Double integrals using polar coordinatesLet’s look at a motivating example:

ExampleEvaluate ZZ

Rex

2+y2 dA

where R is the bounded region in the first quadrant (x � 0, y � 0) between the circles of radius 1 and 2

(1 x2+ y2 4).

Analysis: Sketch R.

Figure: Region R : x � 0, y � 0, 1 x2+ y2 4

Horizontal and vertical slices won’t work, since we cannot do

Zex

2+constant dx or

Zey

2+constant dy .

Slicing radially simplifies the problem.

We use polar coordinates (r , ✓).

Math 267 (University of Calgary) Fall 2015, Winter 2016 1 / 12

Area Element in Polar CoordinatesQuestion: How can we write

Z

Rf (x, y) dA in terms of r and ✓?

We cannot just use dA = drd✓, since the equality is simply not true: dA 6= drd✓.

It turns out that the correct expression for dA is

dA = rdrd✓

.

Let’s see why.

Figure: Area Element in Polar Coordinates dA = r dr d✓

Math 267 (University of Calgary) Fall 2015, Winter 2016 2 / 12

Area element in polar coordinatesSuppose a region R is described in polar coordinates by 0 g(✓) r h(✓),↵ ✓ �.

Chop up R into a mesh of radial lines (✓ = constants) and arcs (r = constants)

Figure: Area Element in Polar Coordinates dA = r dr d✓

In the Cartesian coordinate system, �A = �x�y and the area element is dA = dxdy .

For the small pieces in the mesh of radial lines and arcs, we need to find �A.

Consider the small piece bounded by radial lines ✓ = ✓j�1 and ✓ = ✓j and arcs r = ri�1 and r = ri .

By the arc length formula L = r✓, the outer and inner circular arcs of the piece have lengths ri�✓j andri�1�✓j , where �✓j = ✓j � ✓j�1.

As we chop R into a finer mesh, the lengths of the outer and inner circular arcs are approximately the

same, and the area of the piece is approximately the area of a rectangle with base ri�✓j and height

�ri = ri � ri�1.

So, �Aij = ri�ri�✓j . The area element in polar coordinates is, therefore, dA = r dr d✓.

Math 267 (University of Calgary) Fall 2015, Winter 2016 3 / 12

Double integrals and polar coordinates

Let f be a continuous function on a bounded region R (in the xy -plane) that admits a polar coordinate

description

R = {(x, y) | 0 g(✓) r h(✓), ↵ ✓ �} ,

where 0 < � � ↵ 2⇡. Then

ZZ

Rf (x, y) dA =

Z ✓=�

✓=↵

Z r=h(✓)

r=g(✓)f (r cos ✓, r sin ✓) r dr d✓.

Proof (Outline): We follow the same limiting process in the definition

ZZ

Rf (x, y) dA = lim

max�xi ,�yj!0

X

i

X

j

f (x⇤ij , y

⇤ij )�Aij .

The sample point (x⇤ij , y

⇤ij ) in polar coordinates is (r⇤ij cos ✓⇤

ij , r⇤ij sin ✓⇤

ij ).

The area of the small pieces in the mesh is �Aij = ri �ri �✓j .

To describe the diminishing dimensions of the small pieces, the polar coordinates version of

max�xi ,�yj ! 0 is max�ri ,�✓j ! 0.

Putting all these together in a limiting process:

ZZ

Rf (x, y) dA = lim

max�ri ,�✓j!0

mX

i=1

nX

j=1

f (r⇤j cos ✓⇤ij , r

⇤ij sin ✓⇤

ij ) ri �ri �✓j

=

Z ✓=�

✓=↵

Z r=h(✓)

r=g(✓)f (r cos ✓, r sin ✓) r dr d✓

Math 267 (University of Calgary) Fall 2015, Winter 2016 4 / 12

Setting up iterated integrals in polar coordinatesTo set up the integral, we find the polar description of the region:

Holding ✓ fixed, draw a radial line from the origin through R at an angle of ✓.

The radial line crosses R on its boundary. The inner function is r = g(✓) where the radial slice enters R,

and the outer function is r = h(✓) where the radial slice leaves R.

Imagine rotating the radial line about the origin sweeping out the region R. The least angle is ↵, and the

greatest angle to cover R is �.

Let’s look at our original example:

ExampleEvaluate ZZ

Rex

2+y2 dA

where R is the bounded region in the first quadrant (x � 0, y � 0) between the circles of radius 1 and 2

(1 x2+ y2 4).

Figure: Region R : x � 0, y � 0, 1 x2+ y2 4

Math 267 (University of Calgary) Fall 2015, Winter 2016 5 / 12

Solution:

Sketch R.

Figure: Region R : 1 r 2, 0 ✓ ⇡/2

Slice radially at ✓

✓0 ✓

⇡

2

◆.

The inner boundary curve is r = 1 and the outer boundary curve is r = 2.

Using x2+ y2

= r2,ZZ

Rex

2+y2 dA =

Z ✓=⇡/2

✓=0

Zr=2

r=1

er2

r dr d✓

(using u = r2, du = 2rdr) =

Z ✓=⇡/2

✓=0

er2

2

������

r=2

r=1

d✓ =

Z ⇡/2

0

e4

2�

e1

2

!d✓

=

e4 � e

2

!✓

�����

⇡/2

0

=

e4 � e

2

!⇡

2

Math 267 (University of Calgary) Fall 2015, Winter 2016 6 / 12

Example

Find

ZZ

Rxey dA, where R is the region in the first quadrant bounded inside the circle x2

+ y2= 9.

