If 2
F = (x + y + 2 + z )i + (e
2
+ y )j + (3 + x)k,
then di vF = 2x + 2y. By the Divergence Theorem, S
ˆ dS + F•N
D
F • (−k) d x d y =
j ∂ ∂y x3
k ∂ ∂z z3
= 3(x 2 + y 2 )k,
F • k d A = 3µ
2π 0
dθ
0
1
r 3 dr =
3π µ . 2
The flux of F out of S is 10π + (3π µ)/2. ˆ dS = curl F • N div curl F = 0, by Theorem S D 3(g). 1 ˆ dS = 1 r•N 3 dV = V. 3 S 3 D
17. The part of the sphere S: x 2 + y 2 + (z − a)2 = 4a 2
2
(2 + µ × 0) d V = 2(5π ) = 10π.
20. If r = xi + yj + zk, then div r = 3 and
Fig. 16.4.16
x2
R
D
P
div F d V .
D
R
and ∇φ • k = 2z = 0 on D, so
(−1,0,1)
H
x
D
i 1 ∂ curl G = 3 ∂x −y 3
z
(0,−1,1)
F • (−k) d A =
In addition,
(This can also be seen directly, since F radiates from the origin, so is everywhere tangent to the plane of the hexagonal face, the plane x + y + z = 0.) T
Now div curl G = 0 by Theorem 3(g). Also div ∇φ = div (2xi−2yj+2zk) = 2−2+2 = 2. Therefore
Thus the flux of F out of the remaining hexagonal face H is 1 7 + = 0. 12 − 3 × 2 2
(−1,−1,1)
ˆ dS + F•N
R
div F d V = 0
because R is symmetric about x = 0 and y = 0. Thus the flux of F outward across S is ˆ dS = F•N (3 + x) d x d y = 3π(3a2 ) = 9π a 2 . S D
21. We use Theorem 7(b), the proof of which is given in
Exercise 29. Taking φ(x, y, z) = x 2 + y 2 + z 2 , we have 1 ˆ d S = 1 φN ˆ dS (x 2 + y 2 + z 2 )N 2V S 2V S 1 grad φ d V = 2V D 1 (xi + yj + zk) d V = V = r¯ , since
x d V = M x=0 = V x. ¯
611
SECTION 16.4 (PAGE 873)
R. A. ADAMS: CALCULUS
22. Taking F = ∇φ in the first identity in Theorem 7(a), we have
ˆ dS = − ∇φ × N curl ∇φ d V = 0, S D
so φ is constant on any connected component of S, and u and v can only differ by a constant on S.
27. Apply the Divergence Theorem to F = ∇φ:
since ∇ × ∇φ = 0 by Theorem 3(h).
23. div (φF) = φ div F + ∇φ • F by Theorem 3(b). Thus D
φ div F d V +
D
D
∇ 2φ d V =
∇ • ∇φ d V D ˆ d S = ∂φ d S. = ∇φ • N S S ∂n
div (φF) d V D ˆ dS = φF • N S
∇φ • F d V =
by the Divergence Theorem.
28. By Theorem 3(b),
24. If F = ∇φ in the previous exercise, then div F = ∇ 2 φ and
D
φ∇ 2 φ d V +
ˆ d S. |∇φ|2 d V = φ∇φ • N D S
If ∇ 2 φ = 0 in D and φ = 0 on S, then D
|∇φ|2 d V = 0.
div (φ∇ψ − ψ∇φ) = ∇φ • ∇ψ + φ∇ 2 ψ − ∇ψ • ∇φ − ψ∇ 2 φ = φ∇ 2 ψ − ψ∇ 2 φ. Hence, by the Divergence Theorem,
2
D
2
div (φ∇ψ − ψ∇φ) d V D ˆ dS = (φ∇ψ − ψ∇φ) • N S ∂φ ∂ψ = φ d S. −ψ ∂n ∂n S
(φ∇ ψ − ψ∇ φ) d V =
Since φ is assumed to be smooth, ∇φ = 0 throughout D, and therefore φ is constant on each connected component of D. Since φ = 0 on S, these constants must all be 0, and φ = 0 on D.
25. If u and v are two solutions of the given Dirichlet problem, and φ = u − v, then
∇ 2 φ = ∇ 2 u − ∇ 2 v = f − f = 0 on D φ = u − v = g − g = 0 on S. By the previous exercise, φ = 0 on D, so u = v on D. That is, solutions of the Dirichlet problem are unique.
29. If F = φc, where c is an arbitrary, constant vector, then div F = ∇φ • c, and by the Divergence Theorem, c•
26. Re-examine the solution to Exercise 24 above. If
D
div F d V D ˆ dS = F•N S ˆ d S = c • φN ˆ d S. = φc • N S S
∇φ d V =
ˆ = 0 on S, then we ∇ 2 φ = 0 in D and ∂φ/∂n = ∇φ • N can again conclude that D
|∇φ| d V = 0
and ∇φ = 0 throughout D. Thus φ is constant on the connected components of D. (We can’t conclude the constant is 0 because we don’t know the value of φ on S.) If u and v are solutions of the given Neumann problem, then φ = u − v satisfies 2
2
2
∇ φ = ∇ u − ∇ v = f − f = 0 on D ∂φ ∂u ∂v = − = g − g = 0 on S, ∂n ∂n ∂n
612
Thus
c•
ˆ d S = 0. ∇φ d V − φ N D S
Since c is arbitrary, the vector in the large parentheses must be the zero vector. Hence
ˆ d S. ∇φ d V = φ N D S
INSTRUCTOR’S SOLUTIONS MANUAL
30.
SECTION 16.5 (PAGE 878)
1 1 ˆ dS = div F d V F•N vol(D ) S vol(D ) D 1 = div F(P0 ) d V vol(D ) D
+ div F − div F(P0 ) d V D 1 = div F(P0 ) + div F − div F(P0 ) d V . vol(D ) D Thus 1 ˆ d S − div F(P0 ) F • N vol(D ) S 1 |div F − div F(P0 )| d V ≤ vol(D ) D
z 1
C
T 1 x
y
1
Fig. 16.5.1
2. Let S be the part of the surface z = y 2 lying inside the
ˆ cylinder x 2 + y 2 = 4, and having upward normal N. Then C is the oriented boundary of S. Let D be the disk x 2 + y 2 ≤ 4, z = 0, that is, the projection of S onto the x y-plane.
≤ max |div F − div F(P0 )| P in D
z
→ 0 as → 0 + assuming div F is continuous. 1 ˆ d S = div F(P0 ). lim F•N →0+ vol(D ) S
C
ˆ N
S
D
Section 16.5
Stokes’s Theorem
y
(page 878) x
1. The triangle T lies in the plane x + y + z = 1. We use the downward normal
If F = yi −
ˆ = − i +√j + k N 3 on T , because of the given orientation of its boundary. If F = x yi + yzj + zxk, then curl F =
i ∂ ∂x xy
j ∂ ∂y yz
k ∂ ∂z zx
= −yi − zj − xk.
Therefore C
x y d x + yz dz + zx dz = F • dr C y+z+x = curl F • Nˆ d S = dS √ 3 T T 1 1 d S = √ × (area of T ) = √ 3 3 T √ √ 1 3 1 1 = √ × × 2× √ = . 2 2 3 2
xj + z 2 k,
Fig. 16.5.2 then
i ∂ curl F = ∂x y
j ∂ ∂y −x
k ∂ ∂z z2
= −2k.
dx dy on S, we have ˆ k•N y d x − x d y + z 2 dz = F • dr = curl F • Nˆ d S C C S ˆ d x d y = −8π. = −2k • N ˆ k•N D
Since d S =
3. Let C be the circle x 2 + y 2 = a 2 , z = 0, oriented
counterclockwise as seen from the positive z-axis. Let D be the disk bounded by C, with normal k. We have F = 3yi − 2x zj + (x 2 − y 2 )k j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 2 3y −2x z x − y = 2(x − y)i − 2xj − (2z + 3)k.
613
SECTION 16.5 (PAGE 878)
R. A. ADAMS: CALCULUS
Applying Stokes’s Theorem (twice) we calculate
F • dr = curl F • k d A S C D =− 3 d A = −3π a 2 . =
D
z
since D has area π a 2 .
6. The curve C:
S
r = cos ti + sin tj + sin 2tk, D
y
C
x
Fig. 16.5.3
4. The surface S with equation x 2 + y 2 + 2(z − 1)2 = 6,
z ≥ 0,
ˆ is that part of an ellipsoid of with outward normal N, revolution about the z-axis, centred at (0, 0, 1), and lying above the x y-plane. The boundary of S is the circle C: x 2 + y 2 = 4, z = 0, oriented counterclockwise as seen from the positive z-axis. C is also the oriented boundary ˆ = k. of the disk x 2 + y 2 ≤ 4, z = 0, with normal N 2 2 2 If F = (x z − y 3 cos z)i + x 3 e z j + x yze x +y +z k, then, on z = 0, we have
∂ 3 z ∂ x e − (x z − y 3 cos z) ∂x ∂y z=0 2 z 2 2 = 3x e + 3y cos z = 3(x + y 2 ).
curl F • k =
z=0
Thus S
F • dr = curl F • k d A C D 2 2π dθ 3r 2 r dr = 24π. =
curl F • Nˆ d S =
0
0
5. The circle C of intersection of x 2 + y 2 + z 2 = a 2 and
x + y + z = 0 is the boundary of a circular disk of radius a in the plane x + y + z = 0. If F = yi + zj + xk, then i ∂ curl F = ∂x y
614
ˆ = − i +√j + k , N 3 √ ˆ = 3 on D, so then curl F • N y d x + z d y + x dz = F • dr = curl F • Nˆ d S C C D √ √ = 3 d S = 3π a 2 , D
ˆ N
k
If C is oriented so that D has normal
j ∂ ∂y z
k ∂ ∂z x
= −(i + j + k).
0 ≤ t ≤ 2π,
lies on the surface z = 2x y, since sin 2t = 2 cos t sin t. It also lies on the cylinder x 2 +y 2 = 1, so it is the boundary of that part of z = 2x y lying inside that cylinder. Since C is oriented counterclockwise as seen from high on the z-axis, S should be oriented with upward normal, −2yi − 2xj + k ˆ = N , 1 + 4(x 2 + y 2 ) and has area element d S = 1 + 4(x 2 + y 2 ) d x d y. If F = (e x − y 3 )i + (e y + x 3 )j + e z k, then i j k ∂ ∂ ∂ curl F = = 3(x 2 + y 2 )k. ∂y ∂z ∂x x e − y 3 e y + x 3 ez If D is the disk x 2 + y 2 ≤ 1 in the x y-plane, then F • dr = curl F • Nˆ d S = 3(x 2 + y 2 ) d x d y C S D 1 2π 3π . dθ r 2 r dr = =3 2 0 0
7. The part of the paraboloid z = 9 − x 2 − y 2 lying above
ˆ has boundary the the x y-plane having upward normal N 2 2 circle C: x + y = 9, oriented counterclockwise as seen from above. C is also the oriented boundary of the plane disk x 2 + y 2 ≤ 9, z = 0, oriented with normal field ˆ = k. N If F = −yi + x 2 j + zk, then j i ∂ ∂ curl F = ∂x ∂y −y x 2
k ∂ ∂z z
= (2x + 1)k.
INSTRUCTOR’S SOLUTIONS MANUAL
By Stokes’s Theorem, the circulation of F around C is
C
F • dr =
D
=
D
(curl F • k) d A (2x + 1) d A = 0 + π(32 ) = 9π.
8. The closed curve r = (1 + cos t)i + (1 + sin t)j + (1 − cos t − sin t)k, (0 ≤ t ≤ 2π ), lies in the plane x + y + z = 3 and is oriented counterclockwise as seen from above. Therefore it is the boundary of a√ region S in that plane with normal ˆ = (i + j + k)/ 3. The projection of S onto the field N x y-plane is the circular disk D of radius 1 with centre at (1, 1). If F = ye x i + (x 2 + e x )j + z 2 e z k, then i ∂ curl F = ∂x x ye
j ∂ ∂y x 2 + ex
k ∂ = 2xk. ∂z 2 z z +e
By Stokes’s Theorem,
C
F • dr =
S
curl F • Nˆ d S
2x √ dS = S 3 = 2x¯ A = 2π,
=
D
2x √ √ ( 3) d x d y 3
where x¯ = 1 is the x-coordinate of the centre of D, and A = π 12 = π is the area of D.
9. If S1 and S2 are two surfaces joining C1 to C2 , each hav-
ing upward normal, then the closed surface S3 consisting of S1 and −S2 (that is, S2 with downward normal) bound a region R in 3-space. Then S1
ˆ dS − F•N
S2
ˆ dS F•N
ˆ dS F•N S −S2 1 ˆ dS = ± = F•N div F d V = 0, S3 R
=
ˆ dS + F•N
provided that div F = 0 identically. Since F = (αx 2 − z)i + (x y + y 3 + z)j + βy 2 (z + 1)k,
SECTION 16.5 (PAGE 878)
we have div F = 2αx + x + 3y 2 + βy 2 = 0 if α = −1/2 ˆ d S for and β = −3. In this case we can evaluate S F• N any such surface S by evaluating the special case where S is the half-disk H : x 2 + y 2 ≤ 1, z = 0, y ≥ 0, with ˆ = k. We have upward normal N ˆ d S = −3 F•N y2 d x d y S H 1 π 3π sin2 θ dθ r 3 dr = − . = −3 8 0 0
10. The curve C: (x − 1)2 + 4y 2 = 16, 2x + y + z = 3,
oriented counterclockwise as seen from above, bounds an elliptic disk S on the √ plane 2x + y + z = 3. S has normal ˆ = (2i + j + k)/ 6. Since its projection onto the x yN plane is an elliptic disk with centre at√(1, 0, 0) and area π(4)(2) = 8π , therefore S has area 8 6π and centroid (1, 0, 1). If F = (z 2 + y 2 + sin x 2 )i + (2x y + z)j + (x z + 2yz)k, then curl F =
i ∂ ∂x z 2 + y 2 + sin x 2 = (2z − 1)i + zj.
j ∂ ∂y 2x y + z
k ∂ ∂z x z + 2yz
By Stokes’s Theorem, F • dr = curl F • Nˆ d S C S 1 (2(2z − 1) + z) d S =√ 6 S 5¯z − 2 √ = √ (8 6π ) = 24π. 6
11. As was shown in Exercise 13 of Section 7.2, ∇ × (φ∇ψ) = −∇ × (ψ × φ) = ∇φ × ∇ψ. Thus, by Stokes’s Theorem, ˆ dS φ∇ψ = ∇ × (φ∇ψ) • N C S ˆ dS = (∇φ × ∇ψ) • N S ˆ dS − ψ∇φ = −∇ × (ψ∇φ) • N C S ˆ d S. = (∇φ × ∇ψ) • N S ∇φ × ∇ψ is solenoidal, with potential φ∇ψ, or −ψ∇φ.
615
SECTION 16.5 (PAGE 878)
12. We are given that C bounds a region R in a plane P ˆ = ai + bj + ck. Therefore, with unit normal N a 2 + b2 + c2 = 1. If F = (bz − cy)i + (cx − az)j + (ay − bx)k, then i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z bz − cy cx − az ay − bx = 2ai + 2bj + 2ck. ˆ = 2(a 2 + b2 + c2 ) = 2. We have Hence curl F • N 1 (bz − cy) d x + (cx − az) d y + (ay − bx) dz 2 C 1 1 F • dr = curl F • Nˆ d S = 2 C 2 R 1 2 d S = area of R. = 2 R
13. The circle C of radius centred at P is the oriented
boundary of the disk S of area π 2 having constant norˆ By Stokes’s Theorem, mal field N. F • dr = curl F • Nˆ d S C S = curl F(P) • Nˆ d S S + curl F − curl F(P) • Nˆ d S S 2 ˆ = π curl F(P) • N + curl F − curl F(P) • Nˆ d S. S Since F is assumed smooth, its curl is continuous at P. Therefore 1 ˆ F • dr − curl F(P) • N π 2 C 1 ˆ d S ≤ curl F − curl F(P) • N 2 π S ≤ max |curl F(Q) − curl F(P)| Q on S → 0 as → 0+. Thus lim
→0+
616
C
ˆ F • dr = curl F(P) • N.
R. A. ADAMS: CALCULUS
Section 16.6 Some Physical Applications of Vector Calculus (page 885) 1. a) If we measure depth in the liquid by −z, so that the zaxis is vertical and z = 0 at the surface, then the pressure at depth −z is p = −δgz, where δ is the density of the liquid. Thus ∇ p = −δgk = δg,
where g = −gk is the constant downward vector acceleration of gravity. The force of the liquid on surface element d S of the ˆ is solid with outward (from the solid) normal N ˆ d S = −(−δgz)N ˆ d S = δgz N ˆ d S. dB = − pN Thus, the total force of the liquid on the solid (the buoyant force) is ˆ dS B = δgz N S = ∇(δgz) d V (see Theorem 7) R =− δg d V = −Mg, R
where M =
R
δ d V is the mass of the liquid
which would occupy the same space as the solid. Thus B = −F, where F = Mg is the weight of the liquid displaced by the solid. z
y x
z dS R
ˆ N S
Fig. 16.6.1 b) The above argument extends to the case where the solid is only partly submerged. Let R∗ be the part of the region occupied by the solid that is below the surface of the liquid. Let S∗ = S1 ∪ S2 be the boundary of R∗ , with S1 ⊂ S and S2 in the plane of the surface of the liquid. Since p = −δgz = 0 on S2 , we have ˆ d S = 0. δgz N S2
INSTRUCTOR’S SOLUTIONS MANUAL
Therefore the buoyant force on the solid is ˆ dS B= δgz N S1 ˆ dS + ˆ dS = δgz N δgz N S2 S1 ˆ dS = ∗ δgz N S =− δg d V = −M ∗ g, where M ∗ =
SECTION 16.6 (PAGE 885)
where Q =
R
ρ d V is the total charge in R.
