Polar Coordiantes and Change
Double Integrals
Advanced Calculus Lecture 2 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 9, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Polar coordinates In polar coordinates, we have x = r cos θ, r 2 = x 2 + y 2 y = r sin θ, tan θ = y/x and the area element is given by dxdy = dA = r drd θ . Double integrals in polar coordinates Suppose that f is a continuous function on a polar rectangle R given by 0 ≤ a ≤ r ≤ b , α ≤ θ ≤ β where 0 ≤ β − α ≤ 2π , then ZZ
R
f (x, y)dA =
Z
β
dθ
α
Dr. Lahcen Laayouni
Z
b
f (r cos θ, r sin θ)r dr .
a
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Double integrals in polar coordinates (cont.) If the region D on a plane can be represented in the form D = {(r , θ)|α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)} , then
ZZ
f (x, y)dA =
Z
β
dθ
α
D
Z
h2 (θ)
f (r cos θ, r sin θ)r dr .
h1 (θ)
Example
5
Evaluate the integral I =
RR
R
xydA where
3
R is the region in the first quadrant that lies between the circles
x2
4
+
y2
= 4 and
x 2 + y 2 = 25.
2
1
0 0
Dr. Lahcen Laayouni
Advanced Calculus
1
2
3
4
5
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution The region of integration R in the polar coordinates is n πo R = (r , θ)|2 ≤ r ≤ 5, 0 ≤ θ ≤ . 2 So that
I = =
=
Z
π 2
0
Z
5
(r cos θ)(r sin θ)r drd θ = 2
π 2
Z
0
Z
5
r 3 cos θ sin θdrd θ ! Z π
2
Z 5 2 1 r sin(2θ)drd θ = r 3 dr sin(2θ)d θ 2 0 2 2 0 5 ! π ! 1 r 4 cos(2θ) 2 54 − 24 609 − = = . 2 4 2 2 8 8 0 1 2
Z
Z
5
π 2
3
Thus, the value of the double integral I = Dr. Lahcen Laayouni
RR
R
xydA is
Advanced Calculus
609 . 8
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Example Find the volume of the solid lying in the first octant, inside the cylinder
1.0 z
x 2 + y 2 = 4 , and under the plane
0.0 2.0
2.0 1.5
0.5 2.0 1.5
1.5 1.0 y
z=y .
1.0 0.5
0.5 0.0
x
0.0
Solution The base of the solid is a quarter disk, in polar coordinates we obtain 0 ≤ θ ≤ π2 and 0 ≤ r ≤ 2 . The height is given by z = y = r sin θ . The required volume is V
= =
Z
π/2
Z0 π/2 0
dθ
Z
2
(r sin θ)rdr Z 2 8 sin θd θ r 2 dr = units3 . 3 0 0
Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Example Find the volume of the solid region lying inside both the sphere x 2 + y 2 + z 2 = 4a2 and the cylinder x 2 + z 2 = 2az , where a > 0 . Solution We interchange the roles of y and z , so that the equation of the cylinder becomes x 2 + y 2 = 2ay in the new coordinates. We rewrite the equation of the cylinder as follows x 2 + (y − a)2 = a2 . Which is the equation of a vertical circular cylinder or radius a having its axis along the vertical line through (0, a, 0) . The sphere is centered at the origin and has radius 2a . Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) 6
Using polar coordinates the equation of
5
the sphere becomes r 2 + z 2 = 4a2 , and the cylinder has equation
4 z3 2
r2
= 2ra sin θ or, r = 2a sin θ . One-quarter of the required volume is shown in the next figure.
1 5
0 6
5
3 4
2 3 y
2
4 x
1 1
0 0
Solution (cont.) The base of the solid in the first octant in polar coordinates is specified by 0 ≤ θ ≤ π2 and 0 ≤ r ≤ 2a sin θ . Thus the volume is Z π/2 Z 2a sin θ p V = 4 dθ 4a2 − r 2 rdr Let u = 4a2 − r 2 0
= 2
Z
0
0
π/2
dθ
Z
4a2
4a2
cos2
√
θ
4 udu = 3
Dr. Lahcen Laayouni
Z
π/2
0
(8a3 − 8a3 cos3 θ)d θ
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) Let v = sin θ , then Z 4 π/2 V = (8a3 − 8a3 cos3 θ)d θ 3 0 Z 16 3 32 3 1 = πa − a (1 − v 2 )dv 3 3 0 16 3 64 3 16 = πa − a = (3π − 4)a3 cubic units. 3 9 9 As a supplemental question, what is the volume of the region inside the sphere and outside the cylinder? The volume of the required region is 4 16 Volume = Volume of the sphere − V = π(2a)3 − (3π − 4)a3 3 9 16 = (3π + 4)a3 cubic units. 3 Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Theorem Let x = x(u, v) , y = y(u, v) be a one-to-one transformation from a domain S in the uv -plane onto a domain D in the xy -plane. Suppose that the functions x and y , and their first partial derivatives with respect to u and v , are continuous in S . If f (x, y) is integrable on D , and if g(u, v) = f (x(u, v), y(u, v)) , then g is integrable on S and ZZ ZZ ∂(x, y) dudv , f (x, y)dxdy = g(u, v) ∂(u, v) D S ∂x ∂y ∂(x, y) ∂u ∂u where = is the Jacobian determinant. ∂(u, v) ∂x ∂y ∂v ∂v Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Remark The jacobian determinant satisfies ∂(x, y) ∂(u, v) −1 ∂(u, v) = ∂(x, y) Example
Use change of variables to find the area of the elliptic disk E given by x2 y2 + 2 ≤1 a2 b
a > 0,
b>0
Solution Setting x = au and y = bv , then the elliptic disk E is the one-to-one image of the circular disk D given by u 2 + v 2 ≤ 1 , and ∂(x, y) a 0 dudv = abdudv . dxdy = dudv = 0 b ∂(u, v) Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) The area of E is given by ZZ ZZ 1dxdy = abdudv = ab × (area of D) = πab square units . E
D
Example Evaluate
RR
E (x
2
− xy + y 2 )dA , where E is given by x 2 − xy + y 2 ≤ 1 ,
using the change of variables x = u +
√
3 3 v
and y = u −
√
3 3 v
.
