chapter 14. multiple integration - RonEducate

SECTION 14.1 (PAGE 759)

R. A. ADAMS: CALCULUS

CHAPTER 14. MULTIPLE INTEGRATION

Section 14.1 1.

Double Integrals

The solid is split by the vertical plane through the zaxis and the point (3, 2, 0) into two pyramids, each with a trapezoidal base; one pyramid’s base is in the plane y = 0 and the other’s is in the plane z = 0. I is the sum of the volumes of these pyramids:

(page 759)

f (x, y) = 5 − x − y
R = 1 × f (0, 1) + f (0, 2) + f (1, 1) + f (1, 2)  + f (2, 1) + f (2, 2)

I =

7.

= 4 + 3 + 3 + 2 + 2 + 1 = 15

2.


R = 1 × f (1, 1) + f (1, 2) + f (2, 1) + f (2, 2)  + f (3, 1) + f (3, 2) =3+2+2+1+1+0= 9

3.




R = 1 × f (0, 0) + f (0, 1) + f (1, 0) + f (1, 1)  + f (2, 0) + f (2, 1) = 5 + 4 + 4 + 3 + 3 + 2 = 21

4.

J=


R = 4 × 1 × 4 + 4 + 4 + 3 + 0] = 60

9.


R = 4 × 1 × 5 + 5 + 4 + 4 + 2] = 80

10.

J = area of disk = π(52 ) ≈ 78.54

11.

R = 1 × (e 1/2 + e1/2 + e3/2 + e3/2 + e5/2 + e5/2 ) ≈ 32.63

12.

f (x, y) = x 2 + y 2


R = 4 × 1 × f ( 12 , 12 ) + f ( 32 , 12 ) + f ( 52 , 12 ) + f ( 72 , 12 ) + f ( 92 , 12 ) + f ( 12 , 32 ) + f ( 32 , 32 ) + f ( 52 , 32 )

+ f ( 72 , 32 ) + f ( 92 , 32 )

+ f ( 12 , 52 ) + f ( 32 , 52 ) + f ( 52 , 52 ) + f ( 72 , 52 )

 + f ( 12 , 72 ) + f ( 32 , 72 ) + f ( 52 , 72 ) + f ( 12 , 92 ) + f ( 32 , 92 )




R = 1 × f ( 12 , 12 ) + f ( 12 , 32 ) + f ( 32 , 12 ) + f ( 32 , 32 )  + f ( 52 , 12 ) + f ( 52 , 32 )

= 918

13.

 R

d A = area of R = 4 × 5 = 20. y 1

 figure.

D

8.

R = 1 × f (1, 0) + f (1, 1) + f (2, 0) + f (2, 1)  + f (3, 0) + f (3, 1)

I =

   5+2 1 5+3 (3)(2) + (2)(3) = 15. 2 3 2

1dA
R = 4 × 1 × 5 + 5 + 5 + 5 + 4] = 96




= 4 + 3 + 3 + 2 + 2 + 1 = 15

6.





= 4 + 3 + 3 + 2 + 2 + 1 = 15

5.

1 3

−1 D

3

(5 − x − y) d A is the volume of the solid in the

x R

z

z =5−x − y

−4

Fig. 14.1.13

5

14.

3

2

3 Fig. 14.1.6

528

 D

(x + 3) d A =

 D

x dA+3

dA D

= 0 + 3(area of D)

2 x



y

π 22 =3× = 6π. 2 The integral of x over D is zero because D is symmetrical about x = 0.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.1 (PAGE 759)

y

y

√

2

y=

x 2 +y 2 =1

4−x 2

1 D

x 2 x

−2

Fig. 14.1.14

15.

Fig. 14.1.17

T is symmetric about the line x + y = 0. Therefore,  (x + y) d A = 0.

18.

x 2 +y 2 ≤a 2

a2 − x 2 − y 2 d A

= volume of hemisphere shown in the figure   2 1 4 3 π a = π a3 . = 2 3 3

T

y



(2,2)

(−1,1)

z T a

x

z=

√

a 2 −x 2 −y 2

(1,−1) (−2,−2) a

Fig. 14.1.15

16.



x

Fig. 14.1.18





y

x 2 +y 2 =a 2

x 3 cos(y 2 ) + 3 sin y − π d A  = 0 + 0 − π area bounded by |x| + |y| = 1 |x|+|y|≤1

1 = −π × 4 × (1)(1) = −2π. 2 (Each of the first two terms in the integrand is an odd function of one of the variables, and the square is symmetrical about each coordinate axis.)

19.



 x 2 +y 2 ≤a 2

a−



 x 2 + y2 d A

= volume of cone shown in the figure 1 = π a3 . 3 z a

y

√

y=a−

x 2 +y 2

1 1

−1

x

a x

−1

y

x 2 +y 2 =a 2

Fig. 14.1.19 Fig. 14.1.16

17.

20. By the symmetry of S with respect to x and y we have 

 x 2 +y 2 ≤1

(4x 2 y 3 − x + 5) d A

= 0 − 0 + 5(area of disk) = 5π.

 S

(by symmetry)

(x + y) d A = 2

x dA S

= 2 × (volume of wedge shown in the figure) 1 = 2 × (a 2 )a = a 3 . 2

529

SECTION 14.1 (PAGE 759)

R. A. ADAMS: CALCULUS

z

2. z=x

S x

y (a,a,0)

3.

Fig. 14.1.20

21.





0 1  x2y



π 0

(0,0,1)

4.

z=1−x−y

(0,1,0)

T

y

(1,0,0)

Fig. 14.1.21  b2 − y 2 d A

x

cos y d y d x

y=x

sin y

dx

−x π

 0

=2

z

22.

(x y + y 2 ) d x d y  x=y 

2

= + xy

dy 2 0 x=0  1 3 3 = y3 d y = . 2 0 8 0



= volume of the tetrahedron shown in the figure  11 1 = (1)(1) (1) = . 3 2 6

x

1 y

=

(1 − x − y) d A

T



5.



y=−x

π

0

0



2

π

sin x d x = −2 cos x

= 4.

y

y 2 e xy d x 0 0 

x=y  2 1 xy

2 e

y dy = y 0 x=0

2  2 2 e y − y 2

e4 − 5 2 . y(e y − 1) d y = = =

2 2 0 0 dy

 R

R

(x 2 + y 2 ) d A =

z=



a

b

dx 0

0

(x 2 + y 2 ) d y

 y=b y 3

3 y=0 0   a 1 bx 2 + b3 d x = 3 0 

a 1 1 3 = bx + b3 x

= (a 3 b + ab3 ). 3 3 0 

= volume of the quarter cylinder shown in the figure 1 1 = (π b2 )a = π ab2 . 4 4 z √ b



=

b2 −y 2



a

dx

x2y +

y b x

y

b

a

Fig. 14.1.22

R

Section 14.2 Iteration of Double Integrals in Cartesian Coordinates (page 766) 1.





1

x

Fig. 14.2.5

x

(x y + y 2 ) d y 0 0  y=x  2  1 y 3

xy = + dx 2 3 y=0 0  1 5 5 . = x3 dx = 6 0 24

a

dx

530

6.

 R

x 2 y2 d A = =



a

0 a3

x2 dx

b3 = 3 3



b

y2 d y

0 a 3 b3

9

.

x

INSTRUCTOR’S SOLUTIONS MANUAL

7.

SECTION 14.2 (PAGE 766)

 (sin x + cos y) d A S  π/2  π/2 dx (sin x + cos y) d y = 0 0  π/2  

y=π/2 = d x y sin x + sin y

0

 =

π/2  π

9.



x y2 d A =

R

= =

2 0  π 

π/2 π π = − cos x + x

= + = π. 2 2 2 0

= =

x dx

0

√



1

x

y2 d y

x2

 y=√x 1 3

x dx y

3 0 y=x 2  1  1 x 5/2 − x 7 d x 3 0   1 1 2 7/2 x 8

x − 3 7 8 0   1 2 1 3 − = . 3 7 8 56



y=0

 sin x + 1 d x





1

y

y

π 2

x=y 2

S

(1,1) y=x 2

R x x

π 2

Fig. 14.2.7

8.



 T

(x − 3y) d A =



a 0

dx

Fig. 14.2.9

b(1−(x/a))

0

(x − 3y) d y

 y=b(1−(x/a))

3 = d x x y − y 2

2 0 y=0     a  2 3 2 x2 x 2x − b 1− + 2 = b x− dx a 2 a a 0  2  a x b x3 3 1 b2 x 3

3 b2 x 2 = b − − b2 x + − 2 a 3 2 2 a 2 a 2 0 2 2 ab a b − . = 6 2 

a



y

b

x

x

10.

 x cos y d A D



=

x dx 1

0



1−x 2

cos y d y

0

0

 =



1

y=1−x 2

x d x (sin y)

y=0

1

x sin(1 − x 2 ) d x

Let u = 1 − x 2 du = −2x d x

0  0

1 1 1 − cos(1) =− . sin u du = cos u

= 2 1 2 2 1

=

0

y

1

x y a + b =1

y=1−x 2 D 1

T

x x

Fig. 14.2.8

a

x

Fig. 14.2.10

531

SECTION 14.2 (PAGE 766)

R. A. ADAMS: CALCULUS

y

11. For intersection: x y = 1, 2x + 2y = 5. 2x 2

Thus − 5x + 2 = 0, or (2x − 1)(x − 2) = 0. The intersections are at x = 1/2 and x = 2. We have 

 D

ln x d A =



2

ln x d x

1/2

(5/2)−x

(a,a) y

dy

 1 5 −x− dx 2 x 1/2

   2 1  2

2 5 = − x dx − ln x ln x

2 2 1/2 1/2 

=

2



y=x

T

1/x

a

x

ln x





5 − x dx 2 5 x2 V = x− 2 2

2    

x2 1 5 1 (ln 2)2 − (ln )2 + x− ln x

=− 2 2 2 2 1/2   2  x 5 − dx − 2 1/2 2   1 15 15 5 1 − ln − + = (5 − 2) ln 2 − 4 8 2 4 16 45 33 ln 2 − . = 8 16 dV =

U = ln x dx dU = x

Fig. 14.2.12

13.



 √y  1 y x y e e dA = dy x dx R y 0 y y  1 1 = (1 − y)e y d y 2 0 U = 1 − y dV = ey dy V = ey dU = −d y  

1  1

1 y

y (1 − y)e + = e dy 2 0

0

e 1 1 = − + (e − 1) = − 1. 2 2 2 y

(1,1)



y



y=x

1 2 ,2

R D xy=1

y=x 2

  2x+2y=5 1 2, 2

x

Fig. 14.2.13 x

14. Fig. 14.2.11

12.

 T



 a2 − y 2 d A =

a



0

 a2 − y 2 d y

a

dx y

(a − y) a 2 − y 2 d y = 0  a  a a2 − y 2 d y − y a2 − y 2 d y =a a

0

T

 x x d x y dy 4 0 1+x 0  1 x3 1 dx = 2 0 1 + x4

1

1 ln 2 = ln(1 + x 4 )

= . 8 8

xy dA = 1 + x4



1

0

y (1,1)

0

Let u = a 2 − y 2 du = −2y d y  1 0 1/2 π a2 + u du =a 4 2 a2

2   π a3 1 3/2

a 1 π = a3 . − u = − 4 3 4 3 0

532



y=x T 1

Fig. 14.2.14

x

INSTRUCTOR’S SOLUTIONS MANUAL

15.





1 0

1

dy 

=

y 1

e

0



1

−x 2

2

e−x d x =  dx



SECTION 14.2 (PAGE 766)

y

2

e−x d x

(R as shown)

R

0

(1,1)

1

x

dy

R y=x

2

xe−x d x

Let u = x 2 du = 2x d x

1   

1 1 1 1 1 −u 1− . e du = − e−u

= = 2 0 2 2 e 0

=

0

x

y

Fig. 14.2.17 (1,1)

1 y=x

18.

R 1

x

Fig. 14.2.15

16.



π/2 0



=

0

π/2

 x 1/3 dx 1 − y4 d y x 0  1 − y 4 d A (R as shown) = R  1  1 y 1 − y4 d y − y3 1 − y4 d y = 1

0

0

Let v = 1 − y 4 Let u = y 2 du = 2y d y dv = −4y 3 d y  1  1 0 1/2 1 1 − u 2 du + v dv = 2 0 4 1

0  1

1 π 1 π = × 12 + v 3/2

= − . 2 4 6 8 6 1

 sin x sin x dx = d A (R as shown) x R x y  x  π/2 sin x dx dy = sin x d x = 1. x 0 0 

dy



π/2

y

y (π/2,π/2)

1

y=x 1/3

(1,1)

R

y=x

y=x R π/2

x

x

Fig. 14.2.18

Fig. 14.2.16

17.



 1 yλ dx dy (λ > 0) 2 2 0 x x +y  yλ d A (R as shown) = 2 2 R x +y  y  1 dx yλ d y = 2 2 0 0 x +y   x=y  1

1 λ −1 x

tan = y dy

y y 0 x=0

 1 π 1 λ−1 π y λ

π . = y dy = = 4 0 4λ 0 4λ 1

19.

 V =

dx 0

 =

20.



1

1

0



0

x

(1 − x 2 ) d y

(1 − x 2 )x d x =

1 1 1 − = cu. units. 2 4 4

 y dy (1 − x 2 ) d x 0 0   1 1 1 5 y3 dy = − = cu. units. y− = 3 2 12 12 0

V =

1

533

SECTION 14.2 (PAGE 766)

21.

 V =

0

 =



1

0

dx 1

1−x

0

R. A. ADAMS: CALCULUS

25.

(1 − x 2 − y 2 ) d y

(1 − x 2 )y −



 y=1−x y 3

dx 3 y=0

=

1

22. z = 1 − y 2 and z = x 2 intersect on the cylinder

x 2 + y 2 = 1. The volume lying below z = 1 − y2 and above z = x 2 is  V =



(1 − y 2 − x 2 ) d A  √



(1−x 2 )/2

y √ 1/ 2

1−x 2

1

(1 − x 2 − y 2 ) d y 0 0 √ 2   1 3 

y= 1−x y 2

=4 d x (1 − x )y − 3 y=0 0  1 8 = (1 − x 2 )3/2 d x Let x = sin u 3 0 d x = cos u du   π/2 2 8 π/2 4 cos u du = (1 + cos 2u)2 du = 3 0 3 0    1 + cos 4u 2 π/2 du 1 + 2 cos 2u + = 3 0 2 π 23π = cu. units. = 32 2 2

x 2 +2y 2 =1

dx



2

x

1 dy x+y

y=x  2

d x ln(x + y)

= 1 y=0   2 (ln 2x − ln x) d x = ln 2 =

V =

1

(1 − x 2 − 2y 2 ) d A  √

x 2 +y 2 ≤1

=4

23.