4. If f is continuous and vanishes outside a bounded region (say the ball of radius R centred at r), then | f (ξ, η, ζ )| ≤ K , and, if (ρ, φ, θ ) denote spherical coordinates centred at r, then 2π π R 2 | f (s)| ρ ≤ K dθ sin φ dφ d V dρ s 3 |r − s| ρ Ê 0 0 0 = 2π K R 2 a constant.
R∗
R∗
δ d V is the mass of the liquid
which would occupy R∗ . Again we conclude that the buoyant force is the negative of the weight of the liquid displaced.
tion for fluid motion given in the text. If S is an (imaginary) surface bounding an arbitrary region D, then the rate of change of total charge in D is ∂ρ ∂ dV, ρ dV = ∂t D D ∂t where ρ is the charge density. By conservation of charge, this rate must be equal to the rate at which charge is crossing S into D, that is, to ˆ dS = − (−J) • N div J d V . S D
S2
R∗
5. This derivation is similar to that of the continuity equa-
S1
Fig. 16.6.1 ˆ is (F1 G) • N. ˆ Applying 2. The first component of F(G • N) the Divergence Theorem and Theorem 3(b), we obtain ˆ (F1 G) • N d S = div (F1 G) d V S D = ∇ F1 • G + F1 ∇ • G d S. D
But ∇ F1 • G is the first component of (G • ∇)F, and F1 ∇ • G is the first component of Fdiv G. Similar results obtain for the other components, so ˆ dS = F(G • N) Fdiv G + (G • ∇)F d V . S D
3. Suppose the closed surface S bounds a region R in which charge is distributed with density ρ. Since the electric field E due to the charge satisfies div E = kρ, the total flux of E out of R through S is, by the Divergence Theorem, ˆ dS = div E d V = k ρ d V = k Q, E•N S R R
ˆ is the outward (from (The negative sign occurs because N D) normal on S.) Thus we have ∂ρ + div J d V = 0. D ∂t Since D is arbitrary and we are assuming the integrand is continuous, it must be 0 at every point: ∂ρ + div J = 0. ∂t
6. Since r = xi + yj + zk and b = b1 i + b2 j + b3 k, we have |r − b|2 = (x − b1 )2 + (y − b2 )2 + (z − b3 )2 ∂ |r − b| = 2(x − b1 ) 2|r − b| ∂x x − b1 ∂ |r − b| = . ∂x |r − b| Similar formulas hold for the other first partials of |r − b|, so 1 ∇ |r − b| −1 ∂ ∂ = |r − b|i + · · · + |r − b|k |r − b|2 ∂ x ∂z −1 (x − b1 )i + (y − b2 )j + (z − b3 )k = |r − b|2 |r − b| r−b . =− |r − b|3
617
SECTION 16.6 (PAGE 885)
R. A. ADAMS: CALCULUS
7. Using the result of Exercise 4 and Theorem 3(d) and (h),
Similarly, the other components have zero line integrals, so (ds • ∇)F(s) = 0. F
we calculate, for constant a,
r−b div a × |r − b|3 1 = −div a × ∇ |r − b| 1 1 = −(∇ × a) • ∇ +a•∇ ×∇ = 0 + 0 = 0. |r − b| |r − b|
11. Using the results of Exercises 7 and 8, we have curl
for r not on F. (Again, this is because the curl is taken with respect to r, so s and ds can be regarded as constant for the calculation of the curl.)
8. For any element ds on the filament F, we have
div ds ×
r−s |r − s|3
12. By analogy with the filament case, the current in volume
=0
element d V at position s is J(s) d V , which gives rise at position r to a magnetic field
by Exercise 5, since the divergence is taken with respect to r, and so s and ds can be regarded as constant. Hence
div
ds × (r − s) = div F |r − s|3 F
ds ×
r−s |r − s|3
r−s ds × (r − s) = curl ds × =0 |r − s|3 F |r − s|3 F
dH(r) =
= 0.
If R is a region of 3-space outside which J is identically zero, then at any point r in 3-space, the total magnetic field is J(s) × (r − s) 1 dV. H(r) = 4π |r − s|3 R
9. By the result of Exercise 4 and Theorem 3(e), we calculate
r−b |r − b|3 1 = −curl a × ∇ |r − b| 1 1 =− ∇ •∇ a− ∇ •∇ a |r − b| |r − b| 1 1 + (a • ∇)∇ . + (∇ • a)∇ |r − b| |r − b|
curl a ×
1 = 0 for r = b, either by direct |r − b| 1 calculation or by noting that ∇ is the field of a |r − b| point source at r = b and applying the result of Example 3 of Section 7.1. 1 Also − ∇ • ∇ a = 0 and ∇ • a = 0, since a is |r − b| constant. Therefore we have r−b 1 curl a × = (a • ∇)∇ 3 |r − b| |r − b| r−b . = −(a • ∇) |r − b|3
Now A(r) was defined to be A(r) =
F is closed and ∇ F1 is conservative,
i•
618
F
(ds • ∇)F(s) =
F
∇ F1 (s) • ds = 0.
1 4π
R
J(s) dV. |r − s|
We have 1 1 J(s) d V ∇r × 4π |r − s| R 1 1 × J(s) d V ∇r = 4π |r − s| R (by Theorem 3(c)) 1 (r − s) × J(s) =− dV 4π |r − s|3 R (by Exercise 4) = H(r).
curl A(r) =
Observe that ∇ • ∇
10. The first component of (ds • ∇)F(s) is ∇ F1 (s) • ds. Since
1 J(s) × (r − s) dV. 4π |r − s|3
13.
I 4π I div A(r) = 4π I = 4π A(r) =
ds |r − s| F
1 ds div r |r − s| F 1 • ds ∇ |r − s| F (by Theorem 3(b)) = 0 for r not on F, since ∇(1/|r − s|) is conservative.
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.7 (PAGE 896)
1 J(s) d V , where R is a region of 34π R |r − s| space such that J(s) = 0 outside R. We assume that J(s) is continuous, so J(s) = 0 on the surface S of R. In the following calculations we use subscripts s and r to denote the variables with respect to which derivatives are taken. By Theorem 3(b),
This heat content increases at (time) rate
14. A(r) =
div s
dH = δc dt
ˆ d S. −k∇T • N
because ∇r |r − s| = −∇s |r − s|, and because ∇ • J = ∇ • (∇ × H) = 0 by Theorem 3(g). Hence
Therefore, the rate at which heat enters R through S is ˆ d S. k ∇T • N S
1 • J(s) d V ∇r |r − s| R 1 J(s) =− ∇s • dV 4π |r − s| R 1 J(s) ˆ dS = 0 =− •N 4π S |r − s|
1 div A(r) = 4π
By conservation of energy and the Divergence Theorem we have ∂T ˆ dS δc d V = k ∇T • N R ∂t S =k ∇ • ∇T d V R ∇2T d V . =k
since J(s) = 0 on S. By Theorem 3(i), J = ∇ × H = ∇ × (∇ × A) = ∇(∇ • A) − ∇ 2 A = −∇ 2 A.
R
Thus,
15. By Maxwell’s equations, since ρ = 0 and J = 0,
R
∂H ∂t
curl H = 0
k ∂T − ∇2T ∂t δc
d V = 0.
Since R is arbitrary, and the temperature T is assumed to be smooth, the integrand must vanish everywhere. Thus
div H = 0
curl E = −µ0
R
∂T dV. ∂t
If heat is not “created” or “destroyed” (by chemical or other means) within R, then the increase in heat content must be due to heat flowing into R across S. The rate of flow of heat into R across surface element ˆ is d S with outward normal N
1 J(s) 1 = ∇s • J(s) + ∇s • J(s) |r − s| |r − s| |r − s| 1 = −∇r • J(s) + 0 |r − s|
div E = 0
∂E ∂t
k ∂T k ∂2T ∂2T ∂2T = ∇2T = . + + ∂t δc δc ∂ x 2 ∂ y2 ∂z 2
Therefore,
curl curl E = grad div E − ∇ 2 E = −∇ 2 E ∇ 2 E = −curl curl E = µ0
∂ ∂ 2E curl H = µ0 0 2 . ∂t ∂t
Similarly,
∂ 2H . ∂t 2 Thus U = E and U = H both satisfy the wave equation ∇ 2 H = µ0 0
∂2U = c2 ∇ 2 U, ∂t 2
where
c2 =
1.
2. S) at time t is
H (t) = δc
R
T (x, y, z, t) d V .
Orthogonal Curvilinear (page 896)
f (r, θ, z) = r θ z 9,
(cylindrical coordinates). By Example ∂f 1 ∂f ˆ ∂f θ+ k rˆ + ∂r r ∂θ ∂z = θ z rˆ + z θˆ + r θ k.
∇f =
1 . µ0 0
16. The heat content of an arbitrary region R (with surface
Section 16.7 Coordinates
f (ρ, φ, θ ) = ρφθ (spherical coordinates). By Example 10, ∂f 1 ∂f ˆ 1 ∂f ˆ ∇f = ρˆ + φ+ θ ∂ρ ρ ∂φ ρ sin φ ∂θ φ ˆ = φθ ρˆ + θ φˆ + θ. sin φ
619
SECTION 16.7 (PAGE 896)
3.
4.
F(r, θ, z) = r rˆ
1 ∂ 2 (r ) = 2 div F = r ∂r rˆ r θˆ k ∂ ∂ 1 ∂ curl F = r ∂r ∂θ ∂z r 0 0 F(r, θ, z) = r θˆ
1 ∂ (r ) = 0 div F = r ∂θ rˆ r θˆ k ∂ ∂ 1 ∂ curl F = r ∂r ∂θ ∂z 0 r2 0
9. Let r = x(u, v) i + y(u, v) j. The scale factors are
7.
F(ρ, φ, θ ) = ρ θˆ
8.
∂ 2 ρ sin φ = cot φ ∂φ ρˆ ρ φˆ ρ sin φ θˆ ∂ ∂ ∂ = 2 θ. ˆ ∂ρ ∂φ ∂θ 0 ρ2 0
1 2 ρ sin φ
2
F(ρ, φ, θ ) = ρ ρˆ
∂ 1 div F = 2 ρ 4 sin φ = 4ρ ρ sin φ ∂ρ ρ φˆ ρ sin φ θˆ ρˆ ∂ ∂ ∂ 1 = 0. curl F = 2 ∂θ ρ sin φ ∂ρ ∂φ 2 ρ 0 0
620
1 ∂r h u ∂u
and vˆ =
1 ∂r . h v ∂v
nates in Ê 3 , with scale factors h u , h v and h z = 1, we have, for a function f (u, v) independent of z, 1 hu 1 = hu
∇ f (u, v) =
∂f 1 uˆ + ∂u hv 1 ∂f uˆ + ∂u hv
∂f 1 ∂f k vˆ + ∂v 1 ∂z ∂f vˆ . ∂v
For F(u, v) = Fu (u, v) uˆ + Fv (u, v) vˆ (independent of z and having no k component), we have
∂ 1 ∂ (h u Fu ) + (h v Fv ) div F(u, v) = h u h v ∂u ∂v h u uˆ ˆ h v k v 1 ∂ ∂ ∂ curl F(u, v) = h u h v ∂u ∂v ∂z h F h F 0 v v u u
∂ ∂ 1 (h v Fv ) − (h u Fu ) k. = h u h v ∂u ∂v
11. We can use the expressions calculated in the text for cylindrical coordinates, applied to functions independent of z and having no k components: ∂f 1 ∂f ˆ θ rˆ + ∂r r ∂θ Fr 1 ∂ Fθ ∂ Fr + + div F(r, θ ) = ∂r r r ∂θ
Fθ 1 ∂ Fr ∂ Fθ k. + − curl F(r, θ ) = ∂r r r ∂θ ∇ f (r, θ ) =
∂ 2 ρ =0 ∂θ ρ φˆ ρ sin φ θˆ ρˆ ∂ ∂ ∂ 1 curl F = 2 ∂θ ρ sin φ ∂ρ ∂φ 0 0 ρ 2 sin φ ˆ = cot φ ρˆ − 2 φ.
div F =
uˆ =
= 2k.
F(ρ, φ, θ ) = ρ φˆ
1 curl F = 2 ρ sin φ
The local basis consists of the vectors
10. Since (u, v, z) constitute orthogonal curvilinear coordi-
6.
∂r and h v = . ∂v
The area element is d A = h u h v du dv.
F(ρ, φ, θ ) = sin φ ρˆ 2 sin φ ∂ 1 2 2 div F = 2 ρ sin φ = ρ sin φ ∂ρ ρ ρˆ ρ φˆ ρ sin φ θˆ ∂ 1 ∂ ∂ curl F = 2 ρ sin φ ∂ρ ∂φ ∂θ sin φ 0 0 cos φ ˆ θ. =− ρ
1 2 ρ sin φ
∂r h u = ∂u
= 0.
5.
div F =
R. A. ADAMS: CALCULUS
12.
x = a cosh u cos v,
y = a sinh u sin v.
a) u-curves: If A = a cosh u and B = a sinh u, then x2 y2 + 2 = cos2 v + sin2 v = 1. 2 A B Since A2 − B 2 = a 2 (cosh2 u − sinh2 u) = a 2 , the u-curves are ellipses with foci at (±a, 0). b) v-curves: If A = a cos v and B = a sin v, then y2 x2 − 2 = cosh2 u − sinh2 u = 1. 2 A B Since A2 + B 2 = a 2 (cos2 v + sin2 v) = a 2 , the v-curves are hyperbolas with foci at (±a, 0).
INSTRUCTOR’S SOLUTIONS MANUAL
c) The u-curve u = u0 has parametric equations x = a cosh u 0 cos v,
REVIEW EXERCISES 16 (PAGE 896)
∇ f (r, θ, z) =
15.
∇ f (ρ, φ, θ ) =
16.
∇ f (u, v, w) =
y = a sinh u 0 sin v,
and therefore has slope at (u0 , v0 ) given by dy a sinh u 0 cos v0 d y d x = . = mu = dx dv dv (u 0 ,v0 ) −a cosh u 0 sin v0 The v-curve v = v0 has parametric equations x = a cosh u cos v0 ,
y = a sinh u sin v0 ,
and therefore has slope at (u0 , v0 ) given by dy a cosh u 0 sin v0 d y d x = . = mv = dx du du (u 0 ,v0 ) a sinh u 0 cos v0 Since the product of these slopes is mu m v = −1, the curves u = u 0 and v = v0 intersect at right angles. d)
r = a cosh u cos v i + a sinh u sin v j ∂r = a sinh u cos v i + a cosh u sin v j ∂u ∂r = −a cosh u sin v i + a sinh u cos v j. ∂v
The scale factors are ∂r h u = = a sinh2 u cos2 v + cosh2 u sin2 v ∂u ∂r h v = = a sinh2 u cos2 v + cosh2 u sin2 v = h u . ∂v The area element is d A = h u h v du dv = a 2 sinh2 u cos2 v + cosh2 u sin2 v du dv.
13.
x = a cosh u cos v y = a sinh u sin v z = z. Using the result of Exercise 12, we see that the coordinate surfaces are u = u 0 : vertical elliptic cylinders with focal axes x = ±a, y = 0. v = v0 : vertical hyperbolic cylinders with focal axes x = ±a, y = 0. z = z 0 : horizontal planes. The coordinate curves are u-curves: the horizontal hyperbolas in which the v = v0 cylinders intersect the z = z0 planes. v-curves: the horizontal ellipses in which the u = u0 cylinders intersect the z = z0 planes. z-curves: sets of four vertical straight lines where the elliptic cylinders u = u0 and hyperbolic cylinders v = v0 intersect.
∂f 1 ∂f ˆ ∂f θ+ k rˆ + ∂r r ∂θ ∂z ∇ 2 f (r, θ, z) = div ∇ f (r, θ, z)
∂f ∂ 1 ∂f ∂ ∂f 1 ∂ r + + r = r ∂r ∂r ∂θ r ∂θ ∂z ∂z ∂2 f 1 ∂f ∂2 f 1 ∂2 f = + + 2. + 2 r ∂r ∂r 2 r ∂θ 2 ∂z
14.
∂f ˆ 1 ∂f ˆ ∂f 1 ρˆ + φ+ θ ∂ρ ρ ∂φ ρ sin φ ∂θ ∇ 2 f (ρ, φ, θ ) = div f (ρ, φ, θ ) ∂ ∂ 1 ∂f ∂f 1 ρ 2 sin φ + ρ sin φ = 2 ρ sin φ ∂ρ ∂ρ ∂φ ρ ∂φ
∂ ρ ∂f + ∂θ ρ sin φ ∂θ ∂2 f 1 ∂2 f 2 ∂f = + 2 + 2 ∂ρ ρ ∂ρ ρ ∂φ 2 cot φ ∂ f 1 ∂2 f + 2 . + 2 2 ρ ∂φ ρ sin φ ∂θ 2 1 ∂f 1 ∂f 1 ∂f ˆ uˆ + vˆ + w h u ∂u h v ∂v h w ∂w ∇ 2 f (u, v, w) = div ∇ f (u, v, w) ∂ hu hw ∂ f ∂ hvhw ∂ f 1 + = h u h v h w ∂u h u ∂u ∂v h v ∂v
∂ hu hv ∂ f + ∂w h w ∂w
1 ∂2 f 1 ∂h w 1 ∂h u ∂ f 1 ∂h v = 2 + − + h u ∂u 2 h v ∂u h w ∂u h u ∂u ∂u
2 1 ∂ f 1 ∂h w 1 ∂h v ∂ f 1 ∂h u + 2 + − + h v ∂v 2 h u ∂v h w ∂v h v ∂v ∂v
1 ∂2 f 1 ∂h v 1 ∂h w ∂ f 1 ∂h u + 2 . + − + h w ∂w2 h u ∂w h v ∂w h w ∂w ∂w
Review Exercises 16
(page 896)
1. The semi-ellipsoid S with upward normal Nˆ specified in
the problem and the disk D given by x 2 + y 2 ≤ 16, z = 0, with downward normal −k together bound the solid region R: 0 ≤ z ≤ 12 16 − x 2 − y 2 . By the Divergence Theorem: S
ˆ dS + F•N
D
F • (−k) d A =
R
div F d V .
