SECTION 12.3

DOUBLE INTEGRALS IN POLAR COORDINATES

P212.3

DOUBLE INTEGRALS IN POLAR COORDINATES

Suppose that we want to evaluate a double

integral , where R is one of the

regions shown in Figure 1.

( , )R

f x y dA

P312.3

DOUBLE INTEGRALS IN POLAR COORDINATES

In either case, the description of R in terms of rectangular coordinates is rather complicated but R is easily described by polar coordinates.

P412.3

DOUBLE INTEGRALS IN POLAR COORDINATES

Recall from Figure 2 that the polar coordinates (r, ) of a point are related to the rectangular coordinates (x, y) by the equations

r2 = x2 + y2

x = r cos y = r sin

P512.3

POLAR RECTANGLE

The regions in the first figure are special cases of a polar rectangle

R = {(r, ) | a ≤ r ≤ b, ≤ ≤ }

shown in Figure 3.

P612.3

POLAR RECTANGLE

To compute the double integral

where R is a polar rectangle, we divide: The interval [a, b] into m subintervals [ri–1, ri] of

equal width ∆r = ri – ri–1.

The interval [ , ] into n subintervals [j–1, i] of equal width ∆ = i – i–1.

( , )R

f x y dA

P712.3

POLAR RECTANGLES

Then, the circles r = ri and the rays = i divide the polar rectangle R into the small polar rectangles shown in Figure 4.

P812.3

POLAR SUBRECTANGLE

The “center” of the polar subrectangle

Rij = {(r, ) | ri–1 ≤ r ≤ ri, j–1 ≤ ≤ j}

has polar coordinates

ri* = ½ (ri–1 + ri)

j * = ½ (j–1 + j)

P912.3

POLAR SUBRECTANGLE

We compute the area of Rij using the fact that the area of a sector of a circle with radius r and central angle is ½r2.

P1012.3

POLAR SUBRECTANGLE

Subtracting the areas of two such sectors, each of which has central angle ∆j = j – j–1, we find that the area of Rij is:

2 21 112 2

2 2112

11 12

*

( )

( )( )

ij i j i j

i i j

i i i i j

i i j

A r r

r r

r r r r

r r

P1112.3

POLAR RECTANGLES

We have defined the double integral

in terms of ordinary rectangles.However, it can be shown that, for continuous

functions f, we always obtain the same answer using polar rectangles.

( , )R

f x y dA

P1212.3

POLAR RECTANGLES

The rectangular coordinates of the center of Rij are (ri* cos j*, ri* sin j*).

So, a typical Riemann sum is:

* * * *

1 1

* * * * *

1 1

( cos , sin )

( cos , sin )

m n

i j i j iji j

m n

i j i j i i ji j

f r r A

f r r r r

P1312.3

POLAR RECTANGLES

If we write g(r, ) = rf(r cos , r sin ), the Riemann sum in Equation 1 can be written as:

This is a Riemann sum for the double integral

* *

1 1

( , )m n

i j i ji j

g r r

( , )b

ag r dr d

P1412.3

POLAR RECTANGLES

Thus, we have:

* * * *

max , 01 1

* *

max , 01 1

( , ) lim ( cos , sin )

lim ( , )

( , )

( cos , sin )

i j

i j

m n

i j i j ijri jR

m n

i j i jri j

b

a

b

a

f x y dA f r r A

g r r

g r dr d

f r r r dr d

P1512.3

CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL

( , ) ( cos , sin )b

aR

f x y dA f r r r dr d

If f is continuous on a polar rectangle R given by

0 ≤ a ≤ r ≤ b, ≤ ≤where 0 ≤ – ≤ 2, then

P1612.3

CHANGE TO POLAR COORDS.

Formula 2 says that we convert from rectangular to polar coordinates in a double integral by: Writing x = r cos and y = r sin Using the appropriate limits of integration for r

and Replacing dA by r dr d

P1712.3

CHANGE TO POLAR COORDS.

Be careful not to forget the additional factor r on the right side of Formula 2.

P1812.3

CHANGE TO POLAR COORDS.

A classical method for remembering the formula is shown in Figure 5. The “infinitesimal” polar rectangle can be thought of

as an ordinary rectangle with dimensions r d and dr.

So, it has “area” dA = r dr d.

P1912.3

Example 1

Evaluate

where R is the region in the upper half-plane bounded by the circles x2 + y2 = 1 and x2 + y2 = 4.