E

dx (1 − x 2 − 2y 2 ) d y 0 0   1 2 (1 − x 2 )3/2 1 √ (1 − x 2 )3/2 − √ dx =4 3 2 2 2 √0  1 4 2 (1 − x 2 )3/2 d x Let x = sin θ = 3 0 d x = cos θ dθ √  √    4 2 π/2 2 π/2 1 + cos 2θ 2 4 = cos4 θ dθ = dθ 3 3 2 0 0 √  π/2   2 1 + cos 4θ = dθ 1 + 2 cos 2θ + 3 0 2 √  π/2

2 3θ 1 π = + sin 2θ + sin 4θ

= √ cu. units. 3 2 8 2 2 0 =4

 (1 − x)3 dx (1 − x 2 )(1 − x) − 3 0   1 2 2 1 1 2 4x 3 = − 2x 2 + d x = − + = cu. units. 3 3 3 3 3 3 0 

 Vol =

dx

1

0

1

2

1

d x = ln 2 cu. units.

1 x

Fig. 14.2.25

26.

 

y x dA 2− − a b T  b(1−(x/a))   a y x dx = 2− − dy a b   0 a  0   x 1 2 x x 2 b 1− − b 1− 2− dx = a a 2b a 0    b a x2 4x = + 2 dx 3− 2 0 a a   a b 2x 2 x 3

2 = 3x − + 2 = ab cu. units. 2 a 3a 3 0

Vol =

y b x y a + b =1

24.

 V = =

0

1 3

π 1/4



1 = 12

534

 dy 0

π 1/4

0



π 0

y

x 2 sin(y 4 ) d x

y 3 sin(y 4 ) d y

T

Let u = y 4 du = 4y 3 d y

1 sin u du = cu. units. 6

a

Fig. 14.2.26

x

INSTRUCTOR’S SOLUTIONS MANUAL

27.

SECTION 14.3 (PAGE 771)

Vol = 8 × part in the first octant  a  √a 2 −x 2  =8 dx a2 − x 2 d y 0 0  a (a 2 − x 2 ) d x =8 0   a x 3

16 3 2 =8 a x− a cu. units. = 3 0 3

30. Since F  (x) = f (x) and G  (x) = g(x) on a ≤ x ≤ b, we have



 f (x)g(x) d A =

T



a



a

=

z

b

b

 f (x) d x

x a

G  (y) d y

  f (x) G(x) − G(a) d x

b

f (x)G(x) d x − G(a)F(b) + G(a)F(a)  b  b  f (x)g(x) d A = g(y) d y F  (x) d x =

a x 2 +z 2 =a 2

T

a



a

y

  g(y) F(b) − F(y) d y = a  = F(b)G(b) − F(b)G(a) − b

y a

Thus

x 2 +y 2 =a 2



Fig. 14.2.27

b

a

 f (x)G(x) d x = F(b)G(b)−F(a)G(a)−

28. The part of the plane z = 8 − x lying inside the elliptic





 x

=

2 +2y 2 ≤8

x 2 +2y 2 ≤8

T

(by symmetry) (a,a)

y2 x2 + =1 = 12 × area of ellipse 8√ 4 √ = 12 × π(2 2)(2) = 48 2π cu. units.

Fig. 14.2.30

problem, we have

a

x

 G(u) du =

=  where C =

c

d



d



c x

du a

 =

x

d

dt c

 d c

a

Section 14.3 Improper Integrals and a Mean-Value Theorem (page 771)

f 1 (u, t) dt f 1 (u, t) du

1.

 Q

 f (x, t) − f (a, t) dt = g(x) − C,

f (a, t) dt is independent of x. Applying

the Fundamental Theorem of Calculus we obtain  x d G(u) du = G(x). g  (x) = dx a

(b,a) x

29. With g(x) and G(x) defined as in the statement of the 

g(y)F(y) d y. a

(b,b)

 8 − x − (y − 4) d A

12 d A

b

y

cylinder x 2 = 2y 2 = 8 lies above z = 0. The part of the plane z = y − 4 inside the cylinder lies below z = 0. Thus the required volume is Vol =

F(y)g(y) d x.

a

a x

b

e−x−y d A =

2.

Q

∞

0

 =





e−x d x



∞

e−y d y

0

R 2

= 1 (converges) lim (−e−x )

R→∞

dA = 2 (1 + x )(1 + y 2 )



0

∞

dx 1 + x2



∞

dy 1 + y2 0 0 

R 2

π2 = lim (tan−1 x)

= R→∞ 4 0 (converges)

535

SECTION 14.3 (PAGE 771)

3.

4.



y dA = 1 + x2





1

R. A. ADAMS: CALCULUS

∞

dx 1 + x2 

R

π 1 −1 = lim tan x)

(converges) = 2 S→−∞ 2 S R→∞

S



1 √ dA = x y

y dy

0





1

2x

dx dy √ x y 0 x √  1 √ 2( 2x − x) dx = x 0  1 dx √ √ √ = 4( 2 − 1) (converges) = 2( 2 − 1) x 0 y

T

which diverges to infinity. Thus the given double integral diverges to infinity by comparison.

−∞

7.



Ê



e−(|x|+|y|) d A = 4 2

x ≥0 y≥0

 =4

∞

e−(x+y) d A

e−x d x

0

∞

e−y d y

0

 =4



lim −e

R→∞

R 2

=4

−x

0

(The integral converges.)

(1, 2)

8. On the strip S between the parallel lines x + y = 0 and

y = 2x

T

x + y = 1 we have e−|x+y| = e−(x+y) ≥ 1/e. Since S has infinite area,  e−|x+y| d A = ∞.

(1, 1)

y=x

S

Since e−|x+y| > 0 for all (x, y) in

x Fig. 14.3.4

5.

6.



x2



Ê2

y2

+ dA (1 + x 2 )(1 + y 2 )  x2 d A =2 (by symmetry) 2 2 Q (1 + x )(1 + y )  ∞ 2  ∞  ∞ 2 x dx dy x dx =2 =π , 2 2 1 + x 1 + y 1 + x2 0 0 0 2 2 which diverges to infinity, since x /(1 + x ) ≥ 1/2 on [1, ∞). Q

e

−|x+y|

 dA >

0

 =

∞



ln

0

2+x 1+x

y=0





dx =

0

 ∞ ln 1 +

1 1+x

y x+y=1

S

Fig. 14.3.8  d x.

ln(1 + u) = 1, we have ln(1 + u) ≥ u/2 on u→0+ u some interval (0, u0 ). Therefore 1 1+x

 ≥

1 2(1 + x)

on some interval (x0 , ∞), and  0

536

∞

 ln 1 +

1 1+x



 dx ≥

∞ x0

x

x+y=0

Since lim

 ln 1 +

e−|x+y| d A, S

and the given integral diverges to infinity.



 ∞  1 dA 1 = dy dx 1 + x + y 1 + x +y H 0 0

y=1  ∞

= ln(1 + x + y)

dx

Ê2 , we have

1 d x, 2(1 + x)

9.



 x  ∞ 1 −y/x dx e d A = e−y/x d y 3 x3 0 T x 1 

y=x  ∞

dx −y/x

−xe =

3 x 1 y=0   ∞ 1 dx = 1− e x2 1 

R   1

1 1 =1− lim −

= 1− e R→∞ x 1 e

(The integral converges.)

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.3 (PAGE 771)

y

13.

 a) I =

S 1



y=x

=

T

(1,1)

0



dA = x+y 



1

dx

0

1 0

y=1

d x ln(x + y)

dy x+y

y=0



1 = lim (x + 1) ln(x + 1) − x ln x

c→0+


x

1

c

= lim 2 ln 2 − 0 − (c + 1) ln(c + 1) + c ln c = 2 ln 2. c→0+

y

y

Fig. 14.3.9

10.

 T

dA = x 2 + y2



∞



x

dy 2 + y2 x 1 0

y=x  ∞

1 −1 y

tan = dx x x y=0 1  ∞ dx π =∞ = 4 1 x (The integral diverges to infinity.) dx

S T

x

Q

e−xy d A >



e−xy d A, R

c→0+ c

where R satisfies 1 ≤ x < ∞, 0 ≤ y ≤ 1/x. Thus  Q

e−xy d A >



∞



1/x

dx

1

0

1 e−xy d y > e



∞ 1

dx = ∞. x

14.

2x y dA + y2 2x y d A (T as in #9(b)) =4 2 2 T x +y  x  1 y dy x dx Let u = x 2 + y 2 =4 2 + y2 x 0 0 du = 2y d y  2x 2  1 du x dx =2 u x2 0  1 x d x = ln 2 cu. units. = 2 ln 2

Vol =

y 1 y= x Q R

12.



 ∞  1/x 1 1 1 1 sin d A = sin d x dy x x R x 2/π x 0  ∞ 1 1 sin d x Let u = 1/x = 2 x x 2/π du = −1/x 2 d x

0  0

=− sin u du = cos u

=1 π/2

(The integral converges.)

S x 

π/2

2

0

x

Fig. 14.3.11

0



The given integral diverges to infinity.

1

x

Fig. 14.3.13a Fig. 14.3.13b  1  x  dA dy = 2 lim c → 0+ dx b) I = 2 x+y T x+y c 0 

y=x  1

d x ln(x + y)

= 2 lim c → 0+ c y=0  1  1 (ln 2x − ln x) d x = 2 ln 2 d x = 2 ln 2. = 2 lim

11. Since e −xy > 0 on Q we have 

(1,1)

(1,1)

15.

16.

17.



 1 1 d x  xk dA = d y = x k−a d x, which cona a Dk x 0 0 x 0 verges if k − a > −1, that is, if k > a − 1. 







1

xk



1

x k(b+1) d x if b+1 Dk 0 0 0 b > −1. This latter integral converges if k(b + 1) > −1. Thus, the given integral converges if b > −1 and k > −1/(b + 1). b

y dA =

 Rk

xa d A =

dx

 1

∞

xa dx

b

y dy =

 0

xk

 dy =

1

∞

x k+a d x, which

converges if k + a < −1, that is, if k < −(a + 1).

537

SECTION 14.3 (PAGE 771)

18.

19.

20.

R. A. ADAMS: CALCULUS



 ∞  xk  ∞ k(1−b) dA dy x d x if b < 1. = d x = b b y y 1−b Rk 0 1 1 This latter integral converges if k(1 − b) < −1. Thus, the given integral converges if b < 1 and k < −1/(1 − b).





1



xk



1

x a+(b+1)k d x, x y dA = x dx y dy = b+1 Dk 0 0 0 if b > −1. This latter integral converges if a + (b + 1)k > −1. Thus, the given integral converges if b > −1 and k > −(a + 1)/(b + 1). a b

 Rk

x a yb d A =



a

∞ 1

xa dx



b

xk 0

yb d y =

 1

∞

These seemingly contradictory results are explained by the fact that the given double integral is improper and does not, in fact, exist, that is, it does not converge. To see this, we calculate the integral over a certain subset of the square S, namely the triangle T defined by 0 < x < 1, 0 < y < x.  T

x−y dA = (x + y)3



1

x

x−y dy 3 0 0 (x + y) Let u = x + y du = d y  2x  1 2x − u dx du = u3 0 x  u=2x   1 x

1 = − 2

dx u u 0 u=x  1 1 dx = 4 0 x

x a+(b+1)k d x, b+1

if b > −1. This latter integral converges if a + (b + 1)k < −1. Thus, the given integral converges if b > −1 and k < −(a + 1)/(b + 1).



dx

which diverges to infinity.

22. The average value of x 2 over the rectangle R is  1 x2 d A (b − a)(d − c) R  b  d 1 x2 dx dy = (b − a)(d − c) a c a 2 + ab + b2 1 b3 − a 3 = . = b−a 3 3

21. One iteration:  S

x−y dA = (x + y)3





1

1

dx 0



0



1

x−y dy (x + y)3

Let u = x + y du = d y

x+1

2x − u du u3   u=x+1  1 x

1 = − 2

dx u u 0 u=x   1 x 1 1 1 dx − + = − x + 1 (x + 1)2 x x 0

1  1 1

1 dx = = − = .

2 x +1 0 2 0 (x + 1) =

dx

0

x

y d R c a

S

x−y dA = (x + y)3





1

1

dy 

0

0



1

x−y dx (x + y)3

y+1

23. The average value of x 2 + y 2 over the triangle T is Let u = x + y du = d x

u − 2y du u3  u=y+1  1  y 1

dy − = u u=y u2 0    1 1 1 y 1 − + dy = − (y + 1)2 y+1 y y 0

 1 1 dx 1

1 =− =

= −2. 2 (y + 1) y + 1 0 0 =

538

x

Fig. 14.3.22

Other iteration: 

b

0

dy

y

2 a2



(x 2 + y 2 ) d A  a  a−x 2 = 2 dx (x 2 + y 2 ) d y a 0 0  y=a−x   a y 3

2 dx x2y + = 2 a 0 3 y=0  a
 2 = 2 3x 2 (a − x) + (a − x)3 d x 3a 0  a
 a2 2 . a 3 − 3a 2 x + 6ax 2 − 4x 3 d x = = 2 3 3a 0 T

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.3 (PAGE 771)

y

y a

√

y=

a

a 2 −x 2

y=a−x Q

T a x

a

x

x+y=0

Fig. 14.3.23

Fig. 14.3.25

26. Let R be the region 0 ≤ x < ∞, 0 ≤ y ≤ 1/(1 + x 2 ). If

24. The area of region R is  0

1

3 R

dA =3 x





1

√

then 

=

(1,1) y=x 2 x

Fig. 14.3.24

25. The distance √ from (x, y) to the line x + y = 0 is

(x + y)/ 2. The average value of this distance over the quarter-disk Q is √  4 2 x+y √ dA = x dA π a2 2 Q Q √ √   a 2 −x 2 4 2 a x d x dy = π a2 0 0 √  a  4 2 = x a 2 − x 2 d x Let u = a 2 − x 2 π a2 0 du = −2x d x √ √  2 4 2a 2 2 a 1/2 . u du = = 3π π a2 0 

f (x, y) d A =

1/(1+x 2 )



dy =

∞

28.

0

1 2

∞



 x dx ∞

0



∞

1/(1+x 2 )

y dy

0

x dx (1 + x 2 )2

Let u = 1 + x 2 du = 2x d x

1 du = u2 4 π dx = . Area = 2 1 + x 2 0 2 1 1 Thus f (x, y) has average value × = on R. π 4 2π The integral in Example 2 reduced to   ∞  1 ln 1 + 2 d x x 1   1 dV = dx U = ln 1 + 2 x V =x 2 dx dU = − x(x 2 + 1)     R  R 1

dx = lim x ln 1 + 2 + 2 2 R→∞ x 1 1+x 1   2) π  ln 1 + (1/R π =2 − − ln 2 + lim R→∞ 2 4 1/R π −(2/R 3 )  = − ln 2 + lim  R→∞ 1 + (1/R 2 ) (−1/R 2 ) 2 =

x

4 π a2

 R

y

R



x dx

27. If f (x, y) = x y on the region R of the previous exercise,

x

dx dy 0 x x2  1  9 x −1/2 − x d x = . =3 2 0

x=y 2

∞

x dx 1 + x2 R 0 0 0 which diverges to infinity. Thus f has no average value on R.