621
REVIEW EXERCISES 16 (PAGE 896)
R. A. ADAMS: CALCULUS
For F = x 2 zi + (y 2 z + 3y)j + x 2 k we have div F d V = (2x z + 2yz + 3) d V R R = 0+0+3 d V = 3 × (volume of R) R
2
2
(3y 2 + 2xe y ) d x + (2x 2 ye y ) d y C 2 2 = [4x ye y − (6y + 4x ye y )] d A P = −6 y d A = −6 y¯ A = −6, P
3 4 2 π 4 2 = 64π. = 2 3
since P has area A = 2 and its centroid has y-coordinate y¯ = 1/2. y
The flux of F across S is ˆ d S = 64π + F•N F •kdA S D
(1, 1)
D 2π
0
P
2
cos θ dθ
4
0
r dr = 128π.
the planes z = 0 and z = b. The oriented boundary of R ˆ 1 = k and consists of S and the disks D1 with normal N ˆ D2 with normal N2 = −k as shown in the figure. For F = xi + cos(z 2 )j + e z k we have div F = 1 + e z and b div F d V = dx dy (1 + e z ) dz R D2 0 = [b + (eb − 1)] d x d y D2 2
Also
2
b
= π a b + π a (e − 1). F • (−k) d A = − e0 d A = −π a 2 D2 D2 F •kdA = eb d A = π a 2 eb . D1
(2, 0)
S
D1
R
Fig. R-16.3
ˆ1 = k N
S
z
i ∂ curl F = ∂x −z
j ∂ ∂y x
k ∂ ∂z y
= i − j + k.
ˆ to a region in the plane The unit normal N 2x + y + 2z = 7 is
If C is the boundary of a disk D of radius a in that plane, then C
F • dr = =±
curl F • Nˆ d S
D
D
2−1+2 d S = ±π a 2 . 3
5. If Sa is the sphere of radius a centred at the origin, then div F(0, 0, 0) = lim 4 3 F • Nˆ d S a→0+ π a Sa 3 3 3 = lim (π a 3 + 2a 4 ) = . a→0+ 4π a 3 4
b
1
R
ˆ N
6. If S is any surface with upward normal Nˆ and boundary
D2 2a
ˆ 2 = −k N Fig. R-16.2
622
4. If F = −zi + xj + yk, then
ˆ d S = π a2 b. F•N
D1
x
x
ˆ = ± 2i + j + 2k . N 3
By the Divergence Theorem ˆ dS + F•N F• kdA + F • (−k) d A S D1 D2 = div F d V = π a 2 b + π a 2 (eb − 1). Therefore,
C
3
2. Let R be the region inside the cylinder S and between
(3, 1)
x2 d A
= 64π + = 64π +
3.
y
the curve C: x 2 + y 2 = 1, z = 2, then C is oriented counterclockwise as seen from above, and it has parametrization r = cos ti + sin tj + 2k (0 ≤ 2 ≤ 2π ).
INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 16 (PAGE 896)
Thus dr = (− sin ti + cos tj) dt, and if F = −yi + x cos(1 − x 2 − y 2 )j + yzk, then the flux of curl F upward through S is S
curl F • Nˆ d S =
0
r λr
where r = xi + yj + zk and r = |r|. Since r 2 = x 2 + y 2 + z 2 , therefore ∂r/∂ x = x/r and x2 ∂ λ (r x) = λr λ−1 + r λ = r λ−2 (λx 2 + r 2 ). ∂x r Similar expressions hold for (∂/∂ y)(r λ y) and (∂/∂z)(r λ z), so
div F(r) = r
λ−2
2
2
λ
Ê3
that excludes the origin if an only if λ = −3. In this case F is not defined at r = 0. There is no value of λ for which F is solenoidal on all of Ê 3 .
then by Green’s Theorem, the circulation of F around C is F • dr C
∂ ∂ = (x − x 3 + x 2 y) − (2y 3 − 3y + x y 2 ) d A ∂x ∂y R 2 2 (1 − 3x + 2x y − 6y + 3 − 2x y) d A = R = (4 − 3x 2 − 6y 2 ) d x d y. The last integral has a maximum value when the region R is bounded by the ellipse 3x 2 + 6y 2 = 4, oriented counterclockwise; this is the largest region in the x yplane where the integrand is nonnegative.
11. Let S be a closed, oriented surface in
1 div curl F = 0 µ
by Theorem 3(g) of Section 7.2. By part (i) of the same theorem, ∇ 2 F = ∇(div F) − curl curl F = 0 − µcurl F = −µ2 F. Thus ∇ 2 F + µ2 F = 0.
9. Apply the variant of the Divergence Theorem given in Theorem 7(b) of Section 7.3, namely grad φ d V = φ Nˆ d S, P S
F = (4x + 2x 3 z)i − y(x 2 + z 2 )j − (3x 2 z 2 + 4y 2 z)k, then by the Divergence Theorem, the flux of F through S is ˆ dS = F•N div F d V = (4−x 2 −4y 2 −z 2 ) d V . S R R The last integral has a maximum value when the region R is bounded by the ellipsoid x 2 + 4y 2 + z 2 = 4 with outward normal; this is the largest region in Ê3 where the integrand is nonnegative.
12. Let C be a simple, closed curve on the plane
x + y + z = 1, oriented counterclockwise as seen from above, and bounding a plane region √ S on x + y + z = 1. ˆ = (i + j + k)/ 3. If Then S has normal N F = x y 2i + (3z − x y 2 )j + (4y − x 2 y)k, then j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 2 2 x y 3z − x y 4y − x y = (1 − x 2 )i + 2x yj − (y 2 + 2x y)k.
to the scalar field φ = 1 over the polyhedron P. Here n S = Fi is the surface of P, oriented with outward i=1
ˆ i on the face Fi . If Ni = Ai N ˆ i , where Ai normal field N is the area of Fi , then, since grad φ = 0, we have n n n Ni Ni ˆ dS = dS = Ai = Ni . 0= N A S Fi Ai i=1 i=1 i i=1
Ê3
bounding a ˆ If region R, and having outward normal field N.
Ê 3 , where µ = 0 is a constant, then
div F =
F = (2y 3 − 3y + x y 2)i + (x − x 3 + x 2 y)j,
R
(λr + 3r ) = (λ + 3)r .
F is solenoidal on any set in
8. If curl F = µF on
a region R. If
F • dr C 2π (sin2 t + cos2 t + 0) dt = 2π. =
7. F(r) =
10. Let C be a simple, closed curve in the x y-plane bounding
By Stokes’s Theorem we have
C
F • dr =
S
curl F • Nˆ d S =
1 − x 2 − y2 d S. √ 3 S
623
REVIEW EXERCISES 16 (PAGE 896)
R. A. ADAMS: CALCULUS
The last integral will be maximum if the projection of S onto the x y-plane is the disk x 2 + y 2 ≤ 1. This maximum value is 1 − x 2 − y2 √ √ 3 dx dy 3 x 2 +y 2 ≤1 2π 1 π 1 1 = − = . dθ (1 − r 2 )r dr = 2π 2 4 2 0 0
Challenging Problems 16
b) If S is the intersection of a smooth surface with the general half-cone K , and is oriented with normal ˆ pointing away from the vertex P of K , and field N if Sa is the intersection with K of a sphere of radius a centred at P, with a chosen so that S and Sa do not intersect in K , then S, Sa , and the walls of K bound a solid region R that does not contain the origin. If F = r/|r|3 , then div F = 0 in R (see ˆ = 0 on the Example 3 in Section 7.1), and F • N walls of K . It follows from the Divergence Theorem applied to F over R that
(page 897)
1. By Theorem 1 of Section 7.1, we have
S
3 ˆ v(r) • N(r) div v(r1 ) = lim d S. →0+ 4π 3 S Here S is the sphere of radius centred at the point (with position vector) r1 and having outward normal field ˆ N(r). If r is (the position vector of) any point on S , ˆ and then r = r1 + N(r), ˆ v(r) • N(r) dS S ˆ = v(r1 ) + v(r) − v(r1 ) • N(r) dS S ˆ = v(r1 ) • N(r) dS S r−r 1 d S. + v(r) − v(r1 ) • S ˆ But N(r) d S = 0 by Theorem 7(b) of Section 7.3 S with φ = 1. Also, since v satisfies v(r2 ) − v(r1 ) = C|r2 − r1 |2 , we have
r−r 1 v(r) − v(r1 ) • dS S C 2 = d S = 4π C 3 . S
Thus
div v(r1 ) = lim
→0+
3 (0 + 4π C 3 ) = 3C. 4π 3
The divergence of the large-scale velocity field of matter in the universe is three times Hubble’s constant C.
2.
a) The steradian measure of a half-cone of semi-vertical angle α is
2π
dθ 0
624
0
α
sin φ dφ = 2π(1 − cos α).
ˆ dS = F•N
a2 = 4 a
F•
Sa
r dS |r| dS =
Sa = area of S1 .
1 (area of Sa ) a2
The area of S1 (the part of the sphere of radius 1 in K ) is the measure (in steradians) of the solid angle subtended by K at its vertex P. Hence this measure is given by r ˆ d S. •N S |r|3
3.
a) Verification of the identity ∂ ∂t
∂r ∂ ∂r G• − G• ∂s ∂s ∂t ∂F ∂r ∂r ∂r = • + (∇ × F) × • . ∂t ∂s ∂t ∂s
can be carried out using the following MapleV commands: > > > > >
with(linalg): F:=(x,y,z,t)-> [F1(x,y,z,t), F2(x,y,z,t),F3(x,y,z,t)]; r:=(s,t)->[x(s,t),y(s,t),z(s,t)];
G:=(s,t)->F(x(s,t),y(s,t),z(s,t),t); > g:=(s,t)-> dotprod(G(s,t), > map(diff,r(s,t),s)); > h:=(s,t)-> dotprod(G(s,t), > map(diff,r(s,t),t)); > LH1:=diff(g(s,t),t); > LH2:=diff(h(s,t),s); > LHS:=simplify(LH1-LH2); >
RH1:=dotprod(subs(x=x(s,t),y=y(s,t), > z=z(s,t),diff(F(x,y,z,t),t)), > diff(r(s,t),s)); >
RH2:=dotprod(crossprod(subs(x=x(s,t), > y=y(s,t),z=z(s,t),
INSTRUCTOR’S SOLUTIONS MANUAL
> > >
curl(F(x,y,z,t),[x,y,z])), diff(r(s,t),t)),diff(r(s,t),s)); RHS:=RH1+RH2; LHS-RHS; simplify(%);
We omit the output here; some of the commands produce screenfulls of output. The output of the final command is 0, indicating that the identity is valid. b) As suggested by the hint, d dt
Ct b
F • dr =
a
b
∂ ∂t
∂r G• ds ∂s
∂r ∂ = G• ∂s ∂t a
∂ ∂r ∂ ∂r + G• − G• ds ∂t ∂s ∂s ∂t s=b ∂r = G • ∂t s=a
b ∂r ∂r ∂F + (∇ × F) × • ds + ∂t ∂t ∂s a = F r(b, t), t • vC (b, t) − F r(a, t), t • vC (a, t) ∂F + • dr + (∇ × F) × vC • dr. Ct ∂t Ct
4.
a) Verification of the identity
∂r ∂r ∂ ∂r ∂r G• × − G• × ∂u ∂v ∂u ∂t ∂v
∂ ∂r ∂r − G• × ∂v ∂u ∂t
∂F ∂r ∂r ∂r ∂r ∂r = • × + (∇ • F) • × . ∂t ∂u ∂v ∂t ∂u ∂v
∂ ∂t
can be carried out using the following MapleV commands: > > > > > > > > > > >
with(linalg): F:=(x,y,z,t)->[F1(x,y,z,t), F2(x,y,z,t),F3(x,y,z,t)]; r:=(u,v,t)->[x(u,v,t),y(u,v,t), z(u,v,t)]; ru:=(u,v,t)->diff(r(u,v,t),u); rv:=(u,v,t)->diff(r(u,v,t),v); rt:=(u,v,t)->diff(r(u,v,t),t); G:=(u,v,t)->F(x(u,v,t), y(u,v,t),z(u,v,t),t);
ruxv:=(u,v,t)->crossprod(ru(u,v,t), > rv(u,v,t)); >
rtxv:=(u,v,t)->crossprod(rt(u,v,t), > rv(u,v,t));
CHALLENGING PROBLEMS 16 (PAGE 897)
>
ruxt:=(u,v,t)->crossprod(ru(u,v,t), > rt(u,v,t)); > LH1:=diff(dotprod(G(u,v,t), > ruxv(u,v,t)),t); > LH2:=diff(dotprod(G(u,v,t), > rtxv(u,v,t)),u); > LH3:=diff(dotprod(G(u,v,t), > ruxt(u,v,t)),v); > LHS:=simplify(LH1-LH2-LH3); > RH1:=dotprod(subs(x=x(u,v,t), > y=y(u,v,t),z=z(u,v,t), >
diff(F(x,y,z,t),t)),ruxv(u,v,t)); > RH2:=(divf(u,v,t))* >
(dotprod(rt(u,v,t),ruxv(u,v,t))); > RHS:=simplify(RH1+RH2); > simplify(LHS-RHS); Again the final output is 0, indicating that the identity is valid. b) If Ct is the oriented boundary of St and L t is the corresponding counterclockwise boundary of the parameter region R in the uv-plane, then ∂r • dr F× ∂t Ct ∂r ∂r ∂r • du + dv G× = ∂t ∂u ∂v L
t ∂r ∂r ∂r ∂r × +G• × dt −G • = ∂u ∂t ∂t ∂v Lt ∂r ∂r ∂ = G• × ∂t ∂v R ∂u
∂ ∂r ∂r + G• × du dv, ∂v ∂u ∂t by Green’s Theorem. c) Using the results of (a) and (b), we calculate
d ∂r ∂r ∂ ˆ dS = G• × du dv F•N dt St ∂u ∂v R ∂t ∂r ∂r ∂F du dv • × = ∂u ∂v R ∂t ∂r ∂r ∂r + du dv × (div F) • ∂t ∂u ∂v R ∂r ∂r ∂ + G• × ∂t ∂v R ∂u
∂ ∂r ∂r + G• × du dv ∂v ∂u ∂t ∂F ˆ ˆ dS • NdS + (div F)v S • N = St ∂t St
625
CHALLENGING PROBLEMS 16 (PAGE 897)
R. A. ADAMS: CALCULUS
+ Ct (F × vC ) • dr.
5. We have 1 t
Dt+t
f (r, t + t) d V −
Dt
f (r, t) d V
f (r, t + t) − f (r, t) dV = t Dt 1 + f (r, t + t) d V t Dt+t −Dt 1 − f (r, t + t) d V t Dt −Dt+t = I1 + I2 − I3 . ∂f Evidently I1 → d V as t → 0. Dt ∂t I2 and I3 are integrals over the parts of Dt where the surface §t is moving outwards and inwards, respectively, ˆ is, respectively, positive and negative. that is, where vS • N ˆ d S T , we have Since d V = |vS • N|
626
I2 − I3 =
St
=
St
+
ˆ dS f (r, t + t)vS • N ˆ dS f (r, t)v S • N
St
ˆ d S. f (r, t + t) − f (r, t) v S • N
The latter integral approaches 0 as t → 0 because ˆ d S • N f (r, t + t) − f (r, t) v S St ∂f ≤ max |vS | (area of St )t. ∂t
CHAPTER 16. VECTOR CALCULUS Section 16.1 (page 858) 1.
2.
F = xi + yj ∂ ∂ ∂ (x) + (y) + (0) = 1 + 1 = 2 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = =0 ∂ x ∂ y ∂z x y 0 F = yi + xj ∂ ∂ ∂ (y) + (x) + (0) = 0 + 0 = 0 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = (1 − 1)k = 0 ∂ x ∂ y ∂z y x 0 F = yi + zj + xk ∂ ∂ ∂ div F = (y) + (z) + (x) = 0 ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = −i − j − k ∂ x ∂ y ∂z y z x
4.
F = yzi + x zj + x yk ∂ ∂ ∂ (yz) + (x z) + (x y) = 0 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = ∂ x ∂ y ∂z yz x z x y = (x − x)i + (y − y)j + (z − z)k = 0
5.
F = xi + xk ∂ ∂ ∂ (x) + (0) + (x) = 1 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = −j ∂ x ∂ y ∂z x 0 x F = x y 2 i − yz 2j + zx 2 k ∂ 2 ∂ ∂ 2 div F = xy + −yz 2 + zx ∂x ∂y ∂z = y2 − z2 + x 2 j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 2 2 x y −yz zx = 2yzi − 2x zj − 2x yk
600
7.
