Section 13.5: Double Integrals in Polar Coordinates
ZZ
Suppose we want to evaluate the double integral
f (x, y)dA, where D is the unit
D
disk x2 + y 2 ≤ 1. The description of D in terms of rectangular coordinates is somewhat
complicated:
√
√
D = {(x, y)| − 1 ≤ x ≤ 1, − 1 − x2 ≤ y ≤ 1 − x2 }.
However, D is easily described using polar coordinates
D = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
The double integral can be easily evaluated using polar coordinates.
Theorem: (Double Integration in Polar Coordinates)
If f is continuous on the polar region D = {(r, θ)|α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)}, then
ZZ
Z β Z h2 (θ)
f (x, y)dA =
f (r cos θ, r sin θ)rdrdθ.
D
α
h1 (θ)
Note: Be careful not to forget the additional factor r in the integrand! The additional factor
comes from the fact that the area of an infinitesimal polar rectangle is dA = rdrdθ.
ZZ
xydA, where D is the portion of the annular
Example: Evaluate the double integral
D
region 4 ≤ x2 + y 2 ≤ 25 that lies in the first quadrant.
The region D can be described as
D = {(r, θ)|2 ≤ r ≤ 5, 0 ≤ θ ≤ π/2}.
Thus,
ZZ
π/2
Z
Z
5
xydA =
D
(r cos θ)(r sin θ)rdrdθ
0
2
π/2
Z
Z
=
0
Z
2
π/2
5
(r3 sin θ cos θ)drdθ
5
1
r4 sin θ cos θ2 dθ
4 0
Z
609 π/2
=
sin θ cos θdθ
4 0
π/2
609
=
sin2 θ0
8
609
=
.
8
=
Example: Find the volume of the solid bounded by the paraboloid z = 16 − 3x2 − 3y 2 and
the plane z = 4.
The intersection of these surfaces is given by
4 = 16 − 3x2 − 3y 2
3x2 + 3y 2 = 12
x2 + y 2 = 4.
The projection of this solid onto the xy-plane is the disk x2 + y 2 ≤ 4. In polar coordinates,
D = {(r, θ)|0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
The volume of this solid is
ZZ
V
=
=
=
=
=
(16 − 3x2 − 3y 2 − 4)dA
D
Z 2π
Z 2
(12 − 3r2 )rdrdθ
0
Z 20
2π
(12r − 3r3 )dr
0
2
3 4 2
2π 6r − r 4
0
24π.
Theorem: (Area in Polar Coordinates)
The area of the polar region D bounded by r = f (θ) for α ≤ θ ≤ β is
Z β Z f (θ)
ZZ
Z
1 β
rdrdθ =
1dA =
[f (θ)]2 dθ.
A=
2
α
0
D
α
Example: Find the area enclosed by one loop of the four-leaved rose r = cos(2θ).
A =
=
=
=
Z
1 1 2π
2
cos (2θ)
4 2 0
Z
1 2π 1 + cos(4θ)
dθ
8 0
2
2π
1
1
θ + sin 4θ 16
4
0
π
.
8