√ 1 ( x − x 2 ) d x = sq. units. 3

The average value of 1/x over R is 

f (x, y) = x, then   f (x, y) d A =

=

1 4 

1 ∞

π − ln 2. 2

29. By the Mean-Value Theorem (Theorem 3), 

Rhk

f (x, y) d A = f (x0 , y0 )hk

539

SECTION 14.3 (PAGE 771)

R. A. ADAMS: CALCULUS

for some point (x0 , y0 ) in Rhk . Since (x0 , y0 ) → (a, b) as (h, k) → (0, 0), and since f is continuous at (a, b), we have  1 f (x, y) d A lim (h,k)→(0,0) hk Rhk =

lim

(h,k)→(0,0)

 f 12 (x, y) d A =

R



=

a

a+h




a+h

b+k

dx a

b

= f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b).  f 12 (x, y) d A =

R





2π

dθ 0

a

r 2r dr

0

π a4 a4 = 4 2   2π  a 2π a 3 x 2 + y2 d A = dθ r r dr = 3 D 0 0   2π  a dA r dr  = 2π a = dθ 2 2 r D x +y 0 0   π/2  a |x| d A = 4 dθ r cos θ r dr = 2π

2. 3. 4.

5.

D



0

 dθ



π/2

a

dθ

0

0

π/2 0

9.

10.

11.

0

r 4 cos2 θ sin2 θr dr

r sin θ r dr

= (− cos θ )

8.

a

a3 a3 = 3 3



2a 3 ; by symmetry, the value is twice 3 Q that obtained in the previous exercise. (x + y) d A =



ex Q

2 +y 2

 dA =

π/2

0

π 2



 dθ

a

0

2

er r dr

 a 2 1 r 2

π(ea − 1) e = 2 4 0



 π/2  a 2 2x y 2r sin θ cos θ d A = dθ r dr 2 2 r2 Q x +y 0 0

π/2  a 2 cos(2θ )

a2 a 2 π/2 sin(2θ ) dθ = − = =

2 0 4 2 0



 S

(x + y) d A =

π/3



a

(r cos θ + r sin θ )r dr  a π/3 (cos θ + sin θ ) dθ r 2 dr = 0 0

π/3

a3 = (sin θ − cos θ )

3 0   √ √ a3 3 1 ( 3 + 1)a 3 − − (−1) = = 2 2 3 6 

0

dθ

0

y √ y= 3x

x 2 +y 2 =a 2

0

π/2 3

a 4a 3 = 4 sin θ

= 3 3 0

π a4 ; by symmetry the value of this inte4 D gral is half of that in Exercise 1.

540

0

f 21 (x, y) d A.

Section 14.4 Double Integrals in Polar Coordinates (page 780) (x 2 + y 2 ) d A =

ydA =

=

Divide both sides of this identity by hk and let (h, k) → (0, 0) to obtain, using the result of Exercise 31, f 12 (a, b) = f 21 (a, b).

D

 Q

π/2

 a 6 π/2 2 sin (2θ ) dθ = 6 0    a 6 π/2 π a6 = 1 − cos(4θ ) dθ = 12 0 24



 R



7.

 f 1 (x, b + k) − f 1 (x, b) d x

b

1.

D



x 2 y2 d A = 4

f 12 (x, y) d y

= f (a + h, b + k) − f (a, b + k) − f (a + h, b) + f (a, b)  b+k  a+h  f 21 (x, y) d A = dy f 21 (x, y) d x R b a  b+k
 = f 2 (a + h, y) − f 2 (a, y) d y

Thus



f (x0 , y0 ) = f (a, b).

30. If R = {(x, y) : a ≤ x ≤ a + h, b ≤ y ≤ b + k}, then 

6.

S π/3 a

x2 d A =

Fig. 14.4.11

x

INSTRUCTOR’S SOLUTIONS MANUAL

12.



 S

x dA = 2

π/4

 dθ

0

√ 2

sec θ

SECTION 14.4 (PAGE 780)

14.

r cos θ r dr

  √  2 π/4 cos θ 2 2 − sec3 θ dθ = 3 0 √

π/4

π/4



2 4 2 sin θ

− tan θ

= 3 3 0 0 2 4 2 = − = 3 3 3

S 1

0



ln(x 2 + y 2 ) d A =



2π

dθ

0

0

1

(ln r 2 )r dr

r ln r dr

U = ln r d V = r dr dr r2 dU = V = r 2  

1  1

1 r2

r dr ln r − = 4π 2 2 0 0 

1 = −π = 4π 0 − 0 − 4 (Note that the integral is improper, but converges since limr→0+ r 2 ln r = 0.)

√ 2 x

x 2 +y 2 ≤1  1

= 4π

y

π/4



15. The average distance from the origin to points in the disk D: x 2 + y 2 ≤ a 2 is 

1 π a2

Fig. 14.4.12

D

x 2 + y2 d A =

1 π a2



area region is

13.

0

 D

Let u = r 2 du = 2r dr





2π

0

dθ

1

0

r −2k r dr = 2π

 0

1

r 1−2k dr

which converges if 1 − 2k > −1, that is, if k < 1. In this case the value of the integral is

1 r 2−2k

π 2π . = 2 − 2k 0 1−k

(1,1)

x=1 r=sec θ T

π/4 x

18.



dA

Ê 2 (1 + x2 + y 2)k =

∞

2π

dθ 0



Fig. 14.4.13

2a . 3

x 2 + y 2 ≤ b has 2 2 The average value of e−(x +y ) over the

dA = (x 2 + y 2 )k

y

1

r 2 dr =

17. If D is the disk x 2 + y 2 ≤ 1, then

0

y=x

0



 1 2 2 e−(x +y ) d A π(b2 − a 2 ) R  2π  b 2 1 dθ e−r r dr = π(b2 − a 2 ) 0 a  b2 1 1 (2π ) e−u du = π(b2 − a 2 ) 2 a2  1  −a 2 −b2 − e = 2 e . b − a2



 π/4  sec θ (x 2 + y 2 ) d A = dθ r 3 dr T 0 0  1 π/4 4 sec θ dθ = 4 0  π/4 1 (1 + tan2 θ ) sec2 θ dθ Let u = tan θ = 4 0 du = sec2 θ dθ  1 1 (1 + u 2 ) du = 4 0   1 1 u 3

1 = u+ =

4 3 3

− a 2 ).

a

dθ

16. The annular region R: 0 < a ≤ π(b2



2π

=π

1

0 ∞

r dr (1 + r 2 )k

u −k du =

Let u = 1 + r 2 du = 2r dr

−π if k > 1. 1−k

541

SECTION 14.4 (PAGE 780)

The integral converges to

19.



 D

xy d A =

π/4

0

1 2

=

π if k > 1. k −1

22. One quarter of the required volume lies in the first octant.

a

r cos θr sin θr dr  a π/4 sin 2θ dθ r 3 dr dθ

0

 0



a4 = 8



R. A. ADAMS: CALCULUS

0

 π/4 a4 cos 2θ

= − .

2 16 0

y

x 2 +y 2 =a 2

(See the figure.) In polar coordinates the cylinder x 2 + y 2 = ax becomes r = a cos θ . Thus, the required volume is  V =4 a2 − x 2 − y 2 d A D  a cos θ   π/2 dθ a 2 − r 2r dr Let u = a 2 − r 2 =4 0 0 du = −2r dr  π/2  a2 =2 dθ u 1/2 du 2 sin2 θ a 0 ⎞ ⎛

a 2 

4 π/2 ⎠ = dθ ⎝u 3/2

3 2 2 0

y=x



π/4 a

x

Fig. 14.4.19

20.

 C

 ydA = =



π

0

1 3

dθ

0

π

1+cos θ 0

r sin θr dr

sin θ (1 + cos θ )3 dθ

2  1 2 3 u 4

4 = u du = = 3 0 12 0 3 y

4 3 a (1 − sin3 θ ) dθ 3 0    π/2 π 4 − sin θ (1 − cos2 θ ) dθ = a3 3 2 0 Let v = cos θ dv = − sin θ dθ  2π a 3 4a 3 1 = − (1 − v 2 ) dv 3 3 0   1 2π a 3 v 3

4a 3 = v− − 3 3 3

=

D

Let u = 1 + cos θ du = − sin θ dθ

a sin θ

π/2

0

8a 3 2 2π a 3 − = a 3 (3π − 4) cu. units. = 3 9 9 z

r=1+cos θ

a x 2 +y 2 +z 2 =a 2

C 2 x

x 2 +y 2 =ax

Fig. 14.4.20

21. The paraboloids z = x 2 + y 2 and 3z = 4 − x 2 − y 2

intersect where 3(x 2 + y 2 ) = 4 − (x 2 + y 2 ), i.e., on the cylinder x 2 + y 2 = 1. The volume they bound is given by

x

542

y

Fig. 14.4.22

V =

0

a

a

 4 − x 2 − y2 − (x 2 + y 2 ) d A 3 x 2 +y 2 ≤1   1  2π 4 − r2 − r 2 r dr dθ = 3 0 0  8π 1 = (r − r 3 ) dr 3 0   1 8π r 2 r 4

2π = − cu. units. = 3 2 4

3 

D

23. The volume inside the sphere x 2 + y 2 + z 2 = 2a 2 and the cylinder x 2 + y 2 = a 2 is  a  π/2 dθ 2a 2 − r 2r dr V =8

Let u = 2a 2 − r 2 du = −2r dr  2a 2  3 √ 4π a = 2π u 1/2 du = 2 2 − 1 cu. units. 3 a2 0

0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.4 (PAGE 780)

z

z √ 2a x 2 +y 2 +z 2 =2a 2

x 2 +y 2 =a 2 a x 2 +z 2 =a 2 x=y

x 2 +y 2 =a 2

a

√ 2a y

a y

x

x

y 2 +z 2 =a 2

Fig. 14.4.25

Fig. 14.4.23

24.

D

a

 Volume =

2

(r cos θ + r sin θ + 4)r dr  2  2π 2 (cos θ + sin θ ) dθ r dr + 8π dθ

0

 =



2π

0

0

0

= 0 + 4π(22 ) = 16π cu. units.

26. One quarter of the required volume V is shown in the figure. We have

2

r dr

0

25. One eighth of the required volume lies in the first octant. This eighth is divided into two equal parts by the plane x = y. One of these parts lies above the circular sector D in the x y-plane specified in polar coordinate by 0 ≤√ r ≤ a, 0 ≤ θ ≤ π/4, and beneath the cylinder z = a 2 − x 2 . Thus, the total volume lying inside all three cylinders is   a2 − x 2 d A V = 16 D  a  π/4 dθ a 2 − r 2 cos2 θr dr = 16 0

 V =4

√

ydA  2 sin θ √ dθ r sin θ r dr =4 0 0 

2 sin θ  π/2 √ 2 5/2

=4 r

sin θ dθ 5 0 0 √  π/2 √ 32 2 2 64 3 = cu. units. sin dθ = 5 15 0 

D π/2

z

0

Let u = a 2 − r 2 cos2 θ du = −2r cos2 θ dr  π/4  a2 dθ =8 u 1/2 du cos2 θ a 2 sin2 θ 0  16a 3 π/4 1 − sin3 θ dθ = 3 cos2 θ 0    1 − cos2 θ 16a 3 π/4 sin θ dθ = sec2 θ − 3 cos2 θ 0   π/4

16a 3 1 = tan θ − − cos θ

3 cos θ 0   √ 16a 3 1 1−0− 2+1− √ +1 = 3 2   1 = 16 1 − √ a 3 cu. units. 2

y=z 2

x 2 +y 2 =2y 2

1

y

D

x

Fig. 14.4.26

27. By symmetry, we need only calculate the average distance from points in the sector S: 0 ≤ θ ≤ π/4,

543

SECTION 14.4 (PAGE 780)

R. A. ADAMS: CALCULUS

0 ≤ r ≤ 1 to the line x = 1. This average value is 8 π

 S

  1 8 π/4 dθ (1 − r cos θ )r dr π 0 0

  π/4  1 8 π − cos θ dθ r 2 dr = π 8 0 √ 0 8 4 2 units. =1− √ =1− 3π 3 2π

(1 − x) d A =

y

S

1 x

Let x = au, y = bv. Then



∂(x, y)

du dv = ab du dv.

dx dy =

∂(u, v)

The region E corresponds to the quarter disk Q: u 2 + v 2 ≤ 1, u, v ≥ 0 in the uv-plane. Thus   1 − u 2 − v 2 du dv V = 8abc Q   1 × volume of ball of radius 1 = 8abc × 8 4 = π abc cu. units. 3

30. We use the same regions and change of variables as in the previous exercise. The required volume is

28. The area of x¯ = = = =

4π 4π 4π 4π

Fig. 14.4.27 √ S is (4π − 3 3)/3 sq. units. Thus  3 x dA √ −3 3 S  2  π/3 6 dθ r cos θ r dr √ −3 3 0 sec θ  π/3 2 cos θ (8 − sec3 θ ) dθ √ − 3 3 0 √

π/3 √

2 6 3

= √ 4 3 − tan θ

√ . −3 3 4π − 3 3 0 

The segment has centroid

√ 6 3 √ ,0 . 4π − 3 3

y

√ 3 π/3 1 S

2

2 x

Fig. 14.4.28

29. Let E be the region in the first quadrant of the x y-plane bounded by the coordinate axes and the ellipse y2 x2 + = 1. The volume of the ellipsoid is a2 b2   x2 y2 1 − 2 − 2 d x d y. V = 8c a b E

544

   y2 x2 1 − 2 − 2 dx dy a b E 2 (1 − u − v 2 ) du dv. = ab

V =

Q

Now transform to polar coordinates in the uv-plane: u = r cos θ , v = r sin θ .  V = ab =

π/2 0

π ab 2



 dθ

1

(1 − r 2 )r dr

0  1 r4

r2 − 2 4

= π ab cu. units.

8 0

u−v u+v , y = , so that x + y = u and 2 2 x − y = v. We have

1 1



2 | du dv = 1 du dv. d x d y = |

21 1

2 2 −2

31. Let x =

Under the above transformation the square |x| + |y| ≤ a corresponds to the square S: −a ≤ u ≤ a, −a ≤ v ≤ a. Thus   1 e x+y d A = eu du dv 2 S |x|+|y|≤a   a 1 a u e du dv = 2 −a −a = a(ea − e−a ) = 2a sinh a.