F = f (x)i + g(y)j + h(z)k ∂ ∂ ∂ f (x) + g(y) + h(z) div F = ∂x ∂y ∂z = f (x) + g (y) + h (z) i j k ∂ ∂ ∂ curl F = =0 ∂x ∂y ∂z f (x) g(y) h(z)
8.
F = f (z)i − f (z)j ∂ ∂ f (z) + div F = − f (z) = 0 ∂x ∂y i j k ∂ ∂ ∂ curl F = = f (z)(i + j) ∂x ∂y ∂z f (z) − f (z) 0
Gradient, Divergence, and Curl
3.
6.
R. A. ADAMS: CALCULUS
9. Since x = r cos θ , and y = r sin θ , we have r 2 = x 2 + y 2 , and so
x ∂r = = cos θ ∂x r y ∂r = = sin θ ∂y r ∂ ∂ y −x y cos θ sin θ sin θ = = 3 =− ∂x ∂x r r r ∂ y 1 y2 ∂ sin θ = = − 3 ∂y ∂y r r r 2 2 x cos θ = 3 = r r ∂ x 1 x2 ∂ cos θ = = − 3 ∂x ∂x r r r y2 sin2 θ = 3 = r r ∂ ∂ x −x y cos θ sin θ cos θ = = 3 =− . ∂y ∂y r r r (The last two derivatives are not needed for this exercise, but will be useful for the next two exercises.) For F = r i + sin θ j, we have ∂ cos2 θ ∂r + sin θ = cos θ + ∂x ∂y r i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z r sin θ 0 sin θ cos θ − sin θ k. = − r
div F =
INSTRUCTOR’S SOLUTIONS MANUAL
10.
F = rˆ = cos θ i + sin θ j sin2 θ
11.
It follows that
cos2 θ
1 1 + = = r r r x 2 + y2 i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z cos θ sin θ 0 cos θ sin θ cos θ sin θ =− − k=0 r r
div F =
SECTION 16.1 (PAGE 858)
F = θˆ = − sin θ i + cos θ j cos θ sin θ cos θ sin θ − =0 div F = r r i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z − sin θ cos θ 0 2 sin θ cos2 θ 1 1 = + k= k= k r r r x 2 + y2
1 ˆ d S = div F(0, 0, 0). F•N a,b,c→0+ 8abc Ba,b,c lim
13. This proof just mimics that of Theorem 1. F can be expanded in Maclaurin series F = F 0 + F1 x + F2 y + · · · , where F0 = F(0, 0)
∂ ∂ F1 F1 = = F(x, y) i+ ∂x ∂x (0,0) ∂ ∂ F1 F(x, y) i+ F2 = = ∂y ∂y
12. We use the Maclaurin expansion of F, as presented in the proof of Theorem 1: F = F 0 + F1 x + F2 y + F3 z + · · · , where F0 = F(0, 0, 0)
∂ F2 ∂ F3 ∂ ∂ F1 F(x, y, z) i+ j+ k = ∂x ∂x ∂x ∂x (0,0,0) (0,0,0) ∂ F ∂ F ∂ ∂ F 1 2 3 F(x, y, z) i+ j+ k F2 = = ∂y ∂y ∂y ∂y (0,0,0) (0,0,0) ∂ F ∂ F ∂ ∂ F 1 2 3 F(x, y, z) i+ j+ k F3 = = ∂z ∂z ∂z ∂z (0,0,0) (0,0,0) F1 =
and where · · · represents terms of degree 2 and higher in x, y, and z. ˆ = k. On the top of the box Ba,b,c , we have z = c and N ˆ = −k. On the bottom of the box, we have z = −c and N On both surfaces d S = d x d y. Thus ˆ dS + F•N
top a
bottom b
d y cF3 • k − cF3 • (−k) + · · · −a −b ∂ = 8abcF3 • k + · · · = 8abc F3 (x, y, z) + ···, ∂z (0,0,0)
=
dx
where · · · represents terms of degree 4 and higher in a, b, and c. Similar formulas obtain for the two other pairs of faces, and the three formulas combine into ˆ d S = 8abcdiv F(0, 0, 0) + · · · . F•N Ba,b,c
(0,0)
∂ F2 j ∂x (0,0) ∂ F2 j ∂y (0,0)
and where · · · represents terms of degree 2 and higher in x and y. On the curve C of radius centred at (0, 0), we have ˆ = 1 (xi + yj). Therefore, N ˆ = 1 F0 • ix + F0 • jy + F1 • ix 2 F•N + F1 • jx y + F2 • ix y + F2 • jy 2 + · · · where · · · represents terms of degree 3 or higher in x and y. Since C C
x ds = x 2 ds =
C
y ds =
C
C y 2 ds =
x y ds = 0
0
2π
2 cos2 θ dθ = π 3 ,
we have 3 1 ˆ ds = 1 π (F1 • i + F2 • j) + · · · F•N 2 2 π C π = div F(0, 0) + · · · where · · · represents terms of degree 1 or higher in . Therefore, taking the limit as → 0 we obtain lim
→0
1 ˆ ds = div F(0, 0). F•N π 2 C
601
SECTION 16.1 (PAGE 858)
R. A. ADAMS: CALCULUS
14. We use the same Maclaurin expansion for F as in Exercises 12 and 13. On C we have r = cos θ i + sin θ j, (0 ≤ θ ≤ 2π ) dr = − sin θ i + cos θ j F • dr = − sin θ F0 • i + cos θ F0 • j − 2 sin θ cos θ F1 • i + 2 cos2 θ F1 • j
− 2 sin2 θ F2 • i + 2 sin θ cos θ F2 • j + · · · ds,
where · · · represents terms of degree 3 or higher in . Since 0
0
2π
2π
sin θ dθ = cos2 θ dθ =
2π
0
0
cos θ dθ =
2π
2π 0
sin θ cos θ dθ = 0
sin2 θ dθ = π,
we have 1 F • dr = F1 • j − F2 • i + · · · , π 2 C where · · · represents terms of degree at least 1 in . Hence 1 F • dr = F1 • j − F2 • i lim →0+ π 2 C ∂ F2 ∂ F1 = − ∂x ∂y ˆ = curl F • k = curl F • N.
Section 16.2 Some Identities Involving Grad, Div, and Curl (page 864) 1. Theorem 3(a): ∂ ∂ ∂ (φψ) + (φψ) + (φψ) ∇(φψ) = ∂x ∂y ∂z ∂ψ ∂φ ∂ψ ∂φ = φ + ψ i + ··· + φ + ψ k ∂x ∂x ∂z ∂z = φ∇ψ + ψ∇φ.
2. Theorem 3(b): ∂ ∂ ∂ (φ F1 ) + (φ F2 ) + (φ F3 ) ∂x ∂y ∂z ∂φ ∂φ ∂ F1 ∂ F3 F1 + φ + ··· + F3 + φ +··· = ∂x ∂x ∂z ∂z = ∇φ • F + φ∇ • F.
∇ • (φF) =
602
3. Theorem 3(d): ∂ (F2 G 3 − F3 G 2 ) + · · · ∂x ∂ F3 ∂G 3 ∂G 2 ∂ F2 G 3 + F2 − G 2 − F3 + ··· = ∂x ∂x ∂x ∂x = (∇ × F) • G − F • (∇ × G).
∇ • (F × G) =
4. Theorem 3(f). The first component of ∇(F • G) is ∂ F1 ∂G 1 ∂ F2 ∂G 2 ∂ F3 ∂G 3 G 1 + F1 + G 2 + F2 + G 3 + F3 . ∂x ∂x ∂x ∂x ∂x ∂x We calculate the first components of the four terms on the right side of the identity to be proved. The first component of F × (∇ × G) is ∂G 1 ∂G 3 ∂G 2 ∂G 1 F2 − − F3 − . ∂x ∂y ∂z ∂x The first component of G × (∇ × F) is ∂ F1 ∂ F3 ∂ F2 ∂ F1 − − G3 − . G2 ∂x ∂y ∂z ∂x The first component of (F • ∇)G is F1
∂G 1 ∂G 1 ∂G 1 + F2 + F3 . ∂x ∂y ∂z
The first component of (G • ∇)F is G1
∂ F1 ∂ F1 ∂ F1 + G2 + G3 . ∂x ∂y ∂z
When we add these four first components, eight of the fourteen terms cancel out and the six remaining terms are the six terms of the first component of ∇(F • G), as calculated above. Similar calculations show that the second and third components of both sides of the identity agree. Thus ∇(F•G) = F×(∇ ×G)+G×(∇ ×F)+(F•∇)G+(G•∇)F.
5. Theorem 3(h). By equality of mixed partials, i j k ∂ ∂ ∂ ∇ × ∇φ = ∂ x ∂ y ∂z ∂φ ∂φ ∂φ ∂ x ∂ y ∂z ∂ ∂φ ∂ ∂φ = − i + · · · = 0. ∂ y ∂z ∂z ∂ y
6. Theorem 3(i). We examine the first components of the terms on both sides of the identity ∇ × (∇ × F) = ∇(∇ • F) − ∇ 2 F.
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.2 (PAGE 864)
The first component of ∇ × (∇ × F) is ∂ F2 ∂ ∂ F1 ∂ F1 ∂ F3 − − − ∂x ∂y ∂z ∂z ∂x ∂ 2 F2 ∂ 2 F1 ∂ 2 F1 ∂ 2 F3 = − . − + ∂ y∂ x ∂ y2 ∂z 2 ∂z∂ x
∂ ∂y
The first component of ∇(∇ • F) is
−
f (r )r
= 0, so that
du + 3u = 0 dr du 3 dr =− u r ln |u| = −3 ln |r | + ln |C|
r
Thus f (r ) = Cr −3 , for some constant C.
10. Given that div F = 0 and curl F = 0, Theorem 3(i)
The first component of −∇ 2 F is −∇ 2 F1 =
u = Cr −3 .
∂ 2 F2 ∂ 2 F1 ∂ 2 F3 ∂ + ∇•F= + . 2 ∂x ∂x ∂ x∂ y ∂ x∂z
∂ 2 F1 − 2 ∂x
If f (r )r is solenoidal then ∇ • u = f (r ) satisfies
∂ 2 F1 ∂ y2
−
implies that ∇ 2 F = 0 too. Hence the components of F are harmonic functions. If F = ∇φ, then
∂ 2 F1 . ∂z 2
Evidently the first components of both sides of the given identity agree. By symmetry, so do the other components.
7. If the field lines of F(x, y, z) are parallel straight lines, in
∇ 2 φ = ∇ • ∇φ = ∇ • F = 0, so φ is also harmonic.
11. By Theorem 3(e) and 3(f),
the direction of the constant nonzero vector a say, then F(x, y, z) = φ(x, y, z)a for some scalar field φ, which we assume to be smooth. By Theorem 3(b) and (c) we have
div F = div (φa) = ∇φ • a curl F = curl (φa) = ∇φ × a.
8. If r = xi + yj + zk and r = |r|, then ∇ × r = 0,
∇r =
r . r
If c is a constant vector, then its divergence and curl are both zero. By Theorem 3(d), (e), and (f) we have ∇ • (c × r) = (∇ × c) • r − c • (∇ × r) = 0 ∇ × (c × r) = (∇ • r)c + (r • ∇)c − (∇ • c)r − (c • ∇)r = 3c + 0 − 0 − c = 2c ∇(c • r) = c × (∇ × r) + r × (∇ × c) + (c • ∇)r + (r • ∇)c = 0 + 0 + c + 0 = c.
9.
∇ • f (r )r = ∇ f (r ) • r + f (r )(∇ • r) r•r + 3 f (r ) = f (r ) r = r f (r ) + 3 f (r ).
If r = xi + yj + zk, then ∇ • r = 3 and ∇ × r = 0. Also, (F • ∇)r = F1
Since ∇φ is an arbitrary gradient, div F can have any value, but curl F is perpendicular to a, and thereofore to F.
∇ • r = 3,
∇ × (F × r) = (∇ • r)F + (r • ∇)F − (∇ • F)r − (F • ∇)r ∇(F • r) = F × (∇ × r) + r × (∇ × F) + (F • ∇)r + (r • ∇)F.
∂r ∂r ∂r + F2 + F3 = F. ∂x ∂y ∂z
Combining all these results, we obtain ∇ × (F × r) − ∇(F • r) = 3F − 2(F • ∇)r − (∇ • F)r − r × (∇ × F) = F − (∇ • F)r − r × (∇ × F). In particular, if ∇ • F = 0 and ∇ × F = 0, then ∇ × (F × r) − ∇(F • r) = F.
12. If ∇ 2 φ = 0 and ∇ 2 ψ = 0, then ∇ • (φ∇ψ − ψ∇φ) = ∇φ • ∇ψ + φ∇ 2 ψ − ∇ψ • ∇φ − ψ∇ 2 φ = 0, so φ∇ψ − ψ∇φ is solenoidal.
13. By Theorem 3(c) and (h), ∇ × (φ∇ψ) = ∇φ × ∇ψ + φ∇ × ∇ψ = ∇φ × ∇ψ −∇ × (ψ∇φ) = −∇ψ × ∇φ − ψ∇ × ∇φ = ∇φ × ∇ψ.
603
SECTION 16.2 (PAGE 864)
14. By Theorem 3(b), (d), and (h), we have
∇ • f (∇g × ∇h) = ∇ f • (∇g × ∇h) + f ∇ • (∇g × ∇h) = ∇ f • (∇g × ∇h) + f (∇ × ∇g) • ∇h − ∇g • (∇ × ∇h) = ∇ f • (∇g × ∇h) + 0 − 0 = ∇ f • (∇g × ∇h).
R. A. ADAMS: CALCULUS
Look for a solution with G 2 = 0. We have G3 =
Try M(x, z) = 0. Then G 3 = x ye2z , and ∂G 1 ∂G 3 = ye2z + = 2ye2z . ∂z ∂x
15. If F = ∇φ and G = ∇ψ, then ∇ × F = 0 and ∇ × G = 0 by Theorem 3(h). Therefore, by Theorem 3(d) we have ∇ • (F × G) = (∇ × F) • G + F • (∇ × G) = 0.
Thus
Thus F × G is solenoidal. By Exercise 13, ∇ × (φ∇ψ) = ∇φ × ∇ψ = F × G, so φ∇ψ is a vector potential for F × G. (So is −ψ∇φ.)
16. If ∇ × G = F = −yi + xj, then
xe2z d y = x ye2z + M(x, z).
G1 =
2ye2z dz = ye2z + N (x, y).
Since −e2z = −
∂G 1 ∂N = −e2z − , ∂y ∂y
∂G 3 ∂G 2 − = −y ∂y ∂z ∂G 3 ∂G 1 − =x ∂z ∂x ∂G 1 ∂G 2 − = 0. ∂x ∂y
we can take N (x, y) = 0. Thus G = ye2z i + x ye2z k is a vector potential for F.
As in Example 1, we try to find a solution with G2 = 0. Then y2 G 3 = − y d y = − + M(x, z). 2
18. For (x, y, z) in D let v = xi + yj + zk. The line segment r(t) = tv, (0 ≤ t ≤ 1), lies in D, so div F = 0 on the
y2 Again we try M(x, z) = 0, so G 3 = − . Thus 2 ∂G 3 = 0 and ∂x G 1 = x dz = x z + N (x, y). ∂G 1 = 0 we may take N (x, y) = 0. ∂y 1 G = x zi − y 2 k is a vector potential for F. (Of course, 2 this answer is not unique.) Since
17. If F = xe 2z i + ye2z j − e2z k, then div F = e 2z + e2z − 2e2z = 0, so F is solenoidal. If F = ∇ × G, then ∂G 2 ∂G 3 − = xe2z ∂y ∂z ∂G 3 ∂G 1 − = ye2z ∂z ∂x ∂G 1 ∂G 2 − = −e2z . ∂x ∂y
604
path. We have
G(x, y, z) = =
1
tF r(t) × v dt
1
tF ξ(t), η(t), ζ (t) × v dt
0
0
where ξ = t x, η = t y, ζ = t z. The first component of curl G is (curl G)1 1 = t curl (F × v) dt 1 0 1 ∂ ∂ (F × v)3 − (F × v)2 dt t = ∂y ∂z 0 1 ∂ ∂ = (F1 y − F2 x) − (F3 x − F1 z) dt t ∂y ∂z 0 1 ∂ F ∂ F2 ∂ F3 1 = − t2x − t2x t F1 + t 2 y ∂η ∂η ∂ζ 0 ∂ F1 dt + t F1 + t 2 z ∂ζ 1 ∂ F1 ∂ F1 ∂ F1 = + t2 y + t 2z dt. 2t F1 + t 2 x ∂ξ ∂η ∂ζ 0
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.3 (PAGE 868)
To get the last line we used the fact that di vF = 0 to ∂ F2 ∂ F3 ∂ F1 replace −t 2 x − t2x with t 2 x . Continuing the ∂η ∂ζ ∂ξ calculation, we have
(d) LHS := Del . (F &x G):
RHS := (Del &x F) . G - F . (Del &x G): simplify(LHS - RHS); 0
d2 t F1 (ξ, η, ζ ) dt 0 dt 1 = t 2 F1 (t x, t y, t z) = F1 (x, y, z).
(curl G)1 =
1
(e) LHS := Del &x (F &x G):
RHS1 := (Del . G)*F: RHS2 := G[1]*diff(F,x) +G[2]*diff(F,y)+G[3]*diff(F,z): RHS3 := (Del . F)*G: RHS4 := F[1]*diff(G,x) +F[2]*diff(G,y)+F[3]*diff(G,z): RHS := RHS1 + RHS2 - RHS3 - RHS4: simplify(LHS - RHS);
0
Similarly, (curl G)2 = F2 and (curl G)3 = F3 . Thus curl G = F, as required.