Solution Using the new variables, we obtain √ 3 2 3 v) − (u 2u 2 + 23 v 2 − (u 2 u2 + v 2.
x 2 − xy + y 2 = (u + = =
Dr. Lahcen Laayouni
+ −
√ 3 3 v)(u 1 2 3v )
−
Advanced Calculus
√ 3 3 v)
+ (u −
√ 3 2 3 v)
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) The change of variables maps the ellipse E to the disk D given by u 2 + v 2 ≤ 1 . The Jacobian is given by ∂(x, y) 1√ = √1 3 ∂(u, v) 3 − 33
thus
ZZ
2
E
2
(x − xy + y )dxdy
=
=
√ 2 3 3 ,
√ ZZ 2 3 (u 2 + v 2 )dudv . 3 D
Using polar coordinates with u = r cos θ and v = r sin θ , then √ RR RR 2 2 )dxdy = 2 3 2 2 (x − xy + y E 3 D (u + v )dudv √ R R1 2π = 2 3 3 0 d θ 0 r 2 rdrd θ . =
Dr. Lahcen Laayouni
√ 2 3 1 3 (2π)( 4 )
Advanced Calculus
=
√ 3 3 π.
Advanced Calculus Lecture 2 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 9, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Polar coordinates In polar coordinates, we have x = r cos θ, r 2 = x 2 + y 2 y = r sin θ, tan θ = y/x and the area element is given by dxdy = dA = r drd θ . Double integrals in polar coordinates Suppose that f is a continuous function on a polar rectangle R given by 0 ≤ a ≤ r ≤ b , α ≤ θ ≤ β where 0 ≤ β − α ≤ 2π , then ZZ
R
f (x, y)dA =
Z
β
dθ
α
Dr. Lahcen Laayouni
Z
b
f (r cos θ, r sin θ)r dr .
a
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Double integrals in polar coordinates (cont.) If the region D on a plane can be represented in the form D = {(r , θ)|α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)} , then
ZZ
f (x, y)dA =
Z
β
dθ
α
D
Z
h2 (θ)
f (r cos θ, r sin θ)r dr .
h1 (θ)
Example
5
Evaluate the integral I =
RR
R
xydA where
3
R is the region in the first quadrant that lies between the circles
x2
4
+
y2
= 4 and
x 2 + y 2 = 25.
2
1
0 0
Dr. Lahcen Laayouni
Advanced Calculus
1
2
3
4
5
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution The region of integration R in the polar coordinates is n πo R = (r , θ)|2 ≤ r ≤ 5, 0 ≤ θ ≤ . 2 So that
I = =
=
Z
π 2
0
Z
5
(r cos θ)(r sin θ)r drd θ = 2
π 2
Z
0
Z
5
r 3 cos θ sin θdrd θ ! Z π
2
Z 5 2 1 r sin(2θ)drd θ = r 3 dr sin(2θ)d θ 2 0 2 2 0 5 ! π ! 1 r 4 cos(2θ) 2 54 − 24 609 − = = . 2 4 2 2 8 8 0 1 2
Z
Z
5
π 2
3
Thus, the value of the double integral I = Dr. Lahcen Laayouni
RR
R
xydA is
Advanced Calculus
609 . 8
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Example Find the volume of the solid lying in the first octant, inside the cylinder
1.0 z
x 2 + y 2 = 4 , and under the plane
0.0 2.0
2.0 1.5
0.5 2.0 1.5
1.5 1.0 y
z=y .