32. The parallelogram P bounded by x + y = 1, x + y = 2,

3x + 4y = 5, and 3x + 4y = 6 corresponds to the square S bounded by u = 1, u = 2,v = 5, and v = 6 under the transformation u = x + y,

v = 3x + 4y,

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.4 (PAGE 780)

v

or, equivalently, x = 4u − v,

y = v − 3u. v

y

v=2

v=6

x+y=1 3x+4y=5

u=4

v=1

3x+4y=6 P

u

Fig. 14.4.33

x+y=2 x u

Fig. 14.4.32a We have

R

u=1

u=1 R u=2 v=5

34. Under the transformation u = x 2 − y 2 , v = x y, the

region R in the first quadrant of the x y-plane bounded by y = 0, y = x, x y = 1, and x 2 − y 2 = 1 corresponds to the square S in the uv-plane bounded by u = 0, u = 1, v = 0, and v = 1. Since

Fig. 14.4.32b



∂(x, y)

4 −1

=

= 1, −3 1

∂(u, v)

∂(u, v)

2x =

y ∂(x, y)

so d x d y = du dv. Also x 2 + y 2 = (4u − v)2 + (v − 3u)2 = 25u 2 − 14uv + 2v2 .

we therefore have

Thus we have   (x 2 + y 2 ) d x d y = (25u 2 − 14uv + 2v2 ) du dv P S  6  2 7 du (25u 2 − 14uv + 2v2 ) dv = . = 2 1 5

33. Let u = x y, v = y/x. Then

∂(u, v)

y =

−y/x 2 ∂(x, y)

1 ∂(x, y) = . The region D in the first quadrant ∂(u, v) 2v of the x y-plane bounded by x y = 1, x y = 4, y = x, and y = 2x corresponds to the rectangle R in the uv-plane bounded by u = 1, u = 4, v = 1, and v = 2. Thus the area of D is given by   1 du dv dx dy = 2v D R   2 3 dv 1 4 = ln 2 sq. units. du = 2 1 2 1 v so that

1 du dv. 2

(x 2 + y 2 ) d x d y = Hence,  R

y x

= 2 = 2v, 1/x

x

−2y

= 2(x 2 + y 2 ), x

35.

 I =

(x 2 + y 2 ) d x d y =

 S

1 1 du dv = . 2 2

e(y−x)/(y+x) d A. T



a) I =

π/2

0

1 = 2



 dθ

1/(cos θ +sin θ )

cos θ −sin θ

e sin θ +cos θ r dr

0 cos θ −sin θ e sin θ +cos θ

π/2

dθ (cos θ + sin θ )2 0 cos θ − sin θ Let u = sin θ + cos θ 2 dθ du = − (sin θ + cos θ )2  1 1 u e − e−1 . = e du = 4 −1 4

y v

y y=2x

1 y=x

(−1,1) x+y=1

D

xy=4

1

(1,1)

T

T 1

xy=1

x

u

x

Fig. 14.4.33

Fig. 14.4.35

545

SECTION 14.4 (PAGE 780)

R. A. ADAMS: CALCULUS

b) If u = y − x, v = y + x then

where S is the square 0 ≤ s ≤ x, 0 ≤ t ≤ x. By symmetry,



∂(u, v)

−1 1

= −2, =

1 1

∂(x, y)



1 du dv. Also, T corresponds 2  to the triangle T bounded by u = −v, u = v, and v = 1. Thus so that d A = d x d y =

 2 8 2 2 e−(s +t ) ds dt, Erf(x) = π T

where T is the triangle 0 ≤ s ≤ x, 0 ≤ t ≤ s. t (x,x)

 1 eu/v du dv 2 T   v 1 1 dv eu/v du 2 0 −v

 1 1  u/v 

v dv ve

2 0 −v  1   1 e − e−1 e − e−1 . v dv = 2 4 0

I = = = =

s=t

T x

s

Fig. 14.4.37 Now transform to polar coordinates in the st-plane. We have

36. The region R whose area we must find is shown in part



(a) of the figure. The change of variables x = 3u, y = 2v maps the ellipse 4x 2 + 9y 2 = 36 to the circle u2 + v 2 = 1, and the line 2x + 3y = 1 to the line u + v = 1. Thus it maps R to the region S in part (b) of the figure. Since

3 d x d y = |

0

0

| du dv = 6 du dv, 2

the area of R is  A=

Since cos2 θ ≤ 1, we have e−x

 R

dx dy = 6

  x sec θ 2 8 π/4 2 dθ e−r r dr Erf(x) = π 0 0    x sec θ 4 π/4 2

= dθ −e−r

π 0 0   4 π/4  −x 2 / cos2 θ = 1−e dθ. π 0



du dv. S

But the area of S is (π/4) − (1/2), so A = (3π/2) − 3 square units. v y (a) (b) 1

38.

S

R 3

 a) (x) =



=2

u b) 

2 π 

546

 0

x

2 2 e−t dt = √ π



2

3

Fig. 14.4.36

37. Erf(x) = √

1

x

2

e−s ds. Thus

0

 2 2 2 4 Erf(x) = e−(s +t ) ds dt, π S

t x−1 e−t dt

∞

2

≤ e−x , so

Let t = s 2 dt = 2s ds 2

s 2x−1 e−s ds.

0

 =2

θ

2 2 Erf(x) ≥ 1 − e−x 2 Erf(x) ≥ 1 − e−x .

0

1

x

∞

2 / cos2

∞ 0

2

e−s ds = 2

√ √ π = π 2

1   1√  2 =  12 = π. 2 2  1 t x−1 (1 − t) y−1 dt c) B(x, y) = 0

(x > 0, y > 0)

let t = cos2 θ , dt = −2 sin θ cos θ dθ  π/2 =2 cos2x−1 θ sin2y−1 θ dθ. 0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.5 (PAGE 787)

z

d) If Q is the first quadrant of the st-plane,   ∞    ∞ 2 2 s 2x−1 e−s ds 2 t 2y−1 e−t dt (x)(y) = 2 0 0  2x−1 2y−1 −(s 2 +t 2 ) =4 s t e ds dt

c

x y z a + b + c =1

Q

R

(change to polar coordinates)  π/2  ∞ 2 =4 dθ r 2x−1 cos2x−1 θr 2y−1 sin2y−1 θ e−r r dr 0 0    π/2 cos2x−1 θ sin2y−1 θ dθ = 2 0   ∞  2 × 2 r 2(x+y)−1 e−r dr

b x x

Thus B(x, y) =

Section 14.5 1.

by (a) and (c).

Fig. 14.5.4

5.

(x)(y) . (x + y)

Triple Integrals

(page 787)

R is symmetric about the coordinate planes and has volume 8abc. Thus  (1 + 2x − 3y) d V = volume of R + 0 − 0 = 8abc. R

2.



 B

x yz d V =

1





0

4

x dx y dy z dz −2  1 1 4 16 − 1 15 = − =− . 2 2 2 2 0



3. The hemispherical dome x 2 + y 2 + z 2 ≤ 4, z ≥ 0, is

symmetric about the planes x = 0 and y = 0. Therefore 

 D

4.

(3 + 2x y) d V = 3

D

xy dV D

2 = 3 × π(23 ) + 0 = 16π. 3



 R

 dV + 2

x dV =

 b1− x 

a

a

x dx

 c1− x − y  a

b

dy dz 0    a  b 1− x  a y x =c dy x dx 1− − a b 0 0 

 a  x 2 x 2 b 2  1− dx x b 1− − =c a 2b a 0  a  bc x 2 = x d x Let u = 1 − (x/a) 1− 2 0 a du = −(1/a) d x  a 2 bc a 2 bc 1 2 . u (1 − u) du = = 2 24 0 0

y

a

0

= B(x, y)(x + y)

y

R is the cube 0 ≤ x, y, z ≤ 1. By symmetry,   (x 2 + y 2 ) d V = 2 x2 dV R R  1  1  1 2 x2 dx dy dz = . =2 3 0 0 0

6. As in Exercise 5, 

R

(x 2 + y 2 + z 2 ) d V = 3

 R

x2 dV =

3 = 1. 3

7. The set R: 0 ≤ z ≤ 1 − |x| − |y| is a pyramid, one

quarter of which lies in the first octant and is bounded by the coordinate planes and the plane x + y + z = 1. (See the figure.) By symmetry, the integral of x y over R is 0. Therefore,   (x y + z 2 ) d V = z2 d V R R  1−z  1−z−y  1 z 2 dz dy dx =4 0 0 0  1−z  1 z 2 dz (1 − z − y) d y =4 0 0

  1 1 2 2 2 z (1 − z) − (1 − z) dz =4 2 0  1 1 =2 . (z 2 − 2z 3 + z 4 ) dz = 15 0 z

0

1 z=1−x−y y+z=1 R 1 x

1

x+y=1

y

Fig. 14.5.7

547

SECTION 14.5 (PAGE 787)

8.

R. A. ADAMS: CALCULUS

z

R is the cube 0 ≤ x, y, z ≤ 1. We have

1



2 −xyz

yz e R 1



 =

1

z dz 0



0



1

(0,1,1)

dV

 x=1  d y −e−xyz

(1,0,1)

x=0

1

(1 − e−yz ) d y

 1 1 −yz

y=1 dz z 1+ e

= z 0 y=0  1 1 (e−z − 1) dz = + 2 0

1

1 1 1 −z

= −1−e = − . 2 2 e 0

=

9.



0

z dz 



3

R

sin(π y ) d V =

0

=

y+z=1

0

1



x+y+z=2

R

1

0



3

sin(π y ) d y

0

1 y x (1,1,0)

Fig. 14.5.10

11.



y

dz

R is bounded by z = 1, z = 2, y = 0, y = z, x = 0, and x = y + z. These bounds provide an iteration of the triple integral without our having to draw a diagram. 

dV 3 R (x + y + z)  2  z  = dz dy

y 0

dx

1 cos(π y 3 )

2 3 y sin(π y ) d y = −

3π 0

y+z

dx (x + y + z)3  x=y+z   2  z

−1

= dz dy 2 2(x + y + z) x=0 1 0   z dy 3 2 = dz 8 1 (y + z)2 0  y=z  2

−1

3 = dz 8 1 y + z y=0  2 3 3 dz = = ln 2. 16 1 z 16 1

2 . = 3π z

(0,1,1) z=y (1,1,1)

0

0

R

12. We have

1 y x

(1,1,0)

Fig. 14.5.9

10.



 R

y dV =



1

 =

0

1−y 1

y dy 0

2−y−z

dz 

1



1

y dy

1−y

dx 0

(2 − y − z) dz

 z=1  z 2

y d y (2 − y)z − 2 z=1−y 0  1   1 1 − (1 − y)2 = y (2 − y)y − dy 2 0  1   1 5 = . 2y 2 − y 3 d y = 24 0 2 

=

548

1

 cos x cos y cos z d V Rπ  π −x  π −x−y = cos x d x cos y d y cos z dz 0 0 0

 π −x  π

z=π −x−y cos x d x cos y d y (sin z)

=

x=y



=

0

π 0



cos x d x

0

π −x 0

z=0

cos y sin(x + y) d y

 1 sin(a + b) + sin(a − b) recall that sin a cos b =  π  π −x
2  1 = cos x d x sin(x + 2y) + sin x d y 2 0 0  y=π −x



cos(x + 2y) 1 π + y sin x

cos x d x − = 2 0 2 y=0   cos x cos(2π − x) cos2 x 1 π + − = 2 0 2 2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.5 (PAGE 787)

 + (π − x) cos x sin x d x 

z

π−x sin 2x d x 2 U = π − x d V = sin 2x d x cos 2x dU = −d x V =− 2  

π  π

π −x 1 1

− cos 2x − = cos 2x d x 4 2 2 0 0 

π  π sin 2x

1 = . π− = 8 2 0 8

=

1 2

(0,1,1)

(1,0,1)

π

(1,1,1)

0



13. By Example 4 of Section 5.4,

∞

∞

−∞

(1,1,0)

Fig. 14.5.15

16.

√

−∞ √

√ k > 0, let u = kt, so that du = 

2

e−u du =

y x

z

z = y2

π. If

k dt. Thus

2

e−kt dt =



x = (1 − x)2

π . k

Thus

x 

= =

e−x 3

Ê∞

e 

−∞

√

π

2 −2y 2 −3y 2

−x 2



x+y=1 y

dV

∞

Fig. 14.5.16



−2y 2

dx e dy −∞  π 3/2 π π = √ . 2 3 6

∞ −∞

e

−3z 2

dz 

 R

f (x, y, z) d V =

14. Let E be the elliptic disk bounded by

x2

+ = 4. Then E has area π(2)(1) = 2π square units. The volume of the region of 3-space lying above E and beneath the plane z = 2 + x is V = since

15.

 E

=

E

(2 + x) d A = 2

E

=





1

0



1



1

x dx

=

0

0

1

√

dx

 dz f (x, y, z) d x 0  1−y dy f (x, y, z) d x

z

1

dz 0

0

(1−x)2

1−x √

 dx

0



dz

0  1−√z

0

f (x, y, z) dz

0 1−y

y2



1

y2

z 1−x

√

z

f (x, y, z) d y f (x, y, z) d y.

dz 2−x−y

x dx (x + y − 1) d y 0 1−x   1 1 (x − 1)2 = +x − dx x 2 2 0  1 3 x 1 = dx = . 8 0 2 =

0

f (x, y, z) dz

0

1

dy 1−x  1



1

dz



x d A = 0 by symmetry.

x dV =

0

y2



1−y

dy



d A = 4π cu. units,

dy dx



1



1−x 0



1

0





 T

=

dx dy



4y 2



1

0

 =



(0, 1, 1)



17.



1

dz 0  =



1−z 0

dy

1 0

f (x, y, z) d x

f (x, y, z) d V (R is the prism in the figure)  1  1−y  1 dx dy f (x, y, z) dz. = R

0

0

0

549

SECTION 14.5 (PAGE 787)

R. A. ADAMS: CALCULUS

z 1

20.

(1,0,1)

y+z=1 R 1

x

1



 1 dy dz f (x, y, z) d x 0 0 y 2 +z 2  = f (x, y, z) d V (R is the paraboloid in the figure) R  √  √  =

y

 √1−y 2

1

1

x−y 2

x

dx

dy

0

0

0

(1,1,0)

Fig. 14.5.17



18.



1

dz 0  =  =

0



1

dy z

f (x, y, z) dz.

z

x=y 2 +z 2

y

f (x, y, z) d x

0

R

f (x, y, z) d V (R is the pyramid in the figure)   1 1 y dx dy f (x, y, z) dz.

y

R

1

0

x

z

Fig. 14.5.20 (0,1,1)

z=y

21.

 I =



1



1−z

dz 0

0

x=0

R

 I =

y=1

22.

 I =



1

0



1

0

 dy

y

0 ≤ x ≤ 1.

0

f (x, y, z) dz.

f (x, y, z) d x.

0

z

1−y

dy

0

1

dz



1

dx

z=0

Fig. 14.5.18

0 ≤ y ≤ 1 − z,

Thus 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − 0 = 1, 0 ≤ z ≤ 1 − y, and

y (1,1,0)

f (x, y, z) d x.