19. In the following we suppress output (which for some calculations can be quite lengthy) except for the final check on each inequality. You may wish to use semicolons instead of colons to see what the output actually looks like. >
0 e¯ x (f) LHS := Del(F . G):
RHS1 := F &x (Del &x G): RHS2 := G &x (Del &x F): RHS3 := F[1]*diff(G,x) +F[2]*diff(G,y)+F[3]*diff(G,z): RHS4 := G[1]*diff(F,x) +G[2]*diff(F,y)+G[3]*diff(F,z): RHS := RHS1 + RHS2 + RHS3 + RHS4: simplify(LHS - RHS);
with(VectorCalculus):
>
SetCoordinates(’cartesian’[x,y,z]): > F := VectorField ():
0 e¯ x
> G := VectorField ():
All these zero outputs indicate that the inequalities (a)–(f) of the theorem are valid.
Section 16.3 (page 868)
(a) LHS := Del(phi(x,y,z)*psi(x,y,z)):
RHS := phi(x,y,z)*Del(psi(x,y,z)) + psi(x,y,z)*Del(phi(x,y,z)): simplify(LHS - RHS); 0 e¯ x (b) LHS := Del . (F*phi(x,y,z)):
RHS := (Del(phi(x,y,z))).F + phi(x,y,z)*(Del.F): simplify(LHS - RHS); 0
1.
Green’s Theorem in the Plane
2
(sin x + 3y 2 ) d x + (2x − e−y ) d y C
∂ ∂ 2 = (2x − e−y ) − (sin x + 3y 2 ) d A ∂x ∂y R (2 − 6y) d A = πR a dθ (2 − 6r sin θ )r dr = 0 0 a π sin θ dθ r 2 dr = π a2 − 6 0
0
= π a 2 − 4a 3 . y
(c) LHS := Del &x (phi(x,y,z)*F):
RHS := RHS := (Del(phi(x,y,z))) &x F + phi(x,y,z)*(Del &x F): simplify(LHS - RHS); 0 e¯ x
C
R a
−a
x
Fig. 16.3.1
605
SECTION 16.3 (PAGE 868)
2.
R. A. ADAMS: CALCULUS
(x 2 − x y) d x + (x y − y 2 ) d y C
∂ 2 ∂ =− (x y − y 2 ) − (x − x y) d A ∂x ∂y T (y + x) d A =− T 1 4 = −( y¯ + x) ¯ × (area of T ) = − +1 ×1= − . 3 3
5. By Example 1,
1 x dy − y dx 2 C 2π 1 a cos3 t 3b sin2 t cos t = 2 0
Area =
− b sin3 t (−3a cos2 t sin t) dt
3ab 2π 2 sin t cos2 t dt 2 0 3π ab 3ab 2π sin2 (2t) dt = . = 2 0 4 8
=
y (1,1)
T
C
2
6. Let R, C, and F be as in the statement of Green’s The-
x
orem. As noted in the proof of Theorem 7, the unit ˆ satisfy tangent Tˆ to C and the unit exterior normal N ˆ = Tˆ × k. Let N
Fig. 16.3.2
G = F2 (x, y)i − F1 (x, y)j.
3.
(x sin y 2 − y 2 ) d x + (x 2 y cos y 2 + 3x) d y C = 2x y cos y 2 + 3 − (2x y cos y 2 − 2y) d A T = (3 + 2y) d A = 3 d A + 0 = 3 × 3 = 9. T
ˆ Applying the 2-dimensional DiverThen F • Tˆ = G • N. gence Theorem to G, we obtain ˆ ds F1 d x + F2 d y = F • Tˆ ds = G•N C C C =
T
div G d A
R
=
y 2
R
∂ F1 ∂ F2 − ∂x ∂y
dA
as required
7. r = sin ti + sin 2tj,
(1,1)
T
(0 ≤ t ≤ 2π ) y
C
C x
R1
R2 x
(1,−1)
−2
Fig. 16.3.3 Fig. 16.3.7 2
4. Let D be the region x 2 + y 2 ≤ 9, y ≥ 0. Since C is the clockwise boundary of D, C
x 2 y d x − x y2 d y
∂ 2 ∂ =− (−x y 2 ) − (x y) d x d y ∂y D ∂x π 3 81π 2 2 = . (y + x ) d A = dθ r 3 dr = 4 D 0 0
F = ye x i + x 3 e y j i j k ∂ ∂ ∂ 2 = (3x 2 e y − e x )k. curl F = ∂ x ∂y ∂z x2 ye x 3e y 0 Observe that C bounds two congruent regions, R1 and R2 , one counterclockwise and the other clockwise. ˆ = k; for R2 , N ˆ = −k. Since R1 and R2 For R1 , N are mirror images of each other in the y-axis, and since curl F is an even function of x, we have curl F • Nˆ d S = − curl F • Nˆ d S. R1
606
R2
INSTRUCTOR’S SOLUTIONS MANUAL
Thus C
8.
F • dr =
R1
+
R2
SECTION 16.4 (PAGE 873)
curl F • Nˆ d S = 0.
a) F = x 2 j F • dr = x2 dy = 2x d A = 2 Ax¯ . C C R b) F = x yi F • dr = xy dx = − x d A = − Ax. ¯ C C R c) F = y 2 i + 3x yj F • dr = y 2 d x + 3x y d y C C = (3y − 2y) d A = A y¯ .
ˆ dS = F•N 0 d V = 0. S B
3. If F = (x 2 + y 2 )i + (y 2 − z 2 )j + zk, then div F = 2x + 2y + 1, and 4 ˆ F•N d S = (2x+2y+1) d V = 1 d V = π a3 . 3 S B B
4. If F = x 3 i + 3yz 2 j + (3y 2 z + x 2 )k, then div F = 3x 2 + 3z 2 + 3y 2 , and
R
9. The circle Cr of radius r and centre at r0 has parametrization
r = r0 + r cos ti + r sin tj,
2. If F = ye z i + x 2 e z j + x yk, then div F = 0, and
(0 ≤ t ≤ 2π ).
ˆ dS = 3 F•N (x 2 + y 2 + z 2 ) d V S B π a 2π dθ sin φ dφ ρ 4 dρ =3 0
ˆ the unit normal to Note that dr/dt = cos ti + sin tj = N, Cr exterior to the disk Dr of which Cr is the boundary. The average value of u(x, y) on Cr is u¯ r =
1 2π
2π
0
u(x0 + r cos t, y0 + r sin t) dt,
12 5 πa . = 5
Therefore the flux of F out of any solid region R is
2π ∂u d u¯ r 1 ∂u = cos t + sin t dr 2π 0 ∂x ∂y 1 ˆ ds ∇u • N = 2πr Cr
Flux =
R
dt
since ds = r dt. By the (2-dimensional) divergence theorem, and since u is harmonic, 1 d u¯ r = ∇ • ∇u d x d y dr 2πr Dr 2 1 ∂ u ∂2u = d x d y = 0. + 2πr ∂x2 ∂ y2 Dr Thus u¯ r = limr→0 u¯ r = u(x0 , y0 ).
Section 16.4 The Divergence Theorem in 3-Space (page 873) 1. In this exercise, the sphere S bounds the ball B of radius
a centred at the origin. If F = xi − 2yj + 4zk, then div F = 1 − 2 + 4 = 3. Thus ˆ dS = F•N 3 d V = 4π a 3 . S B
0
5. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z).
and so
0
=2
div F d V R
(x + y + z) d V = 2(x¯ + y¯ + z¯ )V
where (x, ¯ y¯ , z¯ ) is the centroid of R and V is the volume of R. If R is the ball (x − 2)2 + y 2 + (z − 3)2 ≤ 9, then x¯ = 2, y¯ = 0, z¯ = 3, and V = (4π/3)33 = 36π . The flux of F out of R is 2(2 + 0 + 3)(36π ) = 360π .
6. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is Flux = =2
R
div F d V R
(x + y + z) d V = 2(x¯ + y¯ + z¯ )V
where (x, ¯ y¯ , z¯ ) is the centroid of R and V is the volume of R. If R is the ellipsoid x 2 + y 2 + 4(z − 1)2 ≤ 4, then x¯ = 0, y¯ = 0, z¯ = 1, and V = (4π/3)(2)(2)(1) = 16π/3. The flux of F out of R is 2(0 + 0 + 1)(16π/3) = 32π/3.
607
SECTION 16.4 (PAGE 873)
R. A. ADAMS: CALCULUS
7. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z).
can be calculated directly by the methods of Section 6.6. We will do it here by using the Divergence Theorem instead. S is one face of a tetrahedral domain D whose other faces are in the coordinate planes, as shown in the figure. Since φ = x y + z 2 , we have
Therefore the flux of F out of any solid region R is Flux = div F d V R =2 (x + y + z) d V = 2(x¯ + y¯ + z¯ )V
∇ • ∇φ = ∇ 2 φ = 2.
∇φ = yi + xj + 2zk,
R
where (x¯ , y¯ , z¯ ) is the centroid of R and V is the volume of R.
Thus
If R is the tetrahedron with vertices (3, 0, 0), (0, 3, 0), (0, 0, 3), and (0, 0, 0), then x¯ = y¯ = z¯ = 3/4, and V = (1/6)(3)(3)(3) = 9/2. The flux of F out of R is 2((3/4) + (3/4) + (3/4))(9/2) = 81/4.
D
∇ • ∇φ d V = 2 ×
the volume of the tetrahedron D being abc/6 cubic units. z c
8. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is Flux = div F d V R =2 (x + y + z) d V = 2(x¯ + y¯ + z¯ )V
back side
S D
R
x
ˆ = −k and z = 0, so ∇φ • N ˆ = 0, and On the bottom, N the flux out of the bottom face is 0.
region having volume V , then div F d V = 3V .
ˆ = −j, so ∇φ • N ˆ = −x. The On the side, y = 0 and N flux out of the side face is
C
The region C described in the statement of the problem is the part of a solid cone with vertex at the origin that lies inside a ball of radius R with centre at the origin. The surface S of C consists of two parts, the conical wall S1 , and the region D on the spherical boundary of the ball. At any point P on S1 , the outward normal ˆ is perpendicular to the line O P, that is, to F, so field N ˆ = 0. At any point P on D, N ˆ is parallel to F, in F•N ˆ = F/|F| = F/R. Thus fact N ˆ dS = ˆ dS + ˆ dS F•N F•N F•N S S1 D R2 F•F =0+ dS = d S = AR R R D D
side
x d x dz = −
a2 c ac a × =− . 2 3 6
(We used the fact that Mx=0 = area × x¯ and x¯ = a/3 for that face.) ˆ = −i, so the flux out of On the back face, x = 0 and N that face is back
ˆ dS = − ∇φ • N
back
y d y dz = −
b2 c bc b × =− . 2 3 6
Therefore, by the Divergence Theorem I−
a 2 c b2 c abc − +0 = , 6 6 3
so
10. The required surface integral,
608
ˆ dS = − ∇φ • N
side
where A is the area of D. By the Divergence Theorem, 3V = AR, so V = AR/3.
S
bottom
The flux of ∇φ out of D is the sum of its fluxes out of the four faces of the tetrahedron.
9. If F = xi + yj + zk, then div F = 3. If C is any solid
ˆ d S, ∇φ • N
a
Fig. 16.4.10
If R is the cylinder x 2 + y 2 ≤ 2y (or, equivalently, x 2 + (y − 1)2 ≤ 1), 0 ≤ z ≤ 4, then x¯ = 0, y¯ = 1, z¯ = 2, and V = (π 12 )(4) = 4π . The flux of F out of R is 2(0 + 1 + 2)(4π ) = 24π .
y
b
where (x¯ , y¯ , z¯ ) is the centroid of R and V is the volume of R.
I =
abc abc = , 6 3
11.
2 2 ˆ d S = I = abc + c(a + b ) . ∇φ • N 3 6 S
F = (x + y 2 )i + (3x 2 y + y 3 − x 3 )j + (z + 1)k
div F = 1 + 3(x 2 + y 2 ) + 1 = 2 + 3(x 2 + y 2 ).
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.4 (PAGE 873)
z
z
a b
back
ˆ N
side
S
S
D
D B
y
a
a
x
bottom
−k Fig. 16.4.11
Fig. 16.4.12 ˆ = −j, F • N ˆ = 0, so On the side, y = 0, N ˆ d S = 0. F•N
Let D be the conical domain, S its conical surface, and B its base disk, as shown in the figure. We have D
div F d V =
2π
0
dθ a
a 0
r dr
b(1−(r/a)) 0
side
2
(2 + 3r ) dz
ˆ = −i, F • N ˆ = −y, so On the back, x = 0, N π/2 a ˆ dS = − F•N dθ r cos θ r dr
r
dr r (2 + 3r 2 ) 1 − a a 2r 2 3r 4 dr − 2r + 3r 3 − = 2π b a a 0 2π a 2 b 3π a 4 b = + . 3 10 = 2π b
0
B
By the Divergence Theorem 3 3 3 ˆ d S + 0 − a + 2a = π a . F•N 3 3 6 S
ˆ d S = −area of B = −π a2 . F•N
S
ˆ dS + F•N
B
ˆ dS = F•N
D
Hence the flux of F upward through S is 3 3 ˆ dS = πa − a . F•N 6 3 S
div F d V ,
so the flux of F upward through the conical surface S is
3π a 4 b 2π a 2 b + + π a2 . = 3 10 S
0
π/2 a3 a3 = − sin θ × =− . 3 3 0
ˆ = −k, F • N ˆ = 2x, so On the bottom, z = 0, N π/2 a 2a 3 ˆ dS = 2 . F•N dθ r cos θ r dr = 3 bottom 0 0
By the Divergence Theorem,
0
back
ˆ = −k, F • N ˆ = −1, so On B we have z = 0, N
y
a
a
x
13.
F = (x + yz)i + (y − x z)j + (z − e x sin y)k div F = 1 + 1 + 1 = 3. z
ˆ N
12. F = (y + x z)i + (y + yz)j − (2x + z 2 )k div F = z + (1 + z) − 2z = 1. Thus
π a3 , div F d V = volume of D = 6 D
where D is the region in the first octant bounded by the sphere and the coordinate planes. The boundary of D consists of the spherical part S and the four planar parts, called the bottom, side, and back in the figure.
2a a x
ˆ N
S1
S2
D y
Fig. 16.4.13
609
SECTION 16.4 (PAGE 873)
R. A. ADAMS: CALCULUS
z
a) The flux of F out of D through S = S1 ∪ S2 is
back
S front
x
a
4a 2
= 6π
3a 2
1
u 0
S1
Fig. 16.4.14
√ du = 12 3π a 3 .
ˆ = −j, F • N ˆ = x, so On the side, y = 0, N 1 1 1 ˆ dS = F•N x dx dz = . 2 0 0 side ˆ = −k, F • N ˆ = y, so On the bottom, z = 0, N 1 1 1 ˆ dS = F•N y dy dx = . 2 bottom 0 0
ˆ = −i, F • N ˆ = 0, so On the back, x = 0, N ˆ d S = 0. F•N
− 3a
0
back
ˆ = i, F • N ˆ = 3z 2 , so On the front, x = 1, N π/2 1 3π ˆ dS = 3 . F•N dθ r 2 cos2 θ r dr = 16 front 0 0
c) The flux of F out of D through the spherical part S2 is
ˆ dS = F • N ˆ dS − ˆ dS F•N F•N S2 S S1 √ √ √ = 12 3π a 3 + 4 3π a 3 = 16 3π a 3 .
Hence, ˆ d S = 3π − 1 − 1 −0− 3π = −1. (3x z 2 i− xj− yk)• N 16 2 2 16 S
15. 14. Let D be the domain bounded by S, the coordinate planes, and the plane x = 1. If
F = 3x z 2 i − xj − yk, then div F = 3z 2 , so the total flux of F out of D is
bdry of D
ˆ dS = F•N
=3
=3×
D 1
π/2 0
dθ
3π 1 π × = . 4 4 16
0
1
r 2 cos2 θ r dr
The boundary of D consists of the cylindrical surface S and four planar surfaces, the side, bottom, back, and front.
