08 Double Integrals in Polar Coordinates - Handout

Double Integrals in Polar Coordinates

Math 55 - Elementary Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 Double Integrals in Polar Coordinates 1/ 12

Recall

If f is continuous on a Type I region D such that

D = {(x, y) : a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)} ,

then ¨

D

f(x, y) dA =

ˆ b

a

ˆ g2(x)

g1(x)f(x, y) dy dx

If f is continuous on a Type II region D such that

D = {(x, y) : h1(y) ≤ x ≤ h2(y), c ≤ y ≤ d} ,

then ¨

D

f(x, y) dA =

ˆ d

c

ˆ h2(y)

h1(y)f(x, y) dx dy


Double Integrals

Suppose we want to evaluate

¨

R

√x2 + y2 dA where R is the

region given below:

1 2 3

1

2

3 y =√

9− x2

y =√

1− x2

R = R1 ∪R2 where

R1 ={

(x, y) : 0 ≤ x ≤ 1,√

1− x2 ≤ y ≤√

9− x2}

R2 ={

(x, y) : 1 ≤ x ≤ 3, 0 ≤ y ≤√

9− x2}

Then

¨

R

√x2 + y2 dA =

ˆ 1

0

ˆ √9−x2

√1−x2

√x2 + y2 dy dx

+

ˆ 3

1

ˆ √9−x2

0

√x2 + y2 dy dx


Polar Coordinates

The polar coordinates (r, θ) are related to the rectangularcoordinates (x, y) by

(x, y)

θ

r

r2 = x2 + y2

x = r cos θ

y = r sin θ

Remark: In polar coordinates,

R the equation of a line through the pole is θ = k for some0 ≤ k < 2π

R the equation of a circle centered at the pole of radius k is r = k

R the equation of a circle centered at (a, b) passing through theorigin is r = 2a cos θ + 2b sin θ


Double Integral in Polar Coordinates

Define a polar rectangle as R = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β} .

Subdivide [a, b] into m subintervals[ri−1, ri] of equal width ∆r and [α, β]into n subintervals [θj−1, θj ] of equalwidth ∆θ. This divides R into polarsubrectangles Rij .

Choose a point (r∗i , θ∗j ) ∈ R and ap-

proximate the area of Rij by

∆Aij = r∗i ∆θ∆r

Therefore, we have¨

R

f(x, y) dA = limm,n→∞

m∑i=1

n∑j=1

f(r∗i cos θ∗j , r∗i sin θ∗j )∆Aij

= limm,n→∞

m∑i=1

n∑j=1

f(r∗i cos θ∗j , r∗i sin θ∗j )r∗i ∆r∆θ

=

ˆ β

α

ˆ b

a

f(r cos θ, r sin θ) r dr dθ



Change to Polar Coordinates in a Double Integral

If f is continuous on a polar rectangleR = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β}, where 0 ≤ α− β ≤ 2π,then

¨

R

f(x, y) dA =

ˆ β

α

ˆ b

af(r cos θ, r sin θ) r dr dθ



Example

Evaluate

¨

R

√x2 + y2 dA where R is the region given below.

Solution.

1 2 3

1

2

3 r = 3

r = 1

In polar coordinates,

R ={

(r, θ) : 1 ≤ r ≤ 3, 0 ≤ θ ≤ π

2

}.

Hence,

¨

R

√x2 + y2 dA =

ˆ π2

0

ˆ 3

1

r · r dr dθ =

ˆ π2

0

ˆ 3

1

r2 dr dθ

=

ˆ π2

0

r3

3

∣∣∣∣r=3

r=1

dθ =

ˆ π2

0

26

3dθ

=26

3θ

∣∣∣∣π20

=13π

3



Example

Evaluate

ˆ 2√2

0

ˆ √8−x2

0

1√x2 + y2 + 1

dy dx.

Solution: Note that the region of integration is

R ={

(x, y) : 0 ≤ x ≤ 2√

2, 0 ≤ y ≤√

8− x2}

1 2 3

1

2

3

0

In polar coordinates,

R ={

(r, θ) : 0 ≤ r ≤ 2√

2, 0 ≤ θ ≤ π

2

}.

ˆ 2√2

0

ˆ √8−x2

0

1√x2 + y2 + 1

dy dx =

ˆ π2

0

ˆ 2√2

0

1√r2 + 1

r dr dθ

=

ˆ π2

0

√r2 + 1

∣∣∣∣r=2√2

r=0

dθ

=

ˆ π2

0

(3− 1) dθ = 2θ

∣∣∣∣π20

= π



Double Integral over General Polar Region

If f is continuous on a polar region of the form

D = {(r, θ) : α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)}

then

¨

D

f(x, y) dA =

ˆ β

α

ˆ h2(θ)

h1(θ)f(r cos θ, r sin θ) r dr dθ.


Double Integral over General Polar Region

Example

Evaluate

¨

R

x dA, where R is the region in the first quadrant between

x2 + (y − 3)2 = 9 and x2 + y2 = 36.

Solution: R ={

(r, θ) : 0 ≤ θ ≤ π2 , 6 sin θ ≤ r ≤ 6

}. Hence,

1 2 3 4 5 6

1

2

3

4

5

6

r = 6 sin θ

r = 6

¨

R

x dA =

ˆ π2

0

ˆ 6

6 sin θ

(r cos θ)r dr dθ

=

ˆ π2

0

ˆ 6

6 sin θ

r2 cos θ dr dθ

=

ˆ π2

0

r3

3cos θ

∣∣∣∣r=6

r=6 sin θ

dθ

=

ˆ π2

0

72 cos θ − 72 sin3 θ cos θ dθ

= 72 sin θ − 18 sin4 θ

∣∣∣∣π20

= 54


Exercises

1 Evaluate the following double integrals by converting them to polar

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2 + y2) dx dy

b.

ˆ 2

0

ˆ 0

−√

4−y2x2y dx dy

c.

ˆ √2

2

0

ˆ √1−y2

y

(x+ y) dx dy

d.

ˆ 2

0

ˆ √2x−x2

0

√x2 + y2 dy dx

2 Evaluate

¨

R

y3 dA, where R is the region enclosed by the line y = x,

the circle (x− 1)2 + y2 = 1 and the x-axis.

3 Find the volume of the solid that lies inside the spherex2 + y2 + z2 = 16 and outside the cylinder x2 + y2 = 4.


References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Leithold, L., The Calculus 7, Harper Collins College Div., 1995

3 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/


08 Double Integrals in Polar Coordinates - Handout

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