0

0 ≤ z ≤ 1,

y=x

1

dy

The given iteration corresponds to

(1,1,1)

x

x

The given iteration corresponds to 

19.

1

dz 0 





1

dx z

x−z

0

0 ≤ z ≤ 1,

f (x, y, z) d y

0

 I =

z

23.

 I =



1

0

0

x



x−z

f (x, y, z) dz.

f (x, y, z) d y.

dx 0

z

y

The given iteration corresponds to

z=x−y

0 ≤ z ≤ 1,

z ≤ x ≤ 1,

0 ≤ y ≤ x − z.

Thus 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ x − y, and

1



x y (1,1,0)

Fig. 14.5.19

550



1

dy

0

1

dz



1

dx

0

(1,0,1)

0 ≤ x ≤ y.

Thus 0 ≤ x ≤ 1, x ≤ y ≤ 1, 0 ≤ z ≤ y, and

f (x, y, z) d V (R is the tetrahedron in the figure) R  x  x−y  1 dx dy f (x, y, z) dz. =

=

0

z ≤ y ≤ 1,

I =

0



1

dx

0



x

dy

0

x−y

f (x, y, z) dz.

INSTRUCTOR’S SOLUTIONS MANUAL

24.

 I =

 √1−y 2

1

dy

 dz

27.

f (x, y, z) d x.

y 2 +z 2

0

0

1

SECTION 14.5 (PAGE 787)

The given iteration corresponds to 0 ≤ y ≤ 1,



0≤z≤

Thus 0 ≤ x ≤ 1, 0 ≤ y ≤  I =

25.

 I =

0

x, 0 ≤ z ≤

 √x−y 2



(1,0,1)

0

The given iteration corresponds to 0 ≤ y ≤ 1,

y ≤ z ≤ 1,

z=x (1,1,1)

y=0

f (x, y, z) d x.

dz y

x − y 2 , and

f (x, y, z) dz.

0

z



 1  x 3 dz dx ex d y 0 z 0  3 ex d V (R is the pyramid in the figure) = R  x  x  1 x3 e dx dy dz = 0 0 0  1 e−1 3 . x 2ex d x = = 3 0 1

z

dy 0

1

dy

√

x

dx 0



1

√



1

y 2 + z 2 ≤ x ≤ 1.

1 − y 2,



y=x

R

x=1 (1,0,0) x

0 ≤ x ≤ z.

y

Thus 0 ≤ x ≤ 1, x ≤ z ≤ 1, 0 ≤ y ≤ z, and  I =



1



1

dx

z

dz

0

0

x

26.

z=0

(1,1,0)

Fig. 14.5.27 f (x, y, z) d y.

28. z

(0, 1, 1) (1, 1, 1)





1

dx 0 



1−x 0

1

dy y

sin(π z) = dV (R is the pyramid in the figure) R z(2 − z)  z  1−y  1 sin(π z) dz dy dx = 0 z(2 − z) 0 0  z  1 sin(π z) dz (1 − y) d y = 0 z(2 − z) 0    1 sin(π z) z2 z− dz = 2 0 z(2 − z)  1 1 1 = sin(π z) dz = . 2 0 π z

(0, 1, 0)

x

sin(π z) dz z(2 − z)

(0,0,1)

y

z=1

x=0 (0,1,1)

(1,0,1)

Fig. 14.5.26

y=0

 I =

0



1



1

dx

y

dy x

x

 f (x, y, z) dz =

P

f (x, y, z) d V ,

z=y y

where P is the triangular pyramid (see the figure) with vertices at (0, 0, 0), (0, 1, 0), (0, 1, 1), and (1, 1, 1). If we we reiterate I to correspond to the horizontal slice shown then 



1 0



1

dz

dy z

0

z

f (x, y, z) d x.

x

y=1−x

(1,0,0)

Fig. 14.5.28

29. The average value of f (x, y, z) over R is f¯ =

1 volume of R

 R

f (x, y, z) d V .

551

SECTION 14.5 (PAGE 787)

R. A. ADAMS: CALCULUS

If f (x, y, z) = x 2 + y 2 + z 2 and R is the cube 0 ≤ x, y, z ≤ 1, then, by Exercise 6, 1 f¯ = 1

 R

6. φ = 2π/3 represents the lower half of the right-circular cone with vertex at the origin, axis along the z-axis, and semi-vertical angle π/3. Its Cartesian equation is  z = − (x 2 + y 2 )/3.

(x 2 + y 2 + z 2 ) d V = 1.

7. φ = π/2 represents the x y-plane. 30. If the function f (x, y, z) is continuous on a closed,

bounded, connected set D in 3-space, then there exists a point (x0 , y0 , z 0 ) in D such that  D

f (x, y, z) d V = f (x0 , y0 , z 0 ) × (volume of D).

Apply this with D = B (a, b, c), which has volume 4 3 π  , to get 3  4 f (x, y, z) d V = f (x0 , y0 , z 0 ) π  3 3 B (a,b,c)

8. ρ = 4 represents the sphere of radius 4 centred at the origin.

9. r = 4 represents the circular cylinder of radius 4 with axis along the z-axis.

10. ρ = z represents the positive half of the z-axis. 11. ρ = r represents the x y-plane. 12. ρ = 2x represents the half-cone with vertex at the origin, axis along the positive x-axis, and semi-vertical angle  π/3. Its Cartesian equation is x = (y 2 + z 2 )/3.

for some (x0 , y0 , z 0 ) in B (a, b, c). Thus lim

→0

3 4π  3

 B (a,b,c)

f (x, y, z) d V

13. If ρ = 2 cos φ, then ρ 2 = 2ρ cos φ, so x 2 + y 2 + z 2 = 2z

= lim f (x0 , y0 , z 0 ) = f (a, b, c) →0

x 2 + y 2 + z 2 − 2z + 1 = 1

since f is continuous at (a, b, c).

x 2 + y 2 + (z − 1)2 = 1. Thus ρ = 2 cos φ represents the sphere of radius 1 centred at (0, 0, 1).

Section 14.6 Change of Variables in Triple Integrals (page 795) 1. Spherical: [4,√ π/3, 2π/3];

√ Cartesian: (− 3, 3, −2); Cylindrical: [2 3, 2π/3, 2].

14. r = 2 cos θ ⇒ x 2 + y 2 = r 2 = 2r cos θ = 2x, or

(x − 1)2 + y 2 = 1. Thus the given equation represents the circular cylinder of radius 1 with axis along the vertical line x = 1, y = 0.

2. Cartesian: (2, −2, √ 1);

Cylindrical: [2 2, −π/4, 1]; Spherical: [3, cos−1 (1/3), −π/4].

3. Cylindrical: √[2, π/6, −2];

√ Cartesian: ( 3, 1, −2]; Spherical: [2 2, 3π/4, π/6].

15.

 V =



2π

0

2π a = 3

dθ 3 

4. Spherical: [1, φ, θ ]; Cylindrical: [r, π/4, r ].

z

√ x = sin φ cos θ = r cos π/4 = r/ 2 √ y = sin φ sin θ = r sin π/4 = r/ 2 z = cos φ = r. Thus x = y, θ = π/4,√and r = sin φ = cos φ. Hence φ = π/4, √ so r = 1/ 2. Finally: x = y = 1/2, z = 1/ 2. √ Cartesian: (1/2, 1/2, 1/ 2).

5. θ = π/2 represents the half-plane x = 0, y > 0.

552

 a sin φ dφ R2 d R 0 0  1 1− √ cu. units. 2 π/4

π/4

R=a

y x

Fig. 14.6.15

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.6 (PAGE 795)

√ r intersects the sphere r 2 + z 2 = 2 2 where r + r − 2 = 0. This equation has positive root r = 1. The required volume is

16. The surface z =

 V =



2π

dθ 0



r dr 0

 1 

V =

dz

√ r

√  dθ − r r dr = 0 0   1  2 Let u = 2 − r 2 r 2 − r 2 dr − = 2π 5 0 du = −2r dr  2 4π =π u 1/2 du − 5 1 √  4π 2π  √ 4 2π 22π = = − cu. units. 2 2−1 − 3 5 3 15 2π

where r + r − 12 = 0, that is, where r = required volume is 

 √2−r 2

1

18. The paraboloid z = r 2 intersects the sphere r 2√+ z 2 = 12 4 2

2 − r2

√ 3 



2π

dθ 0



= 2π

√

0



12

0

3

3. The

 12 − r 2 − r 2 r dr



r 12 − r 2 dr −

9π 2

Let u = 12 − r 2 du = −2r dr

9π 2 9  9π √ 45π 2π  3/2 12 − 27 − = 16 3π − cu. units. = 3 2 2 =π

u 1/2 du −

2

z r 2 +z 2 =2 z=x 2 +y 2 √ z= r y x 2 +y 2 +z 2 =12

x

Fig. 14.6.18

19. One half of the required volume V lies in the first octant,

Fig. 14.6.16

17. The paraboloids z = 10

− r2

2(r 2

and z = − 1) intersect where r 2 = 4, that is, where r = 2. The volume lying between these surfaces is  V =



2π

2

dθ 0

= 2π

 0

0 2

[10 − r 2 − 2(r 2 − 1)]r dr

(12r − 3r 3 ) dr = 24π cu. units.

inside the cylinder with polar equation r = 2a sin θ . Thus  π/2  2a sin θ V =2 dθ (2a − r )r dr 0 0   π/2 2 π/2 3 3 4a 2 sin2 θ dθ − 8a sin θ dθ = 2a 3 0 0  π/2  16a 3 π/2 3 (1 − cos 2θ ) dθ − sin θ dθ = 4a 3 3 0 0 32a 3 cu. units. = 2π a 3 − 9

z

z

z=10−r 2

z=2a−r 2

r=2a sin θ

z=2(r 2 −1) y

Fig. 14.6.17

2a y

x

Fig. 14.6.19

553

SECTION 14.6 (PAGE 795)

R. A. ADAMS: CALCULUS

20. The required volume V lies above z = 0, below

z

z = 1 − r 2 , and between θ = −π/4 and θ = π/3. Thus  π/3  1 V = dθ (1 − r 2 )r dr −π/4 0   7π 7π 1 1 = − cu. units. = 12 2 4 48

21. Let R be the region in the first octant, inside the ellipsoid

b

a

x2 a2

+

y2 b2

+

z2 c2

x

= 1,

and between the planes y = 0 and y = x. Under the transformation x = au,

y = bv,

Fig. 14.6.22

23. Let x = au, y = bv, z = w. The indicated region R cor-

responds to the region S above the uv-plane and below the surface w = 1 − u 2 − v 2 . We use polar coordinates in the uv-plane to calculate the volume V of R:   d V = ab du dv dw V = R S  1  2π π ab cu. units. dθ (1 − r 2 )r dr = = ab 2 0 0

z = cw,

R corresponds to the region S in the first octant of uvwspace, inside the sphere u 2 + v 2 + w2 = 1, and between the planes v = 0 and bv = au. Therefore, the volume of R is   d x d y dz = abc du dv dw. V = R

24.

S

Using spherical coordinates in uvw-space, S corresponds to 0 ≤ R ≤ 1,

0≤φ≤

π , 2

0 ≤ θ ≤ tan−1

a . b

Thus  V = abc

tan−1 (a/b)

0

 dθ

π/2

0

a 1 = abc tan−1 cu. units. 3 b

 sin φ dφ

1

25. R2 d R

0

22. One eighth of the required volume V lies in the first octant. Call this region R. Under the transformation x = au,

y = bv,

z = cw,

R corresponds to the region S in the first octant of uvwspace bounded by w = 0, w = 1, and u2 + v 2 − w2 = 1. Thus V = 8abc × (volume of S). The volume of S can be determined by using horizontal slices:  V = 8abc

554

1 0

π 8 (1 + w2 ) dw = π abc cu. units. 4 3

y

x 2 y 2 z2 + + =1 a 2 b2 c2

26.

27.



(x 2 + y 2 + z 2 ) d V  a  h dθ r dr (r 2 + z 2 ) dz = 0 0 0   a 1 = 2π r 3 h + r h 3 dr 3 0  4  a h π a4 h a2 h 3 π a2 h 3 = 2π = + + . 4 6 2 3  (x 2 + y 2 ) d V B  π  a  2π dθ sin φ dφ R 2 sin2 φ R 2 d R = 0 0 0  a  π sin3 φ dφ R4 d R = 2π 0 0   5 8π a 5 4 a = . = 2π 3 5 15  (x 2 + y 2 + z 2 ) d V B  π  a  2π 4π a 5 . dθ sin φ dφ R4 d R = = 5 0 0 0  (x 2 + y 2 + z 2 ) d V R  2π

R 2π

 tan−1 (1/c)  a dθ sin φ dφ R4 d R 0 0 0 

   2π a 5 c 1 2π a 5 . 1 − cos tan−1 = 1− √ = 5 c 5 c2 + 1 

=

INSTRUCTOR’S SOLUTIONS MANUAL

28.



=

and above the plane y + z = b, then the corresponding region S lies inside the sphere

(x 2 + y 2 ) d V

R  2π



tan−1 (1/c)

dθ 0

2π a 5 = 5 =

SECTION 14.6 (PAGE 795)

2π a 5 5

0 tan−1 (1/c)

 0



1

√

c/

c2 +1



sin3 φ dφ

a

u 2 + v 2 + w2 = 1

R4 d R

0 2

sin φ(1 − cos φ) dφ

and above the plane bv + cw = b. The distance from the origin to this plane is

Let u = cos φ du = − sin φ dφ

b D= √ b2 + c2

(1 − u 2 ) du

by Example 7 of Section 1.4. By symmetry, the volume of S is equal to the volume lying inside the sphere u 2 + v 2 + w2 = 1 and above the plane w = D. We calculate this latter volume by slicing; it is

  1 u 3

= u− √ 5 3 c/ c2 +1   c3 c 2π a 5 2 + −√ = . 5 3 c2 + 1 3(c2 + 1)3/2 √ z = r 2 and z = 2 − r 2 intersect where r 4 + r 2 − 2 = 0, that is, on the cylinder r = 1. Thus 2π a 5

29.



 R

z dV =



2π

dθ 0

r dr

r2

0



=π

1 0

(2 − r 2 − r 4 )r dr =

z dz

R

31.