610
F = (x 2 − x − 2y)i + (2y 2 + 3y − z)j − (z 2 − 4z + x y)k div F = 2x − 1 + 4y + 3 − 2z + 4 = 2x + 4y − 2z + 6. The flux of F out of R through its surface S is ˆ dS = F•N (2x + 4y − 2z + 6) d V . S R Now x d V = M x=0 = V x, ¯ where R has volume R
3z 2 d V dx
0
y
bottom
−x 2 − x yz − y 2 + x yz a dθ dz a S1 √ 2π 3a √ dθ √ dz = −4 3π a 3 . = −a 2
ˆ dS = F•N
1
− r2
ˆ = − xi + yj , d S = a dθ dz. The flux of F b) On S1 , N a out of D through S1 is
D
Let u = du = −2r dr 1/2
1
side
ˆ dS = F•N div F d V S D 2π 2a √4a 2 −r 2 =3 dθ r dr 2 dz 0 0 a 2a r 4a 2 − r 2 dr = 12π
V and centroid (x, ¯ y¯ z¯ ). Similar formulas obtain for the other variables, so the required flux is ˆ d S = 2V x¯ + 4V y¯ − 2V z¯ + 6V . F•N S
16. F = xi + yj + zk implies that div F = 3. The total flux of F out of D is
bdry of D
ˆ dS = 3 F•N
D
d V = 12,
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.4 (PAGE 873)
since the volume of D is half that of a cube of side 2, that is, 4 square units. D has three triangular faces, three pentagonal faces, and a hexagonal face. By symmetry, the flux of F out of each triangular face is equal to that out of the triangular face ˆ = k • k = 1 on that T in the plane z = 1. Since F • N face, these fluxes are d x d y = area of T =
T
18. φ = x 2 − y 2 + z 2 , G = 13 (−y 3 i + x 3 j + z 3 k). F = ∇φ + µcurl G. Let R be the region of 3-space occupied by the sandpile. Then R is bounded by the upper surface S of the sandpile and by the disk D: x 2 + y 2 ≤ 1 in the plane z = 0. The outward (from R) normal on D is −k. The flux of F out of R is given by
1 . 2
Similarly, the flux of F out of each pentagonal face is equal to the flux out of the pentagonal face P in the ˆ = −k • (−k) = 1; that plane z = −1, where F • N flux is 7 1 d x d y = area of P = 4 − = . 2 2 P
S
R
div F d V =
y
19.
above z = 0 and the disk D: x 2 + y 2 = 3a 2 in the x yplane form the boundary of a region R in 3-space. The outward normal from R on D is −k. If 2
F = (x + y + 2 + z )i + (e
2
+ y )j + (3 + x)k,
then di vF = 2x + 2y. By the Divergence Theorem, S
ˆ dS + F•N
D
F • (−k) d x d y =
j ∂ ∂y x3
k ∂ ∂z z3
= 3(x 2 + y 2 )k,
F • k d A = 3µ
2π 0
dθ
0
1
r 3 dr =
3π µ . 2
The flux of F out of S is 10π + (3π µ)/2. ˆ dS = curl F • N div curl F = 0, by Theorem S D 3(g). 1 ˆ dS = 1 r•N 3 dV = V. 3 S 3 D
17. The part of the sphere S: x 2 + y 2 + (z − a)2 = 4a 2
2
(2 + µ × 0) d V = 2(5π ) = 10π.
20. If r = xi + yj + zk, then div r = 3 and
Fig. 16.4.16
x2
R
D
P
div F d V .
D
R
and ∇φ • k = 2z = 0 on D, so
(−1,0,1)
H
x
D
i 1 ∂ curl G = 3 ∂x −y 3
z
(0,−1,1)
F • (−k) d A =
In addition,
(This can also be seen directly, since F radiates from the origin, so is everywhere tangent to the plane of the hexagonal face, the plane x + y + z = 0.) T
Now div curl G = 0 by Theorem 3(g). Also div ∇φ = div (2xi−2yj+2zk) = 2−2+2 = 2. Therefore
Thus the flux of F out of the remaining hexagonal face H is 1 7 + = 0. 12 − 3 × 2 2
(−1,−1,1)
ˆ dS + F•N
R
div F d V = 0
because R is symmetric about x = 0 and y = 0. Thus the flux of F outward across S is ˆ dS = F•N (3 + x) d x d y = 3π(3a2 ) = 9π a 2 . S D
21. We use Theorem 7(b), the proof of which is given in
Exercise 29. Taking φ(x, y, z) = x 2 + y 2 + z 2 , we have 1 ˆ d S = 1 φN ˆ dS (x 2 + y 2 + z 2 )N 2V S 2V S 1 grad φ d V = 2V D 1 (xi + yj + zk) d V = V = r¯ , since
x d V = M x=0 = V x. ¯
611
SECTION 16.4 (PAGE 873)
R. A. ADAMS: CALCULUS
22. Taking F = ∇φ in the first identity in Theorem 7(a), we have
ˆ dS = − ∇φ × N curl ∇φ d V = 0, S D
so φ is constant on any connected component of S, and u and v can only differ by a constant on S.
27. Apply the Divergence Theorem to F = ∇φ:
since ∇ × ∇φ = 0 by Theorem 3(h).
23. div (φF) = φ div F + ∇φ • F by Theorem 3(b). Thus D
φ div F d V +
D
D
∇ 2φ d V =
∇ • ∇φ d V D ˆ d S = ∂φ d S. = ∇φ • N S S ∂n
div (φF) d V D ˆ dS = φF • N S
∇φ • F d V =
by the Divergence Theorem.
28. By Theorem 3(b),
24. If F = ∇φ in the previous exercise, then div F = ∇ 2 φ and
D
φ∇ 2 φ d V +
ˆ d S. |∇φ|2 d V = φ∇φ • N D S
If ∇ 2 φ = 0 in D and φ = 0 on S, then D
|∇φ|2 d V = 0.
div (φ∇ψ − ψ∇φ) = ∇φ • ∇ψ + φ∇ 2 ψ − ∇ψ • ∇φ − ψ∇ 2 φ = φ∇ 2 ψ − ψ∇ 2 φ. Hence, by the Divergence Theorem,
2
D
2
div (φ∇ψ − ψ∇φ) d V D ˆ dS = (φ∇ψ − ψ∇φ) • N S ∂φ ∂ψ = φ d S. −ψ ∂n ∂n S
(φ∇ ψ − ψ∇ φ) d V =
Since φ is assumed to be smooth, ∇φ = 0 throughout D, and therefore φ is constant on each connected component of D. Since φ = 0 on S, these constants must all be 0, and φ = 0 on D.
25. If u and v are two solutions of the given Dirichlet problem, and φ = u − v, then
∇ 2 φ = ∇ 2 u − ∇ 2 v = f − f = 0 on D φ = u − v = g − g = 0 on S. By the previous exercise, φ = 0 on D, so u = v on D. That is, solutions of the Dirichlet problem are unique.
29. If F = φc, where c is an arbitrary, constant vector, then div F = ∇φ • c, and by the Divergence Theorem, c•
26. Re-examine the solution to Exercise 24 above. If
D
div F d V D ˆ dS = F•N S ˆ d S = c • φN ˆ d S. = φc • N S S
∇φ d V =
ˆ = 0 on S, then we ∇ 2 φ = 0 in D and ∂φ/∂n = ∇φ • N can again conclude that D
|∇φ| d V = 0
and ∇φ = 0 throughout D. Thus φ is constant on the connected components of D. (We can’t conclude the constant is 0 because we don’t know the value of φ on S.) If u and v are solutions of the given Neumann problem, then φ = u − v satisfies 2
2
2
∇ φ = ∇ u − ∇ v = f − f = 0 on D ∂φ ∂u ∂v = − = g − g = 0 on S, ∂n ∂n ∂n
612
Thus
c•
ˆ d S = 0. ∇φ d V − φ N D S
Since c is arbitrary, the vector in the large parentheses must be the zero vector. Hence
ˆ d S. ∇φ d V = φ N D S
INSTRUCTOR’S SOLUTIONS MANUAL
30.
SECTION 16.5 (PAGE 878)
1 1 ˆ dS = div F d V F•N vol(D ) S vol(D ) D 1 = div F(P0 ) d V vol(D ) D
+ div F − div F(P0 ) d V D 1 = div F(P0 ) + div F − div F(P0 ) d V . vol(D ) D Thus 1 ˆ d S − div F(P0 ) F • N vol(D ) S 1 |div F − div F(P0 )| d V ≤ vol(D ) D
z 1
C
T 1 x
y
1
Fig. 16.5.1
2. Let S be the part of the surface z = y 2 lying inside the
ˆ cylinder x 2 + y 2 = 4, and having upward normal N. Then C is the oriented boundary of S. Let D be the disk x 2 + y 2 ≤ 4, z = 0, that is, the projection of S onto the x y-plane.
≤ max |div F − div F(P0 )| P in D
z
→ 0 as → 0 + assuming div F is continuous. 1 ˆ d S = div F(P0 ). lim F•N →0+ vol(D ) S
C
ˆ N
S
D
Section 16.5
Stokes’s Theorem
y
(page 878) x
1. The triangle T lies in the plane x + y + z = 1. We use the downward normal
If F = yi −
ˆ = − i +√j + k N 3 on T , because of the given orientation of its boundary. If F = x yi + yzj + zxk, then curl F =
i ∂ ∂x xy
j ∂ ∂y yz
k ∂ ∂z zx
= −yi − zj − xk.
Therefore C
x y d x + yz dz + zx dz = F • dr C y+z+x = curl F • Nˆ d S = dS √ 3 T T 1 1 d S = √ × (area of T ) = √ 3 3 T √ √ 1 3 1 1 = √ × × 2× √ = . 2 2 3 2
xj + z 2 k,
Fig. 16.5.2 then
i ∂ curl F = ∂x y
j ∂ ∂y −x
k ∂ ∂z z2
= −2k.
dx dy on S, we have ˆ k•N y d x − x d y + z 2 dz = F • dr = curl F • Nˆ d S C C S ˆ d x d y = −8π. = −2k • N ˆ k•N D
Since d S =
3. Let C be the circle x 2 + y 2 = a 2 , z = 0, oriented
counterclockwise as seen from the positive z-axis. Let D be the disk bounded by C, with normal k. We have F = 3yi − 2x zj + (x 2 − y 2 )k j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 2 3y −2x z x − y = 2(x − y)i − 2xj − (2z + 3)k.
613
SECTION 16.5 (PAGE 878)
R. A. ADAMS: CALCULUS
Applying Stokes’s Theorem (twice) we calculate
F • dr = curl F • k d A S C D =− 3 d A = −3π a 2 . =
D
z
since D has area π a 2 .
6. The curve C:
S
r = cos ti + sin tj + sin 2tk, D
y
C
x
Fig. 16.5.3
4. The surface S with equation x 2 + y 2 + 2(z − 1)2 = 6,
z ≥ 0,
ˆ is that part of an ellipsoid of with outward normal N, revolution about the z-axis, centred at (0, 0, 1), and lying above the x y-plane. The boundary of S is the circle C: x 2 + y 2 = 4, z = 0, oriented counterclockwise as seen from the positive z-axis. C is also the oriented boundary ˆ = k. of the disk x 2 + y 2 ≤ 4, z = 0, with normal N 2 2 2 If F = (x z − y 3 cos z)i + x 3 e z j + x yze x +y +z k, then, on z = 0, we have
∂ 3 z ∂ x e − (x z − y 3 cos z) ∂x ∂y z=0 2 z 2 2 = 3x e + 3y cos z = 3(x + y 2 ).
curl F • k =
z=0
Thus S
F • dr = curl F • k d A C D 2 2π dθ 3r 2 r dr = 24π. =
curl F • Nˆ d S =
0
0
5. The circle C of intersection of x 2 + y 2 + z 2 = a 2 and
x + y + z = 0 is the boundary of a circular disk of radius a in the plane x + y + z = 0. If F = yi + zj + xk, then i ∂ curl F = ∂x y
614
ˆ = − i +√j + k , N 3 √ ˆ = 3 on D, so then curl F • N y d x + z d y + x dz = F • dr = curl F • Nˆ d S C C D √ √ = 3 d S = 3π a 2 , D
ˆ N
k
If C is oriented so that D has normal
j ∂ ∂y z
k ∂ ∂z x
= −(i + j + k).
0 ≤ t ≤ 2π,
lies on the surface z = 2x y, since sin 2t = 2 cos t sin t. It also lies on the cylinder x 2 +y 2 = 1, so it is the boundary of that part of z = 2x y lying inside that cylinder. Since C is oriented counterclockwise as seen from high on the z-axis, S should be oriented with upward normal, −2yi − 2xj + k ˆ = N , 1 + 4(x 2 + y 2 ) and has area element d S = 1 + 4(x 2 + y 2 ) d x d y. If F = (e x − y 3 )i + (e y + x 3 )j + e z k, then i j k ∂ ∂ ∂ curl F = = 3(x 2 + y 2 )k. ∂y ∂z ∂x x e − y 3 e y + x 3 ez If D is the disk x 2 + y 2 ≤ 1 in the x y-plane, then F • dr = curl F • Nˆ d S = 3(x 2 + y 2 ) d x d y C S D 1 2π 3π . dθ r 2 r dr = =3 2 0 0
7. The part of the paraboloid z = 9 − x 2 − y 2 lying above
ˆ has boundary the the x y-plane having upward normal N 2 2 circle C: x + y = 9, oriented counterclockwise as seen from above. C is also the oriented boundary of the plane disk x 2 + y 2 ≤ 9, z = 0, oriented with normal field ˆ = k. N If F = −yi + x 2 j + zk, then j i ∂ ∂ curl F = ∂x ∂y −y x 2
k ∂ ∂z z
= (2x + 1)k.
INSTRUCTOR’S SOLUTIONS MANUAL
By Stokes’s Theorem, the circulation of F around C is
C
F • dr =
D
=
D
(curl F • k) d A (2x + 1) d A = 0 + π(32 ) = 9π.
8. The closed curve r = (1 + cos t)i + (1 + sin t)j + (1 − cos t − sin t)k, (0 ≤ t ≤ 2π ), lies in the plane x + y + z = 3 and is oriented counterclockwise as seen from above. Therefore it is the boundary of a√ region S in that plane with normal ˆ = (i + j + k)/ 3. The projection of S onto the field N x y-plane is the circular disk D of radius 1 with centre at (1, 1). If F = ye x i + (x 2 + e x )j + z 2 e z k, then i ∂ curl F = ∂x x ye
j ∂ ∂y x 2 + ex
k ∂ = 2xk. ∂z 2 z z +e
By Stokes’s Theorem,
C
F • dr =
S
curl F • Nˆ d S
2x √ dS = S 3 = 2x¯ A = 2π,
=
D
2x √ √ ( 3) d x d y 3
where x¯ = 1 is the x-coordinate of the centre of D, and A = π 12 = π is the area of D.
9. If S1 and S2 are two surfaces joining C1 to C2 , each hav-
ing upward normal, then the closed surface S3 consisting of S1 and −S2 (that is, S2 with downward normal) bound a region R in 3-space. Then S1
ˆ dS − F•N
S2
ˆ dS F•N
ˆ dS F•N S −S2 1 ˆ dS = ± = F•N div F d V = 0, S3 R
=
ˆ dS + F•N
provided that div F = 0 identically. Since F = (αx 2 − z)i + (x y + y 3 + z)j + βy 2 (z + 1)k,
SECTION 16.5 (PAGE 878)
we have div F = 2αx + x + 3y 2 + βy 2 = 0 if α = −1/2 ˆ d S for and β = −3. In this case we can evaluate S F• N any such surface S by evaluating the special case where S is the half-disk H : x 2 + y 2 ≤ 1, z = 0, y ≥ 0, with ˆ = k. We have upward normal N ˆ d S = −3 F•N y2 d x d y S H 1 π 3π sin2 θ dθ r 3 dr = − . = −3 8 0 0
10. The curve C: (x − 1)2 + 4y 2 = 16, 2x + y + z = 3,
oriented counterclockwise as seen from above, bounds an elliptic disk S on the √ plane 2x + y + z = 3. S has normal ˆ = (2i + j + k)/ 6. Since its projection onto the x yN plane is an elliptic disk with centre at√(1, 0, 0) and area π(4)(2) = 8π , therefore S has area 8 6π and centroid (1, 0, 1). If F = (z 2 + y 2 + sin x 2 )i + (2x y + z)j + (x z + 2yz)k, then curl F =
i ∂ ∂x z 2 + y 2 + sin x 2 = (2z − 1)i + zj.
j ∂ ∂y 2x y + z
k ∂ ∂z x z + 2yz
By Stokes’s Theorem, F • dr = curl F • Nˆ d S C S 1 (2(2z − 1) + z) d S =√ 6 S 5¯z − 2 √ = √ (8 6π ) = 24π. 6
11. As was shown in Exercise 13 of Section 7.2, ∇ × (φ∇ψ) = −∇ × (ψ × φ) = ∇φ × ∇ψ. Thus, by Stokes’s Theorem, ˆ dS φ∇ψ = ∇ × (φ∇ψ) • N C S ˆ dS = (∇φ × ∇ψ) • N S ˆ dS − ψ∇φ = −∇ × (ψ∇φ) • N C S ˆ d S. = (∇φ × ∇ψ) • N S ∇φ × ∇ψ is solenoidal, with potential φ∇ψ, or −ψ∇φ.
615
SECTION 16.5 (PAGE 878)
12. We are given that C bounds a region R in a plane P ˆ = ai + bj + ck. Therefore, with unit normal N a 2 + b2 + c2 = 1. If F = (bz − cy)i + (cx − az)j + (ay − bx)k, then i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z bz − cy cx − az ay − bx = 2ai + 2bj + 2ck. ˆ = 2(a 2 + b2 + c2 ) = 2. We have Hence curl F • N 1 (bz − cy) d x + (cx − az) d y + (ay − bx) dz 2 C 1 1 F • dr = curl F • Nˆ d S = 2 C 2 R 1 2 d S = area of R. = 2 R
13. The circle C of radius centred at P is the oriented
boundary of the disk S of area π 2 having constant norˆ By Stokes’s Theorem, mal field N. F • dr = curl F • Nˆ d S C S = curl F(P) • Nˆ d S S + curl F − curl F(P) • Nˆ d S S 2 ˆ = π curl F(P) • N + curl F − curl F(P) • Nˆ d S. S Since F is assumed smooth, its curl is continuous at P. Therefore 1 ˆ F • dr − curl F(P) • N π 2 C 1 ˆ d S ≤ curl F − curl F(P) • N 2 π S ≤ max |curl F(Q) − curl F(P)| Q on S → 0 as → 0+. Thus lim
→0+
616
C
ˆ F • dr = curl F(P) • N.