1.0 0.5
0.5 0.0
x
0.0
Solution The base of the solid is a quarter disk, in polar coordinates we obtain 0 ≤ θ ≤ π2 and 0 ≤ r ≤ 2 . The height is given by z = y = r sin θ . The required volume is V
= =
Z
π/2
Z0 π/2 0
dθ
Z
2
(r sin θ)rdr Z 2 8 sin θd θ r 2 dr = units3 . 3 0 0
Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Example Find the volume of the solid region lying inside both the sphere x 2 + y 2 + z 2 = 4a2 and the cylinder x 2 + z 2 = 2az , where a > 0 . Solution We interchange the roles of y and z , so that the equation of the cylinder becomes x 2 + y 2 = 2ay in the new coordinates. We rewrite the equation of the cylinder as follows x 2 + (y − a)2 = a2 . Which is the equation of a vertical circular cylinder or radius a having its axis along the vertical line through (0, a, 0) . The sphere is centered at the origin and has radius 2a . Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) 6
Using polar coordinates the equation of
5
the sphere becomes r 2 + z 2 = 4a2 , and the cylinder has equation
4 z3 2
r2
= 2ra sin θ or, r = 2a sin θ . One-quarter of the required volume is shown in the next figure.
1 5
0 6
5
3 4
2 3 y
2
4 x
1 1
0 0
Solution (cont.) The base of the solid in the first octant in polar coordinates is specified by 0 ≤ θ ≤ π2 and 0 ≤ r ≤ 2a sin θ . Thus the volume is Z π/2 Z 2a sin θ p V = 4 dθ 4a2 − r 2 rdr Let u = 4a2 − r 2 0
= 2
Z
0
0
π/2
dθ
Z
4a2
4a2
cos2
√
θ
4 udu = 3
Dr. Lahcen Laayouni
Z
π/2
0
(8a3 − 8a3 cos3 θ)d θ
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) Let v = sin θ , then Z 4 π/2 V = (8a3 − 8a3 cos3 θ)d θ 3 0 Z 16 3 32 3 1 = πa − a (1 − v 2 )dv 3 3 0 16 3 64 3 16 = πa − a = (3π − 4)a3 cubic units. 3 9 9 As a supplemental question, what is the volume of the region inside the sphere and outside the cylinder? The volume of the required region is 4 16 Volume = Volume of the sphere − V = π(2a)3 − (3π − 4)a3 3 9 16 = (3π + 4)a3 cubic units. 3 Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Theorem Let x = x(u, v) , y = y(u, v) be a one-to-one transformation from a domain S in the uv -plane onto a domain D in the xy -plane. Suppose that the functions x and y , and their first partial derivatives with respect to u and v , are continuous in S . If f (x, y) is integrable on D , and if g(u, v) = f (x(u, v), y(u, v)) , then g is integrable on S and ZZ ZZ ∂(x, y) dudv , f (x, y)dxdy = g(u, v) ∂(u, v) D S ∂x ∂y ∂(x, y) ∂u ∂u where = is the Jacobian determinant. ∂(u, v) ∂x ∂y ∂v ∂v Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Remark The jacobian determinant satisfies ∂(x, y) ∂(u, v) −1 ∂(u, v) = ∂(x, y) Example
Use change of variables to find the area of the elliptic disk E given by x2 y2 + 2 ≤1 a2 b
a > 0,
b>0
Solution Setting x = au and y = bv , then the elliptic disk E is the one-to-one image of the circular disk D given by u 2 + v 2 ≤ 1 , and ∂(x, y) a 0 dudv = abdudv . dxdy = dudv = 0 b ∂(u, v) Dr. Lahcen Laayouni
Advanced Calculus
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) The area of E is given by ZZ ZZ 1dxdy = abdudv = ab × (area of D) = πab square units . E
D
Example Evaluate
RR
E (x
2
− xy + y 2 )dA , where E is given by x 2 − xy + y 2 ≤ 1 ,
using the change of variables x = u +
√
3 3 v
and y = u −
√
3 3 v
.
Solution Using the new variables, we obtain √ 3 2 3 v) − (u 2u 2 + 23 v 2 − (u 2 u2 + v 2.
x 2 − xy + y 2 = (u + = =
Dr. Lahcen Laayouni
+ −
√ 3 3 v)(u 1 2 3v )
−
Advanced Calculus
√ 3 3 v)
+ (u −
√ 3 2 3 v)
Double Integrals
Polar coordinates
Change of variables formula
Double Integrals
Solution (cont.) The change of variables maps the ellipse E to the disk D given by u 2 + v 2 ≤ 1 . The Jacobian is given by ∂(x, y) 1√ = √1 3 ∂(u, v) 3 − 33
thus
ZZ
2
E
2
(x − xy + y )dxdy
=
=
√ 2 3 3 ,
√ ZZ 2 3 (u 2 + v 2 )dudv . 3 D
Using polar coordinates with u = r cos θ and v = r sin θ , then √ RR RR 2 2 )dxdy = 2 3 2 2 (x − xy + y E 3 D (u + v )dudv √ R R1 2π = 2 3 3 0 d θ 0 r 2 rdrd θ . =
Dr. Lahcen Laayouni
√ 2 3 1 3 (2π)( 4 )
Advanced Calculus
=
√ 3 3 π.