33. By Example 10 of Section 3.5, we know that



∂ 2u 1 ∂ 2u ∂ 2u ∂2u 1 ∂u + 2 2. + 2 = 2 + 2 ∂x ∂y ∂r r ∂r r ∂θ

x dV =

z dV R  π/2  a  π/2 dθ cos φ sin φ dφ R3 d R = 0 0 0   π a4 π 1 a4 = . = 2 2 4 16



 x dV =

π/2



a



h(1−(r/a))

r dr r cos θ dz R 0 0 0  a   π/2 r ha 3 , cos θ dθ r2 1 − =h dr = a 12 0 0   π/2  a  h(1−(r/a)) z dV = dθ r dr z dz R 0 0 0  π h2 a  r 2 r dr = 1− 4 0 a   a π h2 r 2 2r 3 r4

π a2 h 2 = − + 2

= . 4 2 3a 4a 48 0 dθ

32. If x = au,

y = bv,

z = cw,

then the volume of a region R in x yz-space is abc times the volume of the corresponding region S in uvw-space. If R is the region inside the ellipsoid x2 y2 z2 + 2 + 2 =1 2 a b c

 1  w3

(1 − w2 ) dw = π w − 3 D D   2 D3 =π −D+ . 3 3 1

Hence, the volume of R is   2 b b3 π abc cu. units. −√ + 3 3(b2 + c2 )3/2 b2 + c2

7π . 12

30. By symmetry, both integrals have the same value: 

 π

 √2−r 2

1

(assuming b > 0)

∂ 2u to both sides. ∂z 2 Cylindrical and spherical coordinates are related by The required result follows if we add

34.

z = ρ cos φ,

r = ρ sin φ.

(The θ coordinates are identical in the two systems.) Observe that z, r, ρ, and φ play, respectively, the same roles that x, y, r , and θ play in the transformation from Cartesian to polar coordinates in the plane. We can exploit this correspondence to avoid repeating the calculations of partial derivatives of a function u, since the results correspond to calculations made (for a function z) in Example 10 of Section 3.5. Comparing with the calculations in that Example, we have ∂u ∂u ∂u = cos φ + sin φ ∂ρ ∂z ∂r ∂u ∂u ∂u = −ρ sin φ + ρ cos φ ∂φ ∂z ∂r 2u 2 ∂ 2u ∂ ∂ 2u 2 2 ∂ u + sin = cos φ + 2 cos φ sin φ φ ∂ρ 2 ∂z 2 ∂z∂r ∂r 2  2 2 ∂ u ∂u ∂ u + ρ 2 sin2 φ 2 = −ρ ∂ρ ∂φ 2 ∂z  ∂ 2u ∂ 2u + cos2 φ 2 . − 2 cos φ sin φ ∂z∂r ∂r

555

SECTION 14.6 (PAGE 795)

R. A. ADAMS: CALCULUS

Substituting these expressions into the expression for u given in the statement of this exercise in terms of spherical coordinates, we obtain the expression in terms of cylindrical coordinates established in the previous exercise: ∂ 2u 2 ∂u 1 cot φ ∂u 1 ∂ 2u ∂ 2u + + + + ∂ρ 2 ρ ∂ρ ρ 2 ∂φ ρ 2 ∂φ 2 ρ 2 sin2 φ ∂θ 2 2 2 2 ∂ u 1 ∂u ∂ u 1 ∂ u = 2 + + 2 2 + 2 ∂r r ∂r r ∂θ ∂z ∂ 2u ∂ 2u = + 2 = u ∂x2 ∂y

2.

(x/2)2 +y 2 ≤1

3.

by Exercise 33.

35. Consider the transformation x = x(u, v, w),

y = y(u, v, w),

z = z(u, v, w),

and let P be the point in x yz-space corresponding to u = a, v = b, w = c. Fixing v = b, w = c, results in a parametric curve (with parameter u) through P. The vector ∂y ∂z −→ ∂ x PQ = i+ j+ k ∂u ∂u ∂u

4.

∂y ∂z −→ ∂ x i+ j+ k PR = ∂v ∂v ∂v ∂x ∂y ∂z − → PS = i+ j+ k ∂w ∂w ∂w span a parallelepiped in x yz-space corresponding to a rectangular box with volume du dv dw in uvw-space. The parallelepiped has volume



∂(x, y, z)

−→ −→ − →

du dv dw. |( P Q × P R) • P S| =

∂(u, v, w)

Thus



∂(x, y, z)

du dv dw. d V = d x d y dz =

∂(u, v, w)

Section 14.7 Applications of Multiple Integrals (page 803) ∂z ∂z =2= z = 2x + 2y, ∂x ∂y  d S = 1 + 22 + 22 d A = 3 d A  S= 3 d A = 3π(12 ) = 3π sq. units. x 2 +y 2 ≤1

556

z = a2 − x 2 − y 2 x y ∂z ∂z = − = − , 2 2 2 2 ∂x ∂y a −x −y a − x 2 − y2  x 2 + y2 a dS = 1 + 2 dA =  dA 2 2 2 a −x −y a − x 2 − y2  adA S=  (use polars) x 2 +y 2 ≤a 2 a2 − x 2 − y 2  2π  a r dr =a dθ Let u = a 2 − r 2 √ a2 − r 2 0 0 du = −2r dr  a2 = πa u −1/2 du = 2π a2 sq. units. 0

and corresponding vectors

1.

∂z 3 ∂z 4 z = (3x − 4y)/5, = , = ∂x 5 ∂y 5  √ 32 + 42 dA = 2dA dS = 1 + 2 5  √ √ √ 2 d A = 2π(2)(1) = 2 2π sq. units. S=

z = 2 1 − x 2 − y2 2x 2y ∂z ∂z = − = − , 2 2 ∂x ∂y 1−x − y 1 − x 2 − y2   1 + 3(x 2 + y 2 ) 4(x 2 + y 2 ) dS = 1 + d A = dA 1 − x 2 − y2 1 − x 2 − y2  S= dS x 2 +y 2 ≤1   2π  1 1 + 3r 2 = dθ r dr Let u 2 = 1 − r 2 1 − r2 0 0 u du = −r dr  1 √ = 2π 4 − 3u 2 du Let 3u = 2 sin v √ 0 3 du = 2 cos v dv  π/3 2 dv (2 cos2 v) √ = 2π 3 0  4π π/3 = √ (1 + cos 2v) dv 3 0  π/3  4π 4π 2 sin 2v

= √ v+ = √ + π sq. units.

2 3 3 3 0

5.

∂z x ∂z y ∂z = 2x, = , = 3z 2 = x 2 + y 2 , 6z ∂x ∂x 3z ∂y 3z   x 2 + y2 2 9z 2 + 3z 2 dS = 1 + dA = dA = √ dA 2 9z 9z 2 3  2 2 24π S= √ d A = √ π(12) = √ sq. units. 3 3 3 x 2 +y 2 ≤12

INSTRUCTOR’S SOLUTIONS MANUAL

6.

∂z ∂z z = 1 − x 2 − y 2, = −2x, = −2y ∂x ∂y d S = 1 + 4x 2 + 4y 2 d A  S= 1 + 4(x 2 + y 2 ) d A x 2 +y 2 ≤1, x≥0, y≥0  1 π/2

 =

dθ

0

π = 16 π = 16

0



5

9.



√  5 2 3/2

π(5 5 − 1) u sq. units.

= 3 24 1

1

 1 ∂z 1 = √ , dS = 1 + dA ∂x 4x 2 x  1  √x   1 1 4x + 1 √ dy = dx 1+ x dx S= 4x 4x 0 0 0  1 1√ 4x + 1 d x Let u = 4x + 1 = 2 0 du = 4 d x √   5  1 5 1/2 1 2 3/2

5 5−1 = u sq. units. u du =

= 8 1 8 3 12 1 x,

∂z x ∂z = −2x, =− z 2 = 4 − x 2 , 2z ∂ x ∂ x z  2 2 x 2 dA dS = 1 + 2 d A = d A = √ z z 4 − x2 (since z ≥ 0 on the part of the surface whose area we want to find)  2  x 2 S= dx dy √ 4 − x2 0 0  2 2x = d x Let u = 4 − x 2 √ 4 − x2 0 du = −2x d x

4  4

√ −1/2 = u du = 2 u

= 4 sq. units. 0

1 + (2y)2 + (2x)2 d A = 1 + 4x 2 + 4y 2 d x d y, d S2 = 1 + (2x)2 + (2y)2 d A = 1 + 4x 2 + 4y 2 d x d y. d S1 =

u 1/2 du

1

√

tively, are

1

∂z = 2y, d S = 1 + 4y 2 d A z = y 2, ∂y  1  y dy 1 + 4y 2 d x S= 0 0  1 y 1 + 4y 2 d y Let u = 1 + 4y 2 = 0 du = 8y d y √   5  5 1 1 2 3/2

5 5−1 = u sq. units. u 1/2 du = =

8 8 3 12

z=

10. The area elements on z = 2x y and z = x 2 + y 2 , respec-

Let u = 1 + 4r 2 du = 8r dr

1 + 4r 2 r dr

7. The triangle is defined by 0 ≤ y ≤ 1, 0 ≤ x ≤ y.

8.

SECTION 14.7 (PAGE 803)

0

11.

Since these elements are equal, the area of the parts of both surfaces defined over any region of the x y-plane will be equal.  If z = 12 (x 2 + y 2 ), then d S = 1 + x 2 + y 2 d A. Oneeighth of the part of the surface above −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, lies above the triangle T : given by 0 ≤ x ≤ 1, 0 ≤ y ≤ x, or, in polar coordinates, by 0 ≤ θ ≤ π/4, 0 ≤ r ≤ 1/ cos θ = sec θ . Thus  1 + x 2 + y2 d A T  sec θ   π/4 dθ 1 + r 2 r dr =8

S=8

0

 =4

0

π/4

 dθ

0

1+sec2 θ

√

Let u = 1 + r 2 du = 2r dr

u du

0

  8 π/4
(1 + sec2 θ )3/2 − 1 dθ 3 0  2π 8 π/4 . (1 + sec2 θ )3/2 dθ − = 3 0 3

=

Using a TI-85 numerical integration routine, we obtain the numerical value S ≈ 5.123 sq. units.

12. As the figure suggests, the area√of the canopy is the

area of a hemisphere of radius 2 minus four times the area of half of a spherical cap cut off from the sphere x 2 + y 2 + z 2 = 2 by a plane at distance 1 from the origin,say the plane z = 1. Such a spherical cap, z = 2 − x 2 − y 2 , lies above the disk ∂z ∂z = −x/z and = −y/z x 2 + y 2 ≤ 2 − 1 = 1. Since ∂x ∂y on it, the area of the spherical cap is 



1+

x 2 +y 2 ≤1

√  = 2 2π

1

x 2 + y2 dA z2

r dr √ 2 − r2

Let u = 2 − r 2 du = −2r dr √  2 −1/2 √ √ √ = 2π u du = 2 2( 2 − 1) = 4 − 2 2. 0

1

Thus the area of the canopy is √ √ √ 1 S = 2π( 2)2 −4× ×(4−2 2) = 4(π + 2)−8 sq. units. 2

557

SECTION 14.7 (PAGE 803)

R. A. ADAMS: CALCULUS

z

z (0,0,b)

z

a

y

x

y

x

Fig. 14.7.14

Fig. 14.7.12

13.



 π  a Aρ 2 dρ dθ sin φ dφ 2 0 0 B+ρ   a 0 B dρ 1− 2 = 4π A ρ +B 0  √ a = 4π A a − B tan−1 √ units. B

Mass =

2π

15. The force is  F = 2π kmδ

14. A slice of the ball at height z, having thickness dz, is a √ circular disk of radius a 2 − z 2 and areal density δ dz. As calculated in the text, this disk attracts mass m at (0, 0, b) with vertical force 

b−z

h 0

a



h



a 2 + b2 − v b2 − a 2 + v then b − z = b − = 2b  2b   (b+a)2 2 2 b −a +v 1 dv = 2π kmδ 2a − 2 √ 4b (b−a)2 v

 b2 − a 2  = 2π kmδ 2a − b + a − (b − a) 2 2b  1  − 2 (b + a)3 − (b − a)3 6b 4π kmδa 3 km M = = , 3b2 b2 where M = (4/3)π a3 δ is the mass of the ball. Thus the ball attracts the external mass m as though the ball were a point mass M located at its centre.

558

dz

(0,0,b)

b−z dz 1− √ 2 a + b2 − 2bz −a let v = a 2 + b2 − 2bz, dv = −2b dz

F = 2π kmδ

1−  a 2 + (b − z)2



z

.

Thus the ball attracts m with vertical force 

b−z

Let u = a 2 + (b − z)2 du = −2(b − z) dz   2 2 du 1 a +b √ = 2π kmδ h − 2 a 2 +(b−h)2 u     = 2π kmδ h − a 2 + b2 + a 2 + (b − h)2 .



d F = 2π kmδdz 1 −  a 2 − z 2 + (b − z)2



z

a x

y

Fig. 14.7.15

16. The force is 

b



b−z

1−  a 2 (b − z)2 + (b − z)2   b 1 dz = 2π kmδ 1− √ a 2 +1 0  1 . = 2π kmδb 1 − √ 2 a +1

F = 2π kmδ

0

dz

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.7 (PAGE 803)

7a . 12

Thus x¯ = M x=0 /m =

z b

 7a 7a 7a , , . 12 12 12   1 1 , , 0 , the cenSince the base triangle has centroid 3 3   1 1 1 troid of the prism is , , . 3 3 2 

By symmetry, the centre of mass is

r z=b− a

19.

z

ab x

1 y

Fig. 14.7.16

P

17. The force is   b−z dz F = 2π kmδ 1− √ a 2 + b2 − 2bz 0 use the same substitution as in Exercise 2)   a 2 +b2 2 b − a2 + v 1 dv = 2π kmδ a − 2 √ 4b (b−a)2 v   b 2 − a 2  2 2 − (b − a) = 2π kmδ a − a + b 2b2   1 2 2 3/2 3 − 2 (a + b ) − (b − a) 6b   2π kmδ  3 3 2 2 2 + b2 . + a − (2b − a ) a 2b = 3b2 

z

z=

√

a 2 −x 2 −y 2

y

x

Fig. 14.7.17 

 a dy (x 2 + y 2 + z 2 ) dz 0 0  0  a a a x2 dx dy dz = a 5 =3 0 0 0  a  a  a = x dx dy (x 2 + y 2 + z 2 ) dz 0 0 0   a  a a3 dy a(x 2 + y 2) + x dx = 3 0 0   a 4 2a 7a 6 = + a2 x 2 x d x = . 3 12 0

m=

M x=0



a

y

1

1



20. Volume of region =

Fig. 14.7.19  ∞ 2π 2 dθ e−r r dr = π . By

0

0

symmetry, the moments about x = 0 and y = 0 are both zero. We have  Mz=0 =



2π

dθ

0



=π

21. (0,0,b)

18.

x

a

∞

0

∞ 0

 r dr

e−r

0

2

r e −2r dr =

2

z dz

π . 4

The centroid is (0, 0, 1/4).   π a3 1 4 3 πa . By symmetry, the = The volume is 8 3 6 moments about all three coordinate planes are equal. We have  π/2  π/2  a Mz=0 = dθ sin φ dφ ρ cos φ ρ 2 dρ 0 0 0  π a4 π a 4 π/2 . sin φ cos φ dφ = = 8 0 16 Thus z¯ = Mz=0 /volume = 3a/8.   3a 3a 3a . , , The centroid is 8 8 8

a

z

dx

1

r=a 2 z

y x

Fig. 14.7.21

559

SECTION 14.7 (PAGE 803)

R. A. ADAMS: CALCULUS

z

22. The cube has centroid (1/2, 1/2, 1/2). The tetrahedron

lying above the plane x + y + x = 2 has centroid (3/4, 3/4, 3/4) and volume 1/6. Therefore the part of the cube lying below the plane has centroid (c, c, c) and volume 5/6, where

h z r h + a =1

5 3 1 1 c + × = × 1. 6 4 6 2   9 9 9 Thus c = 9/20; the centroid is , , . 20 20 20 1

27.