R. A. ADAMS: CALCULUS
Section 16.6 Some Physical Applications of Vector Calculus (page 885) 1. a) If we measure depth in the liquid by −z, so that the zaxis is vertical and z = 0 at the surface, then the pressure at depth −z is p = −δgz, where δ is the density of the liquid. Thus ∇ p = −δgk = δg,
where g = −gk is the constant downward vector acceleration of gravity. The force of the liquid on surface element d S of the ˆ is solid with outward (from the solid) normal N ˆ d S = −(−δgz)N ˆ d S = δgz N ˆ d S. dB = − pN Thus, the total force of the liquid on the solid (the buoyant force) is ˆ dS B = δgz N S = ∇(δgz) d V (see Theorem 7) R =− δg d V = −Mg, R
where M =
R
δ d V is the mass of the liquid
which would occupy the same space as the solid. Thus B = −F, where F = Mg is the weight of the liquid displaced by the solid. z
y x
z dS R
ˆ N S
Fig. 16.6.1 b) The above argument extends to the case where the solid is only partly submerged. Let R∗ be the part of the region occupied by the solid that is below the surface of the liquid. Let S∗ = S1 ∪ S2 be the boundary of R∗ , with S1 ⊂ S and S2 in the plane of the surface of the liquid. Since p = −δgz = 0 on S2 , we have ˆ d S = 0. δgz N S2
INSTRUCTOR’S SOLUTIONS MANUAL
Therefore the buoyant force on the solid is ˆ dS B= δgz N S1 ˆ dS + ˆ dS = δgz N δgz N S2 S1 ˆ dS = ∗ δgz N S =− δg d V = −M ∗ g, where M ∗ =
SECTION 16.6 (PAGE 885)
where Q =
R
ρ d V is the total charge in R.
4. If f is continuous and vanishes outside a bounded region (say the ball of radius R centred at r), then | f (ξ, η, ζ )| ≤ K , and, if (ρ, φ, θ ) denote spherical coordinates centred at r, then 2π π R 2 | f (s)| ρ ≤ K dθ sin φ dφ d V dρ s 3 |r − s| ρ Ê 0 0 0 = 2π K R 2 a constant.
R∗
R∗
δ d V is the mass of the liquid
which would occupy R∗ . Again we conclude that the buoyant force is the negative of the weight of the liquid displaced.
tion for fluid motion given in the text. If S is an (imaginary) surface bounding an arbitrary region D, then the rate of change of total charge in D is ∂ρ ∂ dV, ρ dV = ∂t D D ∂t where ρ is the charge density. By conservation of charge, this rate must be equal to the rate at which charge is crossing S into D, that is, to ˆ dS = − (−J) • N div J d V . S D
S2
R∗
5. This derivation is similar to that of the continuity equa-
S1
Fig. 16.6.1 ˆ is (F1 G) • N. ˆ Applying 2. The first component of F(G • N) the Divergence Theorem and Theorem 3(b), we obtain ˆ (F1 G) • N d S = div (F1 G) d V S D = ∇ F1 • G + F1 ∇ • G d S. D
But ∇ F1 • G is the first component of (G • ∇)F, and F1 ∇ • G is the first component of Fdiv G. Similar results obtain for the other components, so ˆ dS = F(G • N) Fdiv G + (G • ∇)F d V . S D
3. Suppose the closed surface S bounds a region R in which charge is distributed with density ρ. Since the electric field E due to the charge satisfies div E = kρ, the total flux of E out of R through S is, by the Divergence Theorem, ˆ dS = div E d V = k ρ d V = k Q, E•N S R R
ˆ is the outward (from (The negative sign occurs because N D) normal on S.) Thus we have ∂ρ + div J d V = 0. D ∂t Since D is arbitrary and we are assuming the integrand is continuous, it must be 0 at every point: ∂ρ + div J = 0. ∂t
6. Since r = xi + yj + zk and b = b1 i + b2 j + b3 k, we have |r − b|2 = (x − b1 )2 + (y − b2 )2 + (z − b3 )2 ∂ |r − b| = 2(x − b1 ) 2|r − b| ∂x x − b1 ∂ |r − b| = . ∂x |r − b| Similar formulas hold for the other first partials of |r − b|, so 1 ∇ |r − b| −1 ∂ ∂ = |r − b|i + · · · + |r − b|k |r − b|2 ∂ x ∂z −1 (x − b1 )i + (y − b2 )j + (z − b3 )k = |r − b|2 |r − b| r−b . =− |r − b|3
617
SECTION 16.6 (PAGE 885)
R. A. ADAMS: CALCULUS
7. Using the result of Exercise 4 and Theorem 3(d) and (h),
Similarly, the other components have zero line integrals, so (ds • ∇)F(s) = 0. F
we calculate, for constant a,
r−b div a × |r − b|3 1 = −div a × ∇ |r − b| 1 1 = −(∇ × a) • ∇ +a•∇ ×∇ = 0 + 0 = 0. |r − b| |r − b|
11. Using the results of Exercises 7 and 8, we have curl
for r not on F. (Again, this is because the curl is taken with respect to r, so s and ds can be regarded as constant for the calculation of the curl.)
8. For any element ds on the filament F, we have
div ds ×
r−s |r − s|3
12. By analogy with the filament case, the current in volume
=0
element d V at position s is J(s) d V , which gives rise at position r to a magnetic field
by Exercise 5, since the divergence is taken with respect to r, and so s and ds can be regarded as constant. Hence
div
ds × (r − s) = div F |r − s|3 F
ds ×
r−s |r − s|3
r−s ds × (r − s) = curl ds × =0 |r − s|3 F |r − s|3 F
dH(r) =
= 0.
If R is a region of 3-space outside which J is identically zero, then at any point r in 3-space, the total magnetic field is J(s) × (r − s) 1 dV. H(r) = 4π |r − s|3 R
9. By the result of Exercise 4 and Theorem 3(e), we calculate
r−b |r − b|3 1 = −curl a × ∇ |r − b| 1 1 =− ∇ •∇ a− ∇ •∇ a |r − b| |r − b| 1 1 + (a • ∇)∇ . + (∇ • a)∇ |r − b| |r − b|
curl a ×
1 = 0 for r = b, either by direct |r − b| 1 calculation or by noting that ∇ is the field of a |r − b| point source at r = b and applying the result of Example 3 of Section 7.1. 1 Also − ∇ • ∇ a = 0 and ∇ • a = 0, since a is |r − b| constant. Therefore we have r−b 1 curl a × = (a • ∇)∇ 3 |r − b| |r − b| r−b . = −(a • ∇) |r − b|3
Now A(r) was defined to be A(r) =
F is closed and ∇ F1 is conservative,
i•
618
F
(ds • ∇)F(s) =
F
∇ F1 (s) • ds = 0.
1 4π
R
J(s) dV. |r − s|
We have 1 1 J(s) d V ∇r × 4π |r − s| R 1 1 × J(s) d V ∇r = 4π |r − s| R (by Theorem 3(c)) 1 (r − s) × J(s) =− dV 4π |r − s|3 R (by Exercise 4) = H(r).
curl A(r) =
Observe that ∇ • ∇
10. The first component of (ds • ∇)F(s) is ∇ F1 (s) • ds. Since
1 J(s) × (r − s) dV. 4π |r − s|3
13.
I 4π I div A(r) = 4π I = 4π A(r) =
ds |r − s| F
1 ds div r |r − s| F 1 • ds ∇ |r − s| F (by Theorem 3(b)) = 0 for r not on F, since ∇(1/|r − s|) is conservative.
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 16.7 (PAGE 896)
1 J(s) d V , where R is a region of 34π R |r − s| space such that J(s) = 0 outside R. We assume that J(s) is continuous, so J(s) = 0 on the surface S of R. In the following calculations we use subscripts s and r to denote the variables with respect to which derivatives are taken. By Theorem 3(b),
This heat content increases at (time) rate
14. A(r) =
div s
dH = δc dt
ˆ d S. −k∇T • N
because ∇r |r − s| = −∇s |r − s|, and because ∇ • J = ∇ • (∇ × H) = 0 by Theorem 3(g). Hence
Therefore, the rate at which heat enters R through S is ˆ d S. k ∇T • N S
1 • J(s) d V ∇r |r − s| R 1 J(s) =− ∇s • dV 4π |r − s| R 1 J(s) ˆ dS = 0 =− •N 4π S |r − s|
1 div A(r) = 4π
By conservation of energy and the Divergence Theorem we have ∂T ˆ dS δc d V = k ∇T • N R ∂t S =k ∇ • ∇T d V R ∇2T d V . =k
since J(s) = 0 on S. By Theorem 3(i), J = ∇ × H = ∇ × (∇ × A) = ∇(∇ • A) − ∇ 2 A = −∇ 2 A.
R
Thus,
15. By Maxwell’s equations, since ρ = 0 and J = 0,
R
∂H ∂t
curl H = 0
k ∂T − ∇2T ∂t δc
d V = 0.
Since R is arbitrary, and the temperature T is assumed to be smooth, the integrand must vanish everywhere. Thus
div H = 0
curl E = −µ0
R
∂T dV. ∂t
If heat is not “created” or “destroyed” (by chemical or other means) within R, then the increase in heat content must be due to heat flowing into R across S. The rate of flow of heat into R across surface element ˆ is d S with outward normal N
1 J(s) 1 = ∇s • J(s) + ∇s • J(s) |r − s| |r − s| |r − s| 1 = −∇r • J(s) + 0 |r − s|
div E = 0
∂E ∂t
k ∂T k ∂2T ∂2T ∂2T = ∇2T = . + + ∂t δc δc ∂ x 2 ∂ y2 ∂z 2
Therefore,
curl curl E = grad div E − ∇ 2 E = −∇ 2 E ∇ 2 E = −curl curl E = µ0
∂ ∂ 2E curl H = µ0 0 2 . ∂t ∂t
Similarly,
∂ 2H . ∂t 2 Thus U = E and U = H both satisfy the wave equation ∇ 2 H = µ0 0
∂2U = c2 ∇ 2 U, ∂t 2
where
c2 =
1.
2. S) at time t is
H (t) = δc
R
T (x, y, z, t) d V .
Orthogonal Curvilinear (page 896)
f (r, θ, z) = r θ z 9,
(cylindrical coordinates). By Example ∂f 1 ∂f ˆ ∂f θ+ k rˆ + ∂r r ∂θ ∂z = θ z rˆ + z θˆ + r θ k.
∇f =
1 . µ0 0
16. The heat content of an arbitrary region R (with surface
Section 16.7 Coordinates
f (ρ, φ, θ ) = ρφθ (spherical coordinates). By Example 10, ∂f 1 ∂f ˆ 1 ∂f ˆ ∇f = ρˆ + φ+ θ ∂ρ ρ ∂φ ρ sin φ ∂θ φ ˆ = φθ ρˆ + θ φˆ + θ. sin φ
619
SECTION 16.7 (PAGE 896)
3.
4.
F(r, θ, z) = r rˆ
1 ∂ 2 (r ) = 2 div F = r ∂r rˆ r θˆ k ∂ ∂ 1 ∂ curl F = r ∂r ∂θ ∂z r 0 0 F(r, θ, z) = r θˆ
1 ∂ (r ) = 0 div F = r ∂θ rˆ r θˆ k ∂ ∂ 1 ∂ curl F = r ∂r ∂θ ∂z 0 r2 0
9. Let r = x(u, v) i + y(u, v) j. The scale factors are
7.
F(ρ, φ, θ ) = ρ θˆ
8.
∂ 2 ρ sin φ = cot φ ∂φ ρˆ ρ φˆ ρ sin φ θˆ ∂ ∂ ∂ = 2 θ. ˆ ∂ρ ∂φ ∂θ 0 ρ2 0
1 2 ρ sin φ
2
F(ρ, φ, θ ) = ρ ρˆ
∂ 1 div F = 2 ρ 4 sin φ = 4ρ ρ sin φ ∂ρ ρ φˆ ρ sin φ θˆ ρˆ ∂ ∂ ∂ 1 = 0. curl F = 2 ∂θ ρ sin φ ∂ρ ∂φ 2 ρ 0 0
620
1 ∂r h u ∂u
and vˆ =
1 ∂r . h v ∂v
nates in Ê 3 , with scale factors h u , h v and h z = 1, we have, for a function f (u, v) independent of z, 1 hu 1 = hu
∇ f (u, v) =
∂f 1 uˆ + ∂u hv 1 ∂f uˆ + ∂u hv
∂f 1 ∂f k vˆ + ∂v 1 ∂z ∂f vˆ . ∂v
For F(u, v) = Fu (u, v) uˆ + Fv (u, v) vˆ (independent of z and having no k component), we have
∂ 1 ∂ (h u Fu ) + (h v Fv ) div F(u, v) = h u h v ∂u ∂v h u uˆ ˆ h v k v 1 ∂ ∂ ∂ curl F(u, v) = h u h v ∂u ∂v ∂z h F h F 0 v v u u
∂ ∂ 1 (h v Fv ) − (h u Fu ) k. = h u h v ∂u ∂v
11. We can use the expressions calculated in the text for cylindrical coordinates, applied to functions independent of z and having no k components: ∂f 1 ∂f ˆ θ rˆ + ∂r r ∂θ Fr 1 ∂ Fθ ∂ Fr + + div F(r, θ ) = ∂r r r ∂θ
Fθ 1 ∂ Fr ∂ Fθ k. + − curl F(r, θ ) = ∂r r r ∂θ ∇ f (r, θ ) =
∂ 2 ρ =0 ∂θ ρ φˆ ρ sin φ θˆ ρˆ ∂ ∂ ∂ 1 curl F = 2 ∂θ ρ sin φ ∂ρ ∂φ 0 0 ρ 2 sin φ ˆ = cot φ ρˆ − 2 φ.
div F =
uˆ =
= 2k.
F(ρ, φ, θ ) = ρ φˆ
1 curl F = 2 ρ sin φ
The local basis consists of the vectors
10. Since (u, v, z) constitute orthogonal curvilinear coordi-
6.
∂r and h v = . ∂v
The area element is d A = h u h v du dv.
F(ρ, φ, θ ) = sin φ ρˆ 2 sin φ ∂ 1 2 2 div F = 2 ρ sin φ = ρ sin φ ∂ρ ρ ρˆ ρ φˆ ρ sin φ θˆ ∂ 1 ∂ ∂ curl F = 2 ρ sin φ ∂ρ ∂φ ∂θ sin φ 0 0 cos φ ˆ θ. =− ρ
1 2 ρ sin φ
∂r h u = ∂u
= 0.
5.
div F =
R. A. ADAMS: CALCULUS
12.
x = a cosh u cos v,
y = a sinh u sin v.
a) u-curves: If A = a cosh u and B = a sinh u, then x2 y2 + 2 = cos2 v + sin2 v = 1. 2 A B Since A2 − B 2 = a 2 (cosh2 u − sinh2 u) = a 2 , the u-curves are ellipses with foci at (±a, 0). b) v-curves: If A = a cos v and B = a sin v, then y2 x2 − 2 = cosh2 u − sinh2 u = 1. 2 A B Since A2 + B 2 = a 2 (cos2 v + sin2 v) = a 2 , the v-curves are hyperbolas with foci at (±a, 0).
INSTRUCTOR’S SOLUTIONS MANUAL
c) The u-curve u = u0 has parametric equations x = a cosh u 0 cos v,
REVIEW EXERCISES 16 (PAGE 896)
∇ f (r, θ, z) =
15.
∇ f (ρ, φ, θ ) =
16.
∇ f (u, v, w) =
y = a sinh u 0 sin v,
and therefore has slope at (u0 , v0 ) given by dy a sinh u 0 cos v0 d y d x = . = mu = dx dv dv (u 0 ,v0 ) −a cosh u 0 sin v0 The v-curve v = v0 has parametric equations x = a cosh u cos v0 ,
y = a sinh u sin v0 ,
and therefore has slope at (u0 , v0 ) given by dy a cosh u 0 sin v0 d y d x = . = mv = dx du du (u 0 ,v0 ) a sinh u 0 cos v0 Since the product of these slopes is mu m v = −1, the curves u = u 0 and v = v0 intersect at right angles. d)
r = a cosh u cos v i + a sinh u sin v j ∂r = a sinh u cos v i + a cosh u sin v j ∂u ∂r = −a cosh u sin v i + a sinh u cos v j. ∂v
The scale factors are ∂r h u = = a sinh2 u cos2 v + cosh2 u sin2 v ∂u ∂r h v = = a sinh2 u cos2 v + cosh2 u sin2 v = h u . ∂v The area element is d A = h u h v du dv = a 2 sinh2 u cos2 v + cosh2 u sin2 v du dv.
13.
x = a cosh u cos v y = a sinh u sin v z = z. Using the result of Exercise 12, we see that the coordinate surfaces are u = u 0 : vertical elliptic cylinders with focal axes x = ±a, y = 0. v = v0 : vertical hyperbolic cylinders with focal axes x = ±a, y = 0. z = z 0 : horizontal planes. The coordinate curves are u-curves: the horizontal hyperbolas in which the v = v0 cylinders intersect the z = z0 planes. v-curves: the horizontal ellipses in which the u = u0 cylinders intersect the z = z0 planes. z-curves: sets of four vertical straight lines where the elliptic cylinders u = u0 and hyperbolic cylinders v = v0 intersect.
∂f 1 ∂f ˆ ∂f θ+ k rˆ + ∂r r ∂θ ∂z ∇ 2 f (r, θ, z) = div ∇ f (r, θ, z)
∂f ∂ 1 ∂f ∂ ∂f 1 ∂ r + + r = r ∂r ∂r ∂θ r ∂θ ∂z ∂z ∂2 f 1 ∂f ∂2 f 1 ∂2 f = + + 2. + 2 r ∂r ∂r 2 r ∂θ 2 ∂z
14.