2(3 4 )R

x y dA

P2012.3

Example 1 SOLUTION

The region R can be described as:

R = {(x, y) | y ≥ 0, 1 ≤ x2 + y2 ≤ 4}It is the half-ring shown in Figure 1(b).In polar coordinates, it is given by:

1 ≤ r ≤ 2, 0 ≤ ≤

P2112.3

Example 1 SOLUTION

Hence, by Formula 2,

2

2 2 2

0 1

2 2 3 2

0 1

3 4 2 210

(3 4 )

(3 cos 4 sin )

(3 cos 4 sin )

[ cos sin ]

R

rr

x y dA

r r r dr d

r r dr d

r r d

P2212.3

Example 1 SOLUTION

2

0

1520

0

(7cos 15sin ]

[7cos (1 cos 2 )]

15 157sin sin 2

2 4

15

2

d

d

P2312.3

Example 2

Find the volume of the solid bounded by: The plane z = 0 The paraboloid z = 1 – x2 – y2

P2412.3

Example 2 SOLUTION

If we put z = 0 in the equation of the paraboloid, we get x2 + y2 = 1. This means that the plane intersects the paraboloid

in the circle x2 + y2 = 1.

P2512.3

Example 2 SOLUTION

So, the solid lies under the paraboloid and above the circular disk D given by x2 + y2 ≤ 1.

P2612.3

Example 2 SOLUTION

In polar coordinates, D is given by 0 ≤ r ≤ 1, 0 ≤ ≤ 2. As 1 – x2 – y2 = 1 – r2, the volume is:

2 12 2 2

0 0

2 1 3

0 0

12 4

0

(1 ) (1 )

( )

22 4 2

D

V x y dA r r dr d

d r r dr

r r

P2712.3

Example 2 SOLUTION

Had we used rectangular coordinates instead, we would have obtained:

This is not easy to evaluate because it involves finding ∫(1 – x2)3/2 dx

2

2

2 2

1 1 2 2

1 1

(1 )

(1 )

D

x

x

V x y dA

x y dy dx

P2812.3

CHANGE TO POLAR COORDS.

What we have done so far can be extended to the more complicated type of region shown in Figure 7. It’s similar to the type II

rectangular regions considered in Section 12.2.

P2912.3

CHANGE TO POLAR COORDS.

In fact, by combining Formula 2 in this section with Formula 5 in Section 12.2, we obtain the following formula.

P3012.3

Formula 3

2

1

( )

( )( , ) ( cos , sin )

h

hD

f x y dA f r r r dr d

If f is continuous on a polar region of the form

D = {(r, ) | ≤ ≤ , h1( ) ≤ r ≤ h2()} then

P3112.3

CHANGE TO POLAR COORDS.

In particular, taking f(x, y) = 1, h1() = 0, and h2() = h() in the formula, we see that the area of the region D bounded by = , = , and r = h() is:

This agrees with Formula 3 in Section 9.4

( )

0

( )221

2

0

( ) 1

[ ( )]2

h

D

h

A D dA r dr d

rd h d

P3212.3

Example 3

Find the volume of the solid that lies: Under the paraboloid z = x2 + y2

Above the xy-plane Inside the cylinder x2 + y2 = 2x

P3312.3

Example 3 SOLUTION

The solid lies above the disk D whose boundary circle has equation x2 + y2 = 2x. After completing the square, that is: (x – 1)2 + y2 =

1. See Figure 8 and 9.

P3412.3

Example 3 SOLUTION

In polar coordinates, we have:

x2 + y2 = r2 and x = r cos So, the boundary circle becomes:

r2 = 2r cos or r = 2 cos

Thus, the disk D is given by:

D = {(r, ) | –/2 ≤ ≤ /2 , 0 ≤ r ≤ 2 cos }

P3512.3

Example 3 SOLUTION

So, by Formula 3, we have:

2 2

/ 2 2cos 2

/ 2 0

2cos4/ 2

/ 20

/ 2 4

/ 2

( )

4

4 cos

D

V x y dA

r r dr d

rd

d

P3612.3

Example 3 SOLUTION

/ 2 4

0

2/ 2

0

/ 2120

/ 23 102 8

8 cos

1 cos 28

2

2 [1 2cos 2 (1 cos 4 )]

2[ sin 2 sin 4 ]

3 32

2 2 2

d

d

d