1

y

1

Fig. 14.7.22

23. The model still involves angular acceleration to spin the

24.

25.

26.

ball — it doesn’t just fall. Part of the gravitational potential energy goes to producing this spin as the ball falls, even in the limiting case where the fall is vertical.  2π  a  h I =δ dθ r 3 dr dz 0 0 0  4 π δha 4 a = . = 2π δh 4 2  a D¯ = I /m = √ . m = π δa 2 h, 2  2π  a  h I =δ dθ r dr (x 2 + z 2 ) dz 0 0 0   a  2π h3 2 2 r dr dθ hr cos θ + =δ 3 0 0   2π  4 ha h 3 a2 =δ dθ cos2 θ + 4 6 0     2 π ha 4 π h 3a2 h2 a 2 =δ + = π δa h + 4 3 4 3   h2 a2 m = π δa 2 h, + . D¯ = I /m = 4 3  2π  a  h(1−(r/a)) I =δ dθ r 3 dr dz 0 0 0  a  r π δa 4 h , r3 1 − = 2π δh dr = a 10 0   π δa 2 h 3 m= , D¯ = I /m = a. 3 10

560

y

Fig. 14.7.26

z

x

a

a

x



 h(1−(r/a)) r dr (x 2 + z 2 ) dz 0 0 0  a   2π r 2 r cos2 θ h 1− dθ =δ a 0 0  h3  r 3 + r dr 1− 3 a   a  2π δh 3 a  r4 r 3 = π δh dr + r 1− dr r3 − a 3 h 0 0 in the second integral put u = 1 − (r/a)  4 2π δa 2 h 3 1 π δa h + (1 − u)u 3 du = 20 3 0 2π δa 2 h 3 π δa 2 h π δa 4 h + = (3a 2 + 2h 2 ), = 20 60 60 

I =δ



2π

a

dθ

 π δa 2 h , D¯ = I /m = 3  I =δ (x 2 + y 2 ) d V

3a 2 + 2h 2 . 20

m=

28.

 = 2δ

0

m = δa 3 ,

Q a





2δa 5 , dz = 3 0 0   2 a. D¯ = I /m = 3

x2 dx

a

a

dy

z

a a a y

x

Fig. 14.7.28

29. The distance s from (x, y, z) to the line x = y, z = 0

satisfies s 2 = u 2 + z 2 , where u is the distance from (x, y, 0) to the line x = y in √ the x y-plane. By Example 7 of Section 1.4 u = |x − y|/ 2, so s2 =

(x − y)2 + z2. 2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.7 (PAGE 803)

The moment of inertia of the cube about this line is   a  a  a (x − y)2 + z 2 dz dx dy I =δ 2 0 0 0   a  a a3 a =δ (x − y)2 + d y Let u = x − y dx 2 3 0 0 du = −d y   x δa a δa 5 + dx u 2 du = 3 2 0 x−a  δa 5 δa a = + (3ax 2 − 3a 2 x + a 3 ) d x 3 6 0   δa 5 δa 3a 4 5δa 5 = + a4 − + a4 = , 3 6 2 12   5 a. m = δa 3 , D¯ = I /m = 12

32.

 I =δ



2π

dθ

0



c

dz

0

m = π δc(b2 − a 2 ),

π δc(b4 − a 4 ) r 3 dr = , 2 a  b2 + a 2 . D¯ = 2 b

z

c

r=a r=b

y

x

Fig. 14.7.32

30. The line L through the origin parallel to the vector v = i + j + k is a diagonal of the cube Q. By Example 8 of Section 1.4, the distance from the point with position vector r = xi + yj + zk to L is s = |v × r|/|v|. Thus, the square of the distance from (x, y, z) to L is (x − y)2 + (y − z)2 + (z − x)2 3  2 2 2 2 = x + y + z − x y − x z − yz . 3





2

Q

Q

x dV =

2

Q

y dV =

 xy dV =

Q



2

Q

 yz d V =

z dV =

m = 2δ



2π

dθ

0



a

a5 3

a5 . xz dV = 4 Q

Therefore, the moment of inertia of Q about L is  a5 a5 δa 5 2δ 3× −3× = . I = 3 3 4 6

I = 2δ = 4π δ

b

31.





 √a 2 −r 2 0

dz

 r a 2 − r 2 dr

0



a

b

 = 2π δ

a

a 2 −r 2

r 3 dr

b

 r 3 a 2 − r 2 dr

a 2 −b2

dz

0

Let u = a 2 − r 2 du = −2r dr

√ (a 2 − u) u du

0  2 2 2 2 = 2π δ a (a − b2 )3/2 − (a 2 − b2 )5/2 3 5 1 2 2 3/2 1 2 = 4π δ(a − b ) (2a + 3b2 ) = m(2a 2 + 3b2 ). 15 5 z

a I /m = √ . 6

 b  c dx dy (x 2 + y 2 ) dz −a −b −c   a 2b3 2 = 2δc dx 2bx + 3 −a 8δabc 2 (a + b2 ), = 3 

I =δ

r dr b

2π

The mass of Q is m = δa 3 , so the radius of gyration is D¯ =

a

Let u = a 2 − r 2 du = −2r dr  a 2 −b2 √ 4π δ 2 (a − b2 )3/2 , u du = = 2π δ 3 0   √ 

= 4π δ

s2 =

We have 

33.



a

m = 8δabc,

D¯ =



I /m =

a 2 + b2 . 3

b

a y

x

Fig. 14.7.33

561

SECTION 14.7 (PAGE 803)

34. By Exercise 26, the cylinder has moment of inertia

R. A. ADAMS: CALCULUS

36. The kinetic energy of the oscillating pendulum is

ma 2 π δa 4 h = , 2 2

I =

where m is its mass. Following the method of Example 4(b), the kinetic energy of the cylinder rolling down the inclined plane with speed v is 1 2 mv + 2 1 = mv 2 + 2

KE =

1 2 I 2 1 2 v2 3 ma 2 = mv 2 . 4 a 4

The potential energy of the cylinder when it is at height h is mgh, so, by conservation of energy,

KE =

−

dv 2 = g sin α. dt 3

35. By Exercise 35, the ball with hole has moment of inertia I =

m (2a 2 + 3b2 ) 5

about the axis of the hole. The kinetic energy of the rolling ball is



1 I 2



dθ dt

2 .

2 − mga cos θ = constant.

Differentiating with respect to time t, we obtain  I

dθ dt



d 2θ + mga sin θ dt 2



dθ dt

 = 0,

or d 2θ mga sin θ = 0. + 2 dt I For small oscillations we have sin θ ≈ θ , and the above equation is approximated by d 2θ + ω2 θ = 0, dt 2 where ω2 = mga/I . The period of oscillation is 2π = 2π T = ω

1 2 m v2 mv + (2a 2 + 3b2 ) 2 2 10 a   2a 2 + 3b2 7a 2 + 3b2 2 1 + = mv . = mv 2 2 2 10a 10a 2

KE =



I . mga

A θ

a C

By conservation of energy, mv 2

dθ dt

The potential energy is mgh, where h is the distance of C above A. In this case, h = −a cos θ . By conservation of energy,

3 2 mv + mgh = constant. 4 Differentiating this equation with respect to time t, we obtain 3 dv dh 0 = mv + mg 2 dt dt 3 dv = mv + mgv sin α. 2 dt Thus the cylinder rolls down the plane with acceleration

1 I 2

7a 2 + 3b2 + mgh = constant. 10a 2

Differentiating with respect to time, we obtain 7a 2 + 3b2 dv + mgv sin α = 0. mv 2 dt 5a Thus the ball rolls down the plane (with its hole remaining horizontal) with acceleration −

562

dv 5a 2 = 2 g sin α. dt 7a + 3b2

Fig. 14.7.36

37. If the centre of mass of B is at the origin, then  M x=0 =

B

xδ d V = 0.

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 14 (PAGE 804)

If line L 0 is the z-axis, and L k is the line x = k, y = 0, then the moment of inertia Ik of B about L k is    (x − k)2 + y 2 δ d V Ik =  B = (x 2 + y 2 + k 2 − 2kx) δ d V B

2

Review Exercises 14

(page 804)

1. By symmetry, 



2

= I0 + k m − 2k Mx=0 = I0 + k m,

R

(x + y) d A = 2

R 1

 =2

where m is the mass of B and I0 is the moment about L 0.

0



z

=2

L0

 x dA = 2

√



1

x dx 0

x2

x

dy

(x 3/2 − x 3 ) d x

 1   2 5/2 x 4

3 2 1 x − − = = 2 5 4 0 5 4 10

Lk

y B

x

(1, 1)

√ y= x x = y2

k y

R

y = x2

Fig. 14.7.37

x

38. The moment of inertia of the ball about the point where it contacts the plane is, by Example 4(b) and Exercise 39,   8 4 I = π δa 5 + π δa 3 a 2 15 3   2 7 + 1 ma 2 = ma 2 . = 5 5

Fig. R-14.1

2.

(x 2 + y 2 ) d A = 1  x3

 0

 1

3





1

0  x=2+y

+ x y 2

2+y

dy y

(x 2 + y 2 ) d x

dy x=y

 (2 + y)3 y3 2 3 = + y (2 + y) − − y dy 3 3 0   1 4 8 8 = + 4y + 4y 2 d y = + 2 + = 6 3 3 3 0 y

v2 7 7 1 2 I = ma 2 2 = mv 2 . 2 10 a 10

39. By Example 7 of Section 1.4, the distance from the point with position vector r = xi + yj + zk to the straight line L through the origin parallel to the vector a = Ai + Bj + Ck is |a × r| s= . |a| The moment of inertia of the body occupying region R about L is, therefore,  1 I = 2 |a × r|2 δ d V |a| R 
1 = 2 (Bz − C y)2 + (C x − Az)2 A + B2 + C 2 R  + (Ay − Bx)2 δ d V
1 = 2 (B 2 + C 2 )Pxx + (A2 + C 2 )Pyy 2 2 A + B +C  + (A2 + B 2 )Pzz − 2 AB Pxy − 2 AC Pxz − 2BC Pyz .

P

=

The kinetic energy of the ball, regarded as rotating about the point of contact with the plane, is therefore KE =



(1, 1) y=x

P 2 Fig. R-14.2

3.

 D

y dA = x



π/4



2

dθ

0

0

(3, 1) x =2+y x

tan θ r dr

π/4 2 2 √

r

= ln sec θ

= 2 ln 2 = ln 2 2

0

0

563

REVIEW EXERCISES 14 (PAGE 804)

R. A. ADAMS: CALCULUS

y

we need to ensure that   k = 1. 2 1− √ k2 + 1

√ Thus k 2 + 1 = (2k)2 , and so 3k 2 = 1, and k = 1/ 3.

y=x S

4.

√

 a) I =

 √4−y 2

3

dy

√ y/ 3

0

 =

e−x

2 −y 2

e−x

2 −y 2

a

x

2 Fig. R-14.3

Fig. R-14.5

dA

R

6.

 I =



2

y

dy 0 

= √

0 R

 f (x, y) d x +



r =2

e−x  √

dx 0

 c) I =

π 3



2 −y 2

I =

dx

0



e 2

dy

tan−1 (1/k).

φ0 = Thus the volume inside the cone and inside the sphere x 2 + y 2 + z 2 = a 2 is 

a

ρ 2 dρ   k 2π a 3 2π a 3 . (1 − cos φ0 ) = 1− √ = 3 3 k2 + 1 dθ

0

0

To have V =

564

sin φ dφ

1 4



(2, 2)

Fig. R-14.6



V =

f (x, y) d y.

x

2

e−r r dr

2 2 e−r

π(1 − e−4 ) − =

2 6 0

φ0

f (x, y) d x

y=x

0



0

y = 6 − x2

0

dθ

2π

x

R

−x 2 −y 2

6−x 2

dx

y 6

x

5. The cone z = k x 2 + y 2 has semi-vertical angle





2

dy

4−x 2

2 1

π/3 0

0



+

d) I =

√ 3x



1

dy 2

where R is as shown in the figure. Thus

3x

1 2 Fig. R-14.4 

 √6−y

6

f (x, y) d A,

R

b) I =

z = kr

dx

where R is as shown in the figure. y

y=

φ0

4 3 πa 3

0

 =

π a3 , 3

7.

 J=



1

dz 0



z

dy 0

0

y

f (x, y, z) d x

corresponds to the region 0 ≤ z ≤ 1,

0 ≤ y ≤ z,

0 ≤ x ≤ y,

which can also be expressed in the form 0 ≤ x ≤ 1, x ≤ y ≤ 1, y ≤ z ≤ 1.  1  1  1 Thus J = dx dy f (x, y, z) dz. 0

x

y

8. A horizontal slice of the object at height z above the base, and having thickness dz, is a disk of radius r = 12 (10 − z) m. Its volume is dV = π

(10 − z)2 dz m3 . 4

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 14 (PAGE 804)

The density of the slice is δ = kz 2 kg/m3 . Since δ = 3, 000 when z = 10, we have k = 30. a) The mass of the object is 

10

π 30z 2 (10 − z)2 dz 4 0 11.  15π 10 2 3 4 (100z − 20z + z ) dz = 2 0   15π 100, 000 = − 50, 000 + 20, 000 ≈ 78, 540 kg. 2 3

m=

b) The moment of inertia (about its central axis) of the disk-shaped slice at height z is d I = 30z 2 dz





2π

(10−z)/2

dθ 0

r 3 dr.

0

Thus the moment of inertia about the whole solid cone is  I =

9.

 f (t) =

a

10

30z 2 dz

0





2π

dθ 0

(10−z)/2

r 3 dr.