∂f ˆ 1 ∂f ˆ ∂f 1 ρˆ + φ+ θ ∂ρ ρ ∂φ ρ sin φ ∂θ ∇ 2 f (ρ, φ, θ ) = div f (ρ, φ, θ ) ∂ ∂ 1 ∂f ∂f 1 ρ 2 sin φ + ρ sin φ = 2 ρ sin φ ∂ρ ∂ρ ∂φ ρ ∂φ
∂ ρ ∂f + ∂θ ρ sin φ ∂θ ∂2 f 1 ∂2 f 2 ∂f = + 2 + 2 ∂ρ ρ ∂ρ ρ ∂φ 2 cot φ ∂ f 1 ∂2 f + 2 . + 2 2 ρ ∂φ ρ sin φ ∂θ 2 1 ∂f 1 ∂f 1 ∂f ˆ uˆ + vˆ + w h u ∂u h v ∂v h w ∂w ∇ 2 f (u, v, w) = div ∇ f (u, v, w) ∂ hu hw ∂ f ∂ hvhw ∂ f 1 + = h u h v h w ∂u h u ∂u ∂v h v ∂v
∂ hu hv ∂ f + ∂w h w ∂w
1 ∂2 f 1 ∂h w 1 ∂h u ∂ f 1 ∂h v = 2 + − + h u ∂u 2 h v ∂u h w ∂u h u ∂u ∂u
2 1 ∂ f 1 ∂h w 1 ∂h v ∂ f 1 ∂h u + 2 + − + h v ∂v 2 h u ∂v h w ∂v h v ∂v ∂v
1 ∂2 f 1 ∂h v 1 ∂h w ∂ f 1 ∂h u + 2 . + − + h w ∂w2 h u ∂w h v ∂w h w ∂w ∂w
Review Exercises 16
(page 896)
1. The semi-ellipsoid S with upward normal Nˆ specified in
the problem and the disk D given by x 2 + y 2 ≤ 16, z = 0, with downward normal −k together bound the solid region R: 0 ≤ z ≤ 12 16 − x 2 − y 2 . By the Divergence Theorem: S
ˆ dS + F•N
D
F • (−k) d A =
R
div F d V .
621
REVIEW EXERCISES 16 (PAGE 896)
R. A. ADAMS: CALCULUS
For F = x 2 zi + (y 2 z + 3y)j + x 2 k we have div F d V = (2x z + 2yz + 3) d V R R = 0+0+3 d V = 3 × (volume of R) R
2
2
(3y 2 + 2xe y ) d x + (2x 2 ye y ) d y C 2 2 = [4x ye y − (6y + 4x ye y )] d A P = −6 y d A = −6 y¯ A = −6, P
3 4 2 π 4 2 = 64π. = 2 3
since P has area A = 2 and its centroid has y-coordinate y¯ = 1/2. y
The flux of F across S is ˆ d S = 64π + F•N F •kdA S D
(1, 1)
D 2π
0
P
2
cos θ dθ
4
0
r dr = 128π.
the planes z = 0 and z = b. The oriented boundary of R ˆ 1 = k and consists of S and the disks D1 with normal N ˆ D2 with normal N2 = −k as shown in the figure. For F = xi + cos(z 2 )j + e z k we have div F = 1 + e z and b div F d V = dx dy (1 + e z ) dz R D2 0 = [b + (eb − 1)] d x d y D2 2
Also
2
b
= π a b + π a (e − 1). F • (−k) d A = − e0 d A = −π a 2 D2 D2 F •kdA = eb d A = π a 2 eb . D1
(2, 0)
S
D1
R
Fig. R-16.3
ˆ1 = k N
S
z
i ∂ curl F = ∂x −z
j ∂ ∂y x
k ∂ ∂z y
= i − j + k.
ˆ to a region in the plane The unit normal N 2x + y + 2z = 7 is
If C is the boundary of a disk D of radius a in that plane, then C
F • dr = =±
curl F • Nˆ d S
D
D
2−1+2 d S = ±π a 2 . 3
5. If Sa is the sphere of radius a centred at the origin, then div F(0, 0, 0) = lim 4 3 F • Nˆ d S a→0+ π a Sa 3 3 3 = lim (π a 3 + 2a 4 ) = . a→0+ 4π a 3 4
b
1
R
ˆ N
6. If S is any surface with upward normal Nˆ and boundary
D2 2a
ˆ 2 = −k N Fig. R-16.2
622
4. If F = −zi + xj + yk, then
ˆ d S = π a2 b. F•N
D1
x
x
ˆ = ± 2i + j + 2k . N 3
By the Divergence Theorem ˆ dS + F•N F• kdA + F • (−k) d A S D1 D2 = div F d V = π a 2 b + π a 2 (eb − 1). Therefore,
C
3
2. Let R be the region inside the cylinder S and between
(3, 1)
x2 d A
= 64π + = 64π +
3.
y
the curve C: x 2 + y 2 = 1, z = 2, then C is oriented counterclockwise as seen from above, and it has parametrization r = cos ti + sin tj + 2k (0 ≤ 2 ≤ 2π ).
INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 16 (PAGE 896)
Thus dr = (− sin ti + cos tj) dt, and if F = −yi + x cos(1 − x 2 − y 2 )j + yzk, then the flux of curl F upward through S is S
curl F • Nˆ d S =
0
r λr
where r = xi + yj + zk and r = |r|. Since r 2 = x 2 + y 2 + z 2 , therefore ∂r/∂ x = x/r and x2 ∂ λ (r x) = λr λ−1 + r λ = r λ−2 (λx 2 + r 2 ). ∂x r Similar expressions hold for (∂/∂ y)(r λ y) and (∂/∂z)(r λ z), so
div F(r) = r
λ−2
2
2
λ
Ê3
that excludes the origin if an only if λ = −3. In this case F is not defined at r = 0. There is no value of λ for which F is solenoidal on all of Ê 3 .
then by Green’s Theorem, the circulation of F around C is F • dr C
∂ ∂ = (x − x 3 + x 2 y) − (2y 3 − 3y + x y 2 ) d A ∂x ∂y R 2 2 (1 − 3x + 2x y − 6y + 3 − 2x y) d A = R = (4 − 3x 2 − 6y 2 ) d x d y. The last integral has a maximum value when the region R is bounded by the ellipse 3x 2 + 6y 2 = 4, oriented counterclockwise; this is the largest region in the x yplane where the integrand is nonnegative.
11. Let S be a closed, oriented surface in
1 div curl F = 0 µ
by Theorem 3(g) of Section 7.2. By part (i) of the same theorem, ∇ 2 F = ∇(div F) − curl curl F = 0 − µcurl F = −µ2 F. Thus ∇ 2 F + µ2 F = 0.
9. Apply the variant of the Divergence Theorem given in Theorem 7(b) of Section 7.3, namely grad φ d V = φ Nˆ d S, P S
F = (4x + 2x 3 z)i − y(x 2 + z 2 )j − (3x 2 z 2 + 4y 2 z)k, then by the Divergence Theorem, the flux of F through S is ˆ dS = F•N div F d V = (4−x 2 −4y 2 −z 2 ) d V . S R R The last integral has a maximum value when the region R is bounded by the ellipsoid x 2 + 4y 2 + z 2 = 4 with outward normal; this is the largest region in Ê3 where the integrand is nonnegative.
12. Let C be a simple, closed curve on the plane
x + y + z = 1, oriented counterclockwise as seen from above, and bounding a plane region √ S on x + y + z = 1. ˆ = (i + j + k)/ 3. If Then S has normal N F = x y 2i + (3z − x y 2 )j + (4y − x 2 y)k, then j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 2 2 x y 3z − x y 4y − x y = (1 − x 2 )i + 2x yj − (y 2 + 2x y)k.
to the scalar field φ = 1 over the polyhedron P. Here n S = Fi is the surface of P, oriented with outward i=1
ˆ i on the face Fi . If Ni = Ai N ˆ i , where Ai normal field N is the area of Fi , then, since grad φ = 0, we have n n n Ni Ni ˆ dS = dS = Ai = Ni . 0= N A S Fi Ai i=1 i=1 i i=1
Ê3
bounding a ˆ If region R, and having outward normal field N.
Ê 3 , where µ = 0 is a constant, then
div F =
F = (2y 3 − 3y + x y 2)i + (x − x 3 + x 2 y)j,
R
(λr + 3r ) = (λ + 3)r .
F is solenoidal on any set in
8. If curl F = µF on
a region R. If
F • dr C 2π (sin2 t + cos2 t + 0) dt = 2π. =
7. F(r) =
10. Let C be a simple, closed curve in the x y-plane bounding
By Stokes’s Theorem we have
C
F • dr =
S
curl F • Nˆ d S =
1 − x 2 − y2 d S. √ 3 S
623
REVIEW EXERCISES 16 (PAGE 896)
R. A. ADAMS: CALCULUS
The last integral will be maximum if the projection of S onto the x y-plane is the disk x 2 + y 2 ≤ 1. This maximum value is 1 − x 2 − y2 √ √ 3 dx dy 3 x 2 +y 2 ≤1 2π 1 π 1 1 = − = . dθ (1 − r 2 )r dr = 2π 2 4 2 0 0
Challenging Problems 16
b) If S is the intersection of a smooth surface with the general half-cone K , and is oriented with normal ˆ pointing away from the vertex P of K , and field N if Sa is the intersection with K of a sphere of radius a centred at P, with a chosen so that S and Sa do not intersect in K , then S, Sa , and the walls of K bound a solid region R that does not contain the origin. If F = r/|r|3 , then div F = 0 in R (see ˆ = 0 on the Example 3 in Section 7.1), and F • N walls of K . It follows from the Divergence Theorem applied to F over R that
(page 897)
1. By Theorem 1 of Section 7.1, we have
S
3 ˆ v(r) • N(r) div v(r1 ) = lim d S. →0+ 4π 3 S Here S is the sphere of radius centred at the point (with position vector) r1 and having outward normal field ˆ N(r). If r is (the position vector of) any point on S , ˆ and then r = r1 + N(r), ˆ v(r) • N(r) dS S ˆ = v(r1 ) + v(r) − v(r1 ) • N(r) dS S ˆ = v(r1 ) • N(r) dS S r−r 1 d S. + v(r) − v(r1 ) • S ˆ But N(r) d S = 0 by Theorem 7(b) of Section 7.3 S with φ = 1. Also, since v satisfies v(r2 ) − v(r1 ) = C|r2 − r1 |2 , we have
r−r 1 v(r) − v(r1 ) • dS S C 2 = d S = 4π C 3 . S
Thus
div v(r1 ) = lim
→0+
3 (0 + 4π C 3 ) = 3C. 4π 3
The divergence of the large-scale velocity field of matter in the universe is three times Hubble’s constant C.
2.
a) The steradian measure of a half-cone of semi-vertical angle α is
2π
dθ 0
624
0
α
sin φ dφ = 2π(1 − cos α).
ˆ dS = F•N
a2 = 4 a
F•
Sa
r dS |r| dS =
Sa = area of S1 .
1 (area of Sa ) a2
The area of S1 (the part of the sphere of radius 1 in K ) is the measure (in steradians) of the solid angle subtended by K at its vertex P. Hence this measure is given by r ˆ d S. •N S |r|3
3.
a) Verification of the identity ∂ ∂t
∂r ∂ ∂r G• − G• ∂s ∂s ∂t ∂F ∂r ∂r ∂r = • + (∇ × F) × • . ∂t ∂s ∂t ∂s
can be carried out using the following MapleV commands: > > > > >
with(linalg): F:=(x,y,z,t)-> [F1(x,y,z,t), F2(x,y,z,t),F3(x,y,z,t)]; r:=(s,t)->[x(s,t),y(s,t),z(s,t)];
G:=(s,t)->F(x(s,t),y(s,t),z(s,t),t); > g:=(s,t)-> dotprod(G(s,t), > map(diff,r(s,t),s)); > h:=(s,t)-> dotprod(G(s,t), > map(diff,r(s,t),t)); > LH1:=diff(g(s,t),t); > LH2:=diff(h(s,t),s); > LHS:=simplify(LH1-LH2); >
RH1:=dotprod(subs(x=x(s,t),y=y(s,t), > z=z(s,t),diff(F(x,y,z,t),t)), > diff(r(s,t),s)); >
RH2:=dotprod(crossprod(subs(x=x(s,t), > y=y(s,t),z=z(s,t),
INSTRUCTOR’S SOLUTIONS MANUAL
> > >
curl(F(x,y,z,t),[x,y,z])), diff(r(s,t),t)),diff(r(s,t),s)); RHS:=RH1+RH2; LHS-RHS; simplify(%);
We omit the output here; some of the commands produce screenfulls of output. The output of the final command is 0, indicating that the identity is valid. b) As suggested by the hint, d dt
Ct b
F • dr =
a
b
∂ ∂t
∂r G• ds ∂s
∂r ∂ = G• ∂s ∂t a
∂ ∂r ∂ ∂r + G• − G• ds ∂t ∂s ∂s ∂t s=b ∂r = G • ∂t s=a
b ∂r ∂r ∂F + (∇ × F) × • ds + ∂t ∂t ∂s a = F r(b, t), t • vC (b, t) − F r(a, t), t • vC (a, t) ∂F + • dr + (∇ × F) × vC • dr. Ct ∂t Ct
4.
a) Verification of the identity
∂r ∂r ∂ ∂r ∂r G• × − G• × ∂u ∂v ∂u ∂t ∂v
∂ ∂r ∂r − G• × ∂v ∂u ∂t
∂F ∂r ∂r ∂r ∂r ∂r = • × + (∇ • F) • × . ∂t ∂u ∂v ∂t ∂u ∂v
∂ ∂t
can be carried out using the following MapleV commands: > > > > > > > > > > >
with(linalg): F:=(x,y,z,t)->[F1(x,y,z,t), F2(x,y,z,t),F3(x,y,z,t)]; r:=(u,v,t)->[x(u,v,t),y(u,v,t), z(u,v,t)]; ru:=(u,v,t)->diff(r(u,v,t),u); rv:=(u,v,t)->diff(r(u,v,t),v); rt:=(u,v,t)->diff(r(u,v,t),t); G:=(u,v,t)->F(x(u,v,t), y(u,v,t),z(u,v,t),t);
ruxv:=(u,v,t)->crossprod(ru(u,v,t), > rv(u,v,t)); >
rtxv:=(u,v,t)->crossprod(rt(u,v,t), > rv(u,v,t));
CHALLENGING PROBLEMS 16 (PAGE 897)
>
ruxt:=(u,v,t)->crossprod(ru(u,v,t), > rt(u,v,t)); > LH1:=diff(dotprod(G(u,v,t), > ruxv(u,v,t)),t); > LH2:=diff(dotprod(G(u,v,t), > rtxv(u,v,t)),u); > LH3:=diff(dotprod(G(u,v,t), > ruxt(u,v,t)),v); > LHS:=simplify(LH1-LH2-LH3); > RH1:=dotprod(subs(x=x(u,v,t), > y=y(u,v,t),z=z(u,v,t), >
diff(F(x,y,z,t),t)),ruxv(u,v,t)); > RH2:=(divf(u,v,t))* >
(dotprod(rt(u,v,t),ruxv(u,v,t))); > RHS:=simplify(RH1+RH2); > simplify(LHS-RHS); Again the final output is 0, indicating that the identity is valid. b) If Ct is the oriented boundary of St and L t is the corresponding counterclockwise boundary of the parameter region R in the uv-plane, then ∂r • dr F× ∂t Ct ∂r ∂r ∂r • du + dv G× = ∂t ∂u ∂v L
t ∂r ∂r ∂r ∂r × +G• × dt −G • = ∂u ∂t ∂t ∂v Lt ∂r ∂r ∂ = G• × ∂t ∂v R ∂u
∂ ∂r ∂r + G• × du dv, ∂v ∂u ∂t by Green’s Theorem. c) Using the results of (a) and (b), we calculate
d ∂r ∂r ∂ ˆ dS = G• × du dv F•N dt St ∂u ∂v R ∂t ∂r ∂r ∂F du dv • × = ∂u ∂v R ∂t ∂r ∂r ∂r + du dv × (div F) • ∂t ∂u ∂v R ∂r ∂r ∂ + G• × ∂t ∂v R ∂u
∂ ∂r ∂r + G• × du dv ∂v ∂u ∂t ∂F ˆ ˆ dS • NdS + (div F)v S • N = St ∂t St
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CHALLENGING PROBLEMS 16 (PAGE 897)
R. A. ADAMS: CALCULUS
+ Ct (F × vC ) • dr.
5. We have 1 t
Dt+t
f (r, t + t) d V −
Dt
f (r, t) d V
f (r, t + t) − f (r, t) dV = t Dt 1 + f (r, t + t) d V t Dt+t −Dt 1 − f (r, t + t) d V t Dt −Dt+t = I1 + I2 − I3 . ∂f Evidently I1 → d V as t → 0. Dt ∂t I2 and I3 are integrals over the parts of Dt where the surface §t is moving outwards and inwards, respectively, ˆ is, respectively, positive and negative. that is, where vS • N ˆ d S T , we have Since d V = |vS • N|
626
I2 − I3 =
St
=
St
+
ˆ dS f (r, t + t)vS • N ˆ dS f (r, t)v S • N
St
ˆ d S. f (r, t + t) − f (r, t) v S • N
The latter integral approaches 0 as t → 0 because ˆ d S • N f (r, t + t) − f (r, t) v S St ∂f ≤ max |vS | (area of St )t. ∂t
