MULTIPLE INTEGRALSMULTIPLE INTEGRALS

16

2

MULTIPLE INTEGRALS

16.4Double Integrals

in Polar Coordinates

In this section, we will learn:

How to express double integrals

in polar coordinates.

3

DOUBLE INTEGRALS IN POLAR COORDINATES

Suppose that we want to evaluate a double

integral , where R is one of

the regions shown here.

( , )R

f x y dA

Fig. 16.4.1, p. 1010

4

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.

Fig. 16.4.1, p. 1010

5

DOUBLE INTEGRALS IN POLAR COORDINATES

Recall from this figure that the polar

coordinates (r, θ) of a point are related

to the rectangular coordinates (x, y) by

the equations

r2 = x2 + y2

x = r cos θ

y = r sin θ

Fig. 16.4.2, p. 1010

6

POLAR RECTANGLE

The regions in the first figure are special

cases of a polar rectangle

R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β}

shown here.

Fig. 16.4.3, p. 1010

7

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 = (b – a)/m.

The interval [α ,β] into n subintervals [θj–1, θi] of equal width ∆θ = (β – α)/n.

( , )R

f x y dA

8

POLAR RECTANGLES

Then, the circles r = ri and the rays θ = θi

divide the polar rectangle R into the small

polar rectangles shown here.

Fig. 16.4.4, p. 1010

9

POLAR SUBRECTANGLE

The “center” of the polar subrectangle

Rij = {(r, θ) | ri–1 ≤ r ≤ ri, θj–1 ≤ θ ≤ θi}

has polar coordinates

ri* = ½ (ri–1 + ri)

θj* = ½ (θj–1 + θj)

Fig. 16.4.4, p. 1010

10

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 θ is ½r2θ.

11

POLAR SUBRECTANGLE

Subtracting the areas of two such sectors,

each of which has central angle ∆θ = θj – θj–1,

we find that the area of Rij is:

2 21 112 2

2 2112

11 12

*

( )

( )( )

i i i

i i

i i i i

i

A r r

r r

r r r r

r r

12

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

13

POLAR RECTANGLES

The rectangular coordinates of the center

of Rij are (ri* cos θj*, ri* sin θj*).

So, a typical Riemann sum is: * * * *

1 1

* * * * *

1 1

( cos , sin )

( cos , sin )

m n

i j i j ii j

m n

i j i j ii j

f r r A

f r r r r

Equation 1

14

POLAR RECTANGLES

If we write g(r, θ) = r f(r cos θ, r sin θ),

the Riemann sum in Equation 1 can be

written as:

This is a Riemann sum for the double integral

* *

1 1

( , )m n

i ji j

g r r

( , )b

ag r dr d

15

POLAR RECTANGLES

Thus, we have:

* * * *

,1 1

* *

,1 1

( , ) lim ( cos , sin )

lim ( , )

( , )

( cos , sin )

m n

i j i j im ni jR

m n

i jm n

i 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

16

CHANGE TO POLAR COORDS.

If f is continuous on a polar rectangle R

given by

0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β

where 0 ≤ β – α ≤ 2π, then

( , ) ( cos , sin )b

aR

f x y dA f r r r dr d

Formula 2

17

CHANGE TO POLAR COORDS.

Formula 2 says that we convert

from rectangular to polar coordinates

in a double integral by:

Writing x = r cos θ and y = r sin θ Using the appropriate limits of integration

for r and θ Replacing dA by dr dθ

18

CHANGE TO POLAR COORDS.

Be careful not to forget

the additional factor r on

the right side of Formula 2.

19

CHANGE TO POLAR COORDS.

A classical method for remembering

the formula is shown here.

The “infinitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions r dθ and dr.

So, it has “area” dA = r dr dθ.

Fig. 16.4.5, p. 1012

20

CHANGE TO POLAR COORDS.

Evaluate

where R is the region in

the upper half-plane bounded by

the circles x2 + y2 = 1 and x2 + y2 = 4.

Example 1

2(3 4 )R

x y dA

21

CHANGE TO POLAR COORDS.

