MULTIPLE INTEGRALS 16. 2 MULTIPLE INTEGRALS 16.4 Double Integrals in Polar Coordinates In this...

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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


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


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

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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

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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

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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


Be careful not to forget

the additional factor r on

the right side of Formula 2.

19


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


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


The region R can be described as:

R = {(x, y) | y ≥ 0, 1 ≤ x2 + y2 ≤ 4}

Example 1

22


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


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


2

0

1520

0

(7cos 15sin ]

[7cos (1 cos 2 )]

15 157sin sin 2

2 4

15

2

d

d

Example 1

25


Find the volume of the solid bounded

by: The plane z = 0

The paraboloid z = 1 – x2 – y2

Example 2

26


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


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


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


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


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


In fact, by combining Formula 2 in this

section with Formula 5 in Section 16.3,

we obtain the following formula.

32


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


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


Use a double integral to find the area

enclosed by one loop of the four-leaved

rose r = cos 2θ.

Example 3

35


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


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


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


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


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


Thus, the disk D is given by:

D =

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

Example 4

41


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

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