15.4 Double Integrals in Polar Coordinates

15.4 Double Integrals in Polar Coordinates

I- Change to Polar Coordinates in a Double Integral

If

f

is continuous on a polar rectangle

R

given by

a ≤ ≤ r b , α θ β ≤ ≤

^{, where}

0 ≤ − ≤ β α 2 π

then

( )

^b

( )

R

a

f x , y dA f r cos , r sin rdrd

β

α

θ θ θ

∫∫ = ∫∫

Notice that

The polar coordinates

( r , θ )

of a point are related to the rectangular coordinates

( x , y )

by the equations

2 2

x = r cos , θ y = r sin , θ r = x

²

+ y

Ex: Evaluate

∫∫

_R

( 3 x + 4 y

²

) dA

, where is the region in the upper half-plane bounded by the circles

2

1

x

²

+ y =

and

x

²

+ y

²

= 4

. Solution

( )

{ 0 1

^{2 2}

4 }

R = x , y ≤ y , ≤ x + y ≤

It is the half-ring shown in Figure.

In polar coordinates

( )

{ 0 1 2 }

R = r , θ ≤ ≤ θ π , ≤ ≤ r

( ) ( )

( )

( )

2

2 2 2

0 1 2 2

2 3 2 3 4 2

0 1 0 1

2

0 0

0

3 4 3 4

3 4

7 15 7 15 1 2

2

15 15 15

7 2

2 4 2

R

r

r

x y dA r cos r sin rdrd

r cos r sin drd r cos r sin d

cos sin d cos cos d

sin sin

π

π π

π π

π

θ θ θ

θ θ θ θ θ θ

θ θ θ θ θ θ

θ θ θ π

=

=

+ = +

 

= + =  + 

 

 

=  +  =   + −  

 

=   + −   =

∫∫ ∫ ∫

∫ ∫ ∫

∫ ∫

Ex: Find the volume of the solid bounded by the plane

z = 0

and the paraboloid

z = − 1 x

²

− y

². Solution

The plane

z = 0

intersects the paraboloid in the circle

x

²

+ y

²

= 1

^,

so the solid lies under the paraboloid and above the circular disk

D

given by

2 2

1

x + y ≤

. The volume given by

( )

V = ∫∫

D

z x , y dA

In polar coordinates is given by

R = { ( r , θ ) 0 ≤ ≤ r 1 0 , ≤ ≤ θ 2 π }

^,

z = − 1 x

²

− y

²

= − 1 r

²

dA = rdrd θ

, the volume is given by

( ) ( )

( ) ( )

2 1 1 2

2 3

0 0 0 0

2 4 1

1 1 1 1

2 4 0 2 4 2

1

2 2 0

r r

r r

r r

V r rdr d r r dr d

r r

θ π θ π

θ θ

π

θ θ

π π

= = = =

= = = =

=

=

 

=   −   = −

= − =   − −   =

∫ ∫ ∫ ∫

Notice that:

If we had used rectangular coordinates instead of polar coordinates, then we would have obtained

( )

2

2

1 1

2 2

1 1

1

x y x

x y x

V

^{= = −}

x y dy dx

=− =− −

 

= ∫   ∫ − −  

, which is not easy to evaluate.

Change to Polar Coordinates in a Double Integral (general region) If

f

is continuous on a polar region of the form

( ) ( ) ( )

{

^{1 2}

}

D = r , θ α θ β ≤ ≤ , h θ ≤ ≤ r h θ

^then

( )

²_{( )}^{( )}

( )

1

r h

D

f x , y dA

^{θ β} r h ^θ

f r cos , r sin rdr d

θ α₌⁼



₌⁼ θ

θ θ  θ

=    

∫∫ ∫ ∫

Rule: The area of the region

D

bounded by

θ α =

^,

θ β =

^,

r = h

1

( ) θ

^and

r = h

2

( ) θ

^{is given}

by

( )

_{( )}^{( )}

( )

( )

( ) ( )

2 2

1

1

2 2 2

2 1

1

2 2

h h

D h

h

A D dA rdrd r d h h d

β θ β θ β

α θ α α

θ

θ   θ  θ θ  θ

= ∫∫ = ∫ ∫ = ∫     = ∫  − 

Ex: Use a double integral to find the area enclosed by one loop of the four leaved rose

r = cos ( ) 2 θ

^.

Solution: we see that a loop is given by the region

( ) 0 2

4 4

D =   r , θ − ≤ ≤ π θ π , ≤ ≤ r cos θ  

 

^.

( )

( )

2 2

4 2 4

4 0 4 0

2 4

4 4

0 0

0

2

1 1 1

2 1 4 4

2 2 4 8

cos cos

D

A D dA rdrd r d

cos d cos d sin

π θ π θ

π π

π π π

θ θ

θ θ θ θ θ θ π

− −

 

= = =  

 

 

= = + =   +   =

∫∫ ∫ ∫ ∫

∫ ∫

Ex: Find the volume of the solid that lies under the paraboloid z = x

²

+ y

²

,above the xy -plane, and inside the cylinder x

²

+ y

²

= 2 x .

Solution: The volume is given by V = ∫∫

_D

z x , y dA ( ) = ∫∫

_D

( x 2 + y 2) dA ,

where D is the disk whose boundary circle has equation x

²

+ y

²

= 2 x ⇒ ( x − 1 )

²

+ y

²

= 1 .

In polar coordinates we have ( ) 0 2

2 2

D =   r , θ − ≤ ≤ π θ π , ≤ ≤ r cos θ  

 

( )

( )

2 2

2 2 2 2 2 3

0 0

2 2

2 2

4

2 2 4 2

0 0

2 0

2 2 2

0 0

1 2

8 8

4 2

1 4

2 1 2 2 2 2 1 2 2

2

3 1

2 2 2

2 2

cos cos

D

cos

V x y dA r rdr d r dr d

r cos

d cos d d

cos cos d cos cos d

cos c

π θ π θ

π π

π θ π π

π

π π

θ θ

θ θ θ θ θ

θ θ θ θ θ θ

θ

− −

−

   

= + =     =    

   + 

=     = =    

  +  

= + + =   + +      

= + +

∫∫ ∫ ∫ ∫ ∫

∫ ∫ ∫

∫ ∫

2 2

0 0

3 1 3

4 2 2 4

2 8 2

os d sin sin

π

  θ θ   =   θ + θ + θ  

π

= π

   

∫