The region R can be described as:

R = {(x, y) | y ≥ 0, 1 ≤ x2 + y2 ≤ 4}

Example 1

22

CHANGE TO POLAR COORDS.

It is the half-ring shown here.

In polar coordinates,

it is given by:

1 ≤ r ≤ 2, 0 ≤ θ ≤ π

Example 1

Fig. 16.4.1b, p. 1010

23

CHANGE TO POLAR COORDS.

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

Example 1

24

CHANGE TO POLAR COORDS.

2

0

1520

0

(7cos 15sin ]

[7cos (1 cos 2 )]

15 157sin sin 2

2 4

15

2

d

d

Example 1

25

CHANGE TO POLAR COORDS.

Find the volume of the solid bounded

by: The plane z = 0

The paraboloid z = 1 – x2 – y2

Example 2

26

CHANGE TO POLAR COORDS.

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.

Example 2

27

CHANGE TO POLAR COORDS.

So, the solid lies under the paraboloid

and above the circular disk D given by

x2 + y2 ≤ 1.

Example 2

Fig. 16.4.1a, p. 1010 Fig. 16.4.6, p. 1012

28

CHANGE TO POLAR COORDS.

In polar coordinates, D is given by

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.

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

Example 2

29

CHANGE TO POLAR COORDS.

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

Example 2

30

CHANGE TO POLAR COORDS.

What we have done so far can be extended

to the more complicated type of region

shown here.

It’s similar to the type II rectangular regions considered in Section 15.3

Fig. 16.4.7, p. 1013

31

CHANGE TO POLAR COORDS.

In fact, by combining Formula 2 in this

section with Formula 5 in Section 16.3,

we obtain the following formula.

32

CHANGE TO POLAR COORDS.

If f is continuous on a polar region

of the form

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

then

2

1

( )

( )( , ) ( cos , sin )

h

hD

f x y dA f r r r dr d

Formula 3

33

CHANGE TO POLAR COORDS.

In particular, taking f(x, y) = 1, h1(θ) = 0,

and h2(θ) = h(θ) in the formula, we see that

the area of the region D bounded by θ = α,

θ = β, and r = h(θ) is:

This agrees with Formula 3 in Section 10.4

( )2( )

00

212

( ) 12

[ ( )]

hh

D

rA D dA r dr d d

h d

34

CHANGE TO POLAR COORDS.

Use a double integral to find the area

enclosed by one loop of the four-leaved

rose r = cos 2θ.

Example 3

35

CHANGE TO POLAR COORDS.

From this sketch of the curve, we see that

a loop is given by the region

D = {(r, θ) | –π/4 ≤ θ ≤ π/4, 0 ≤ r ≤ cos 2θ}

Example 3

Fig. 16.4.8, p. 1013

36

CHANGE TO POLAR COORDS.

So, the area is:

/ 4 cos2

/ 4 0

/ 4 2 cos2102/ 4

/ 4 212 / 4

/ 414 / 4

/ 41 14 4 / 4

( )

[ ]

cos 2

(1 cos 4 )

sin 48

D

A D dA r dr d

r d

d

d

Example 3

37

CHANGE TO POLAR COORDS.

Find the volume of the solid that

lies: Under the paraboloid z = x2 + y2

Above the xy-plane

Inside the cylinder x2 + y2 = 2x

Example 4

38

CHANGE TO POLAR COORDS.

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

Example 4

Fig. 16.4.9, p. 1014 Fig. 16.4.10, p. 1014

39

CHANGE TO POLAR COORDS.

In polar coordinates, we have:

x2 + y2 = r2 and x = r cos θ

So, the boundary circle becomes:

r2 = 2r cos θ

or r = 2 cos θ

Example 4

40

CHANGE TO POLAR COORDS.

Thus, the disk D is given by:

D =

{(r, θ) | –π/2 ≤ θ ≤ π/2 , 0 ≤ r ≤ 2 cos θ}

Example 4

41

CHANGE TO POLAR COORDS.

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

Example 4

42

CHANGE TO POLAR COORDS./ 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 ]

3 322 2 2

d

d

d

Example 4