0

2

e−x d x

t   a a 1 a 1 2 f (t) dt = dt e−x d x f¯ = a 0 a  t  x 0 1 a −x 2 1 a −x 2 e dx dt = xe dx = a 0 a 0 0 

a 2 2 e−x

1 − e−a 1 − = =

a 2 2a 0

quarter disk Q, as shown in the figure. y 3

2 f =2 f =1 1

=

2π

0

= 2π

2a

dθ √ 2a

 0

0



r 3 dr

6a 2 −r 2

r 2 /a

 r5 r 3 6a 2 − r 2 − a

dz

 dr

Let u = 6a 2 − r 2 du = −2r dr  6a 2 √ π √ 6 =π ( 2a) (6a 2 − u) u du − 2 3a 4a 2  6a 

2 8 = π 4a 2 u 3/2 − u 5/2

− π a 5 5 3 4a 2 √ 8π 5 (18 6 − 41)a = 15 bounded by the circle x 2 + y 2 = 2ay, which has polar equation r = 2a sin θ , (0 ≤ θ ≤ π ). It lies below the cone z = x 2 + y 2 = r . The moment of inertia of S about the z-axis is   π  2a sin θ  r I = (x 2 + y 2 ) d V = dθ r 3 dr dz S 0 0 0   2a sin θ  π 32a 5 π 5 dθ r 4 dr = sin θ dθ = 5 0 0 0  32a 5 π (1 − cos2 θ )2 sin θ dθ Let u = cos θ = 5 0 du = − sin θ dθ  32a 5 1 = (1 − 2u 2 + u 4 ) du 5 −1   64a 5 2 1 512a5 = 1− + = . 5 3 5 75

13. A horizontal slice of D at height z is a right triangle

Q

f =0

The sphere x 2 + y 2 + z 2 = 6a 2 and the paraboloid z = (x 2 + y 2)/a intersect where z 2 + az − 6a 2 = 0, that is, where (z + 3a)(z − 2a) = 0. Only z = 2a is possible; the plane z = −3a does not intersect the sphere. If z = 2a, 2 then x 2 + y 2 = r 2 = 6a 2 − 4a 2 = 2a √ , so the intersection is on the vertical cylinder of radius 2a with axis on the z-axis. We have,  (x 2 + y 2) d V D  √  √ 

12. The solid S lies above the region in the x y-plane

10. If f (x, y) = x + y , then f = 0, 1, or 2, in parts of the

1

Thus      3 5 1 +1 + 2 (π − 2) = 2π − , f (x, y) d A = 0 2 2 2 Q   1 5 5 and f¯ = 2π − =2− . π 2 2π

2

Fig. R-14.10

3

x

with legs (2 − z)/2 and 2 − z. Thus the volume of D is  1 1 7 V = . (2 − z)2 dz = 4 0 12

565

REVIEW EXERCISES 14 (PAGE 804)

R. A. ADAMS: CALCULUS

z

Its moment about z = 0 is 

(0, 0, 1)

1

1 z(2 − z)2 dz 4 0  11 1 1 (4z − 4z 2 + z 3 ) dz = . = 4 0 48

Mz=0 =

y+z =1 S

(0, 1, 0) 2

The z-coordinate of the centroid of D is z¯ =

11 48



x

7 11 = . 12 28

(0, 0, 1)

15.

y+z =2 (0, 1, 1)

2x + z = 2 (0, 0, 0) (1, 0, 0) x

(1, 1, 0)

y

x = 2 − y − 2z Fig. R-14.14

z

( 12 , 0, 1)

(2, 0, 0)

(0, 2, 0)

2x + y + z = 2

y



 1  1+z  1+z−y z dV = z dz dy dx S 0 0 0  1+z  1 z dz (1 + z − y) d y = 0 0   1 (1 + z)2 dz z (1 + z)2 − = 2 0  1 1 17 = (z + 2z 2 + z 3 ) dz = 2 0 24 z (0, 0, 1)

Fig. R-14.13

(0, 2, 1)

14.



 1  1−y  2−y−2z dV = dy dz dx S 0 0 0  1  1−y dy (2 − y − 2z) dz = 0 0  1 [(2 − y)(1 − y) − (1 − y)2 ] d y = 0  1 1 (1 − y) d y = = 2 0   1  1−y  2−y−2z x dV = dy dz x dx M x=0 = S 0 0 0   1−y 1 1 dy [(2 − y)2 − 4(2 − y)z + 4z 2 ] dz = 2 0 0  1 1 (2 − y)2 (1 − y) − 2(2 − y)(1 − y)2 = 2 0  4 3 + (1 − y) d y Let u = 1 − y 3 du = −d y   4 1 1 (u + 1)2 u − 2(u + 1)u 2 + u 3 du = 2 0 3   1 1 1 3 7 = u + u du = 2 0 3 24  1 7 7 = x¯ = 24 2 12 V =

566

(2, 0, 1)

S (1, 0, 0)

x

y =1+z (0, 1, 0) y x +y−z =1

Fig. R-14.15

16. The plane z = 2x intersects the paraboloid z = x 2 + y 2

on the circular cylinder x 2 + y 2 = 2x, (that is, 2 + y 2 = 1), which has radius 1. Since (x − 1)√ √ d S = 1 + 22 d A = 5 d A on the plane, the area of the part of the plane inside √ the paraboloid (and therefore inside the cylinder) √ is 5 times the area of a circle of radius 1, that is, 5π square units.

17. As noted in the previous exercise, the part of the

paraboloid z = x 2 + y 2 that lies below the plane z = 2x is inside the vertical cylinder x 2 + y 2 = 2x, which has polar equation r = 2 cos θ (−π/2 ≤ θ ≤ π/2). On the paraboloid: dS =

 1 + (2x)2 + (2y)2 d A = 1 + 4r 2 r dr dθ.

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 14 (PAGE 805)

to the region S inside the sphere u2 + v 2 + w2 = 1 and above the plane u + v + w = 1. The distance from the 1 origin to this plane is √ , so, by symmetry, the volume 3 of S is equal to the √ volume inside the sphere and above the plane w = 1/ 3, that is,

The area of that part of the paraboloid is  S= =

π/2

−π/2

1 8



 dθ

π/2

2 cos θ

0

 dθ

−π/2  π/2 2

 1 + 4r 2 r dr

1+16 cos2 θ

Let u = 1 + 4r 2 du = 8r dr

u 1/2 du

 1  w3

2 π(1 − w ) dw = π w − √ 3 1√3 1/ 3 √ 2π(9 − 4 3) cu. units. = 27



1

1 [(1 + 16 cos2 θ )3/2 − 1] dθ 4 0 3  1 π/2 [(1 + 16 cos2 θ )3/2 − 1] dθ = 6 0 ≈ 7.904 sq. units.

=

(using a TI-85 numerical integration function). y2 z2 x2 + + = 1 36 9 4 and above the plane x + y + z = 1 is transformed by the change of variables

18. The region R inside the ellipsoid

x = 6u,

y = 3v,

z = 2w

to the region S inside the sphere u2 + v 2 + w2 = 1 and above the plane 6u + 3v + 2w = 1. The distance from the origin to this plane is D= √

1 62

+ 32

+ 22

=

1 , 7

so, by symmetry, the volume of S is equal to the volume inside the sphere and above the plane w = 1/7, that is,  1  w3

180π units3 . π(1 − w2 ) dw = π w − =

3 1/7 343 1/7



1

Since |∂(x, y, z)/∂(u, v, w)| = 6 · 3 · 2 = 18, the volume of R is 18 × (180π/343) = 3240π/343 ≈ 29.68 cu. units.

Challenging Problems 14

(page 805)

1. This problem is similar to Review Exercise 18 above.

y2 z2 x2 The region R inside the ellipsoid 2 + 2 + 2 = 1 and a b c x y z above the plane + + = 1 is transformed by the a b c change of variables x = au,

y = bv,

2.

1

Since |∂(x, √ y, z)/∂(u, v, w)| = abc, the volume of R is 2π(9 − 4 3) abc cu. units. 27 The plane (x/a) + (y/b) + (z/c) = 1 intersects the ellipsoid (x/a)2 + (y/b)2 + (z/c)2 = 1 above the region R in the x y-plane bounded by the ellipse y 2 y2  x x2 − + + 1 − = 1, a2 b2 a b or, equivalently, x y y2 xy x2 − − = 0. + + a2 b2 ab a b Thus the area of the part of the plane lying inside the ellipsoid is   c2 c2 S= 1 + 2 + 2 dx dy a b √R 2 2 2 2 a b + a c + b2 c2 (area of R). = ab Under the transformation x = a(u + v), y = b(u − v), R corresponds to the ellipse in the uv-plane bounded by (u + v)2 + (u − v)2 + (u 2 − v 2 ) − (u + v) − (u − v) = 0 3u 2 + v 2 − 2u = 0   1 2 1 2 + v2 = 3 u − u+ 3 9 3 2 2 v (u − 1/3) + = 1, 1/9 1/3 √ √ an ellipse with area π(1/3)(1/ 3) = π/(3 3) sq. units. Since



a a

| du dv = 2ab du dv, d x d y = |

b −b

we have 2π  S = √ a 2 b2 + a 2 c2 + b2 c2 sq. units. 3 3

z = cw

567

CHALLENGING PROBLEMS 14 (PAGE 805)

3.

R. A. ADAMS: CALCULUS

∞

 1 = 1 + x y + (x y)2 + · · · = (x y)n−1 1 − xy n=1  1 ∞  1  dx dy = x n−1 d x y n−1 d y 1 − xy 0 0

a) 

1 1

0

0

=

n=1 ∞  n=1

1 . n2

Remark: The series for 1/(1 − x y) converges for |x y| < 1. Therefore the outer integral is improper (i.e., c limc→1− 0 d x). We cannot do a detailed analysis of the  convergence here, but the convergence of 1/n 2 shows that the iterated double integral must converge. b) Similarly, 1 1 + xy  1 1 0

0

n=1

dx dy 1 + xy

n=1

1

x n−1 d x

0



1

y n−1 d y

0

∞  (−1)n−1

0

=

0 0 ∞  1 

0

=

dx dy 1 − x yz  x n−1 d x

y n−1 d y

0

1

z n−1 dz

Now the volume of the whole ball is (4π/3)23 = 32π/3, so the volume remaining after the hole is cut is

0

0

n3

0

1

y n−1 d y



1

z n−1 dz

0

.

4. Under the transformation u = a • r, v = b • r, w = c • r,

where r = xi + yj + zk, the parallelepiped P corresponds to the rectangle R specified by 0 ≤ u ≤ d1 , 0 ≤ v ≤ d2 , 0 ≤ w ≤ d3 . If a = a1 i+a2 j+a3 k and similar expressions hold for b and c, then

∂(u, v, w)

a1 = b1

∂(x, y, z) c1

568

32π − V0 3  π/4 16 (3 − 4 sin2 θ )3/2 = cos θ dθ 3 0 (1 − sin2 θ )2  √ 16 1/ 2 (3 − 4v 2 )3/2 dv. = 3 0 (1 − v 2 )2

V =

0

∞  (−1)n−1 n=1



1

n=1

=

1

π/4


4−sec2 θ

 16 8 − (4 − sec2 θ )3/2 dθ 3 0  16 π/4 (4 cos2 θ − 1)3/2 32π − dθ. = 3 3 0 cos3 θ =

dx dy 1 + x yz  1  (−1)n−1 x n−1 d x

0 ∞ 

part in the first octant, which is itself split into two equal parts by the plane x = y:  1  x V0 = 16 dx 4 − x 2 − y2 d y 0 0  sec θ   π/4 dθ 4 − r 2 r dr Let u = 4 − r 2 = 16 0 0 du = −2r dr  π/4  4 =8 dθ u 1/2 du 

n=1 0 ∞ 

1 n3 n=1  1 1 =

and we have  (a • r)(b • r)(c • r) d x d y dz P  uvw = du dv dw R |a • (b × c)|  d1  d2  d3 1 u du v dv w dw = |a • (b × c)| 0 0 0 d12 d22 d32 = . 8|a • (b × c)|

0

n2

n=1 1 1 1





∂(x, y, z)

du dv dw = du dv dw , d x d y dz =

∂(u, v, w)

|a • (b × c)|

5. The volume V0 removed from the ball is eight times the

∞  (−x y)n−1 = 1 − x y + (x y)2 − · · · =

 ∞  (−1)n−1 = =

Therefore

a2 b2 c2

a3

b3 = a • (b × c).

c3

Let v = sin θ dv = cos θ dθ

We submitted this last integral to Mathematica to obtain   4 2 32sin−1 − 23/2 + 11tan−1 (3 − 23/2 ) V = 3 3  3/2 −1 − 11tan (3 + 2 ) ≈ 18.9349.

6. Under the transformation x = u3 , y = v 3 ,

z = w3 , the region R bounded by the surface x 2/3 + y 2/3 + z 2/3 = a 2/3 gets mapped to the ball B bounded by u 2 + v 2 + w2 = a 2/3 . Assume that a > 0. Since ∂(x, y, z) = 27u 2 v 2 w2 , ∂(u, v, w)

INSTRUCTOR’S SOLUTIONS MANUAL

the volume of R is

the required volume is 

V = 27

B

u 2 v 2 w2 du dv dw.

Now switch to polar coordinates [ρ, φ, θ ] in uvw-space. Since uvw = (ρ sin φ cos θ )(ρ sin φ sin θ )(ρ cos φ),

V = 8(63 )

 B

u 5 v 5 w5 du dv dw.

Now switch to polar coordinates [ρ, φ, θ ] in uvw-space. Since uvw = (ρ sin φ cos θ )(ρ sin φ sin θ )(ρ cos φ),

we have 

 π  a 1/3 cos2 θ sin2 θ dθ sin5 φ cos2 φ dφ ρ 8 dρ 0 0 0  π  2π sin2 (2θ ) 3 dθ (1 − cos2 φ)2 cos2 φ sin φ dφ = 3a 4 0 0 Let t = cos φ, dt = − sin φ dφ  2π  1 1 − cos(4θ ) dθ (1 − t 2 )2 t 2 dt = 3a 3 8 0 −1  1 3a 3 4π a 3 2 4 = (2π )2 cu. units. (t − 2t + t 6 ) dt = 8 35 0

V = 27

CHALLENGING PROBLEMS 14 (PAGE 805)

2π

7. One-eighth of the required volume lies in the first octant. Under the transformation x = u6 , y = v 6 , z = w6 , the region first-octant R bounded by the surface x 1/3 + y 1/3 + z 1/3 = a 1/3 and the coordinate planes gets mapped to the first octant part B of the ball bounded by u 2 + v 2 + w2 ≤ a 1/3 . Assume that a > 0. Since

we have  V = 1, 728

π/2 0



×

(cos θ sin θ )5 dθ

a 1/6 0



π/2 0

(sin2 φ cos φ)5 sin φ dφ

ρ 17 dρ

 π/2 sin5 (2θ ) dθ sin11 φ(1 − sin2 φ)2 cos φ dφ 32 0 0 Let s = sin φ, ds = cos φ dφ  π/2  1 3 2 2 (1 − cos (2θ )) sin(2θ ) dθ s 11 (1 − s 2 )2 ds = 3a

= 96a 3



0

π/2

0

Let t = cos(2θ ), dt = −2 sin(2θ ) dθ   1 3a 3 1 (1 − 2t 2 + t 4 ) dt (s 11 − 2s 13 + s 15 ) ds = 2 −1 0    1 1 1 a3 2 1 = 3a 3 1 − + = − + cu. units. 3 5 12 7 16 210

∂(x, y, z) = 63 u 5 v 5 w5 , ∂(u, v, w)

569