Solution: Using the polar coordinate system, the region R is 0 r 3, 0 ✓ ⇡

2.

It turns out to be easier to integrate with respect to ✓ first:

ZZ

Rxey dA =

Z r=3

r=0

✓Z ✓=⇡/2

✓=0

(r cos ✓)er sin ✓r d✓

◆dr

(use u = r sin ✓, du = r cos ✓ d✓) =

Z r=3

r=0

⇣r e

r sin ✓⌘���

✓=⇡/2

✓=0

dr

=

Z r=3

r=0

�r e

r � r�dr

(by parts) =

r er � er �

r2

2

!�����

r=3

r=0

=

✓3e3 � e3 �

9

2

◆� (0 � 1 � 0)

= 2e3 �7

2.

Math 267 (University of Calgary) Fall 2015, Winter 2016 7 / 12

Example

Find

ZZ

R(x + y) dA, where R is the region enclosed inside the circle (x � 1)

2+ y2

= 1.

Solution: Recall that the polar equation for this circle is r = 2 cos ✓, �⇡

2 ✓

⇡

2.

Figure: Curve on the right bounds region R : 0 r 2a cos ✓,�⇡

2 ✓

⇡

2

ZZ

Rx dA =

Z ✓=⇡/2

✓=�⇡/2

✓Zr=2 cos ✓

r=0

(r cos ✓) r dr

◆d✓ =

Z ✓=⇡/2

✓=�⇡/2

(cos ✓)

r3

3

!�����

r=2 cos ✓

r=0

d✓

=8

3

Z ⇡/2

�⇡/2cos

4 ✓ d✓ = ... = ⇡.

Math 267 (University of Calgary) Fall 2015, Winter 2016 8 / 12

Further examples/exercises

Evaluate the following double integrals:

1

ZZ

D(x + y) dA, where D is the region in the first quadrant inside the circle x2

+ y2= 25.

2

ZZ

D

px2 + y2 dA, where D is the region bounded between the circles x2

+ y2= a2 and x2

+ y2= b2

(0 < a < b).

3

ZZ

D

✓tan

�1

✓y

x

◆◆2

dA, where D is the region�(x, y) : 9 x2

+ y2 16, |y | x .

4

ZZ

Dx2y dA, where D is the region bounded inside the curve r = 2 sin ✓.

5

ZZ

Ry dA, where R is the overlap between the disks x2

+ y2 1 and x2+ (y � 1)

2 1.

6

ZZ

Rx dA, where R is the overlap between the disks x2

+ y2 1 and x2+ (y � 1)

2 1.

7 Find the volume of the solid bounded by the paraboloid z = 4 � x2 � y2and the xy -plane.

8 Find the volume of the solid bounded by the cylinder (x � 1)2+ y2

= 1, the plane z = 0 and the cone

z =

px2 + y2.

Math 267 (University of Calgary) Fall 2015, Winter 2016 9 / 12

Extra ExamplesRecall: The area of a bounded region D is equal to

RRD 1 dA.

Proof: The double integral computes the net volume under the constant function f (x, y) = 1 over D, which is

equal to the area of D times the height 1.

Example

Find the area of R =�(x, y)

�� 1 x2+ y2 16

Solution:

Sketch R.

Figure: Region R : 1 r 4, 0 ✓ 2⇡

Slice radially at ✓ (0 ✓ 2⇡). Each slice enters R at r = 1 and leaves at r = 4.

ZZ

R1 dA =

Z ✓=2⇡

✓=0

Zr=4

r=1

1 rdr d✓ =

Z ✓=2⇡

✓=0

r2

2

�����

r=4

r=1

d✓ =

Z2⇡

0

✓15

2

◆d✓ = 15⇡

Math 267 (University of Calgary) Fall 2015, Winter 2016 10 / 12

A Special Case

If r ranges from 0 to f (✓) and ✓ from ↵ to �, then the area of the region is:

A =

Z ✓=�

✓=↵

Zr=f(✓)

r=0

1 rdr d✓ =

Z ✓=�

✓=↵

r2

2

�����

r=f(✓)

r=0

d✓ =1

2

Z ✓=�

✓=↵(f (✓))2 d✓

We have seen this formula previously, but now you know more about where it comes from.

We have also done an example very similar to the following. Can you solve it without looking at the

solution?

Example

Find the area of one petal of the rose r = sin(3✓).

Solution:

Sketch the rose. [Try to do it yourself! Check your sketch by using Wolfram Alpha.]

Slice R radially at ✓. Each slice ranges from r = 0 to r = sin(3✓).

As ✓ increases we travel around the petal.

We want the value of ✓ where we end up back at the origin (i.e., the range of ✓ for the first loop).

For what ✓ is r = 0?

r = sin(3✓) = 0 () 3✓ = n⇡ () ✓ =n⇡

3, n = 0,±1,±2, ...

For ✓ from 0 to⇡

3we travel around the first loop.

Math 267 (University of Calgary) Fall 2015, Winter 2016 11 / 12

Using the area formula for the special case:

A =1

2

Z ✓=⇡/3

✓=0

(sin(3✓))2 d✓

=1

2

Z ⇡/3

0

1 � cos(6✓)

2d✓

=1

4

Z ⇡/3

0

(1 � cos(6✓)) d✓

=1

4

✓✓ �

sin(6✓)

6

◆����⇡/3

0

=⇡

12

Math 267 (University of Calgary) Fall 2015, Winter 2016 12 / 12