Double Integrals in Polar Coordinates
Math 55 - Elementary Analysis III
Institute of MathematicsUniversity of the Philippines
Diliman
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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
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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
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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
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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
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Double Integral in Polar Coordinates
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
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Double Integral in Polar Coordinates
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
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Double Integral in Polar Coordinates
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
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Double Integral in Polar Coordinates
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.
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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
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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.
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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/
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