08 Double Integrals in Polar Coordinates.pdf

Double Integrals in Polar Coordinates

Lucky Galvez

Institute of Mathematics University of the Philippines

Recall

If

f

is continuous on a Type I region

D

such that

D

=

{

(x, y) :

a

≤

x

≤

b, g

(x)

≤

y

≤

g

(x)

}

,

then

¨

D

f

(

x, y

)

dA

=

ˆ

a

ˆ

g1(x)

f

(

x, y

)

dy dx

If

f

is continuous on a Type II region

D

such that

D

=

{

(x, y) :

h

(y)

≤

x

≤

h

(y), c

≤

y

≤

d

}

,

then

¨

D

f

(

x, y

)

dA

=

ˆ

c

ˆ

h1(y)

f

(

x, y

)

dx dy

Double Integrals

Suppose we want to evaluate

¨

R

p

x

+

y

dA

where

R

is the

region given below:

1 2 3 1

2

3 _y=p 9−x2

y=p1−x2

R=R1∪R2 where

R1 =

n

(x, y) : 0≤x≤1, p1−x2_≤y≤p_9−x2o

R2 =

n

(x, y) : 1≤x≤3, 0≤y≤p9−x2o

Then

R p

x2_+y2 _{dA =} ˆ 1

0 ˆ √

9−x2

√ 1−x2

p

x2_+y2_{dy dx}

+ ˆ 3

1 ˆ √

9−x2

0

p

Double Integrals

Suppose we want to evaluate

¨

R

p

x

+

y

dA

where

R

is the

region given below:

1 2 3 1

2

3 _y=p 9−x2

y=p1−x2

R=R1∪R2 where

R1 =

n

(x, y) : 0≤x≤1, p1−x2_≤y≤p_9−x2o

R2 =

n

(x, y) : 1≤x≤3, 0≤y≤p9−x2o

Then

R p

x2_+y2 _{dA =} ˆ 1

0 ˆ √

9−x2

√ 1−x2

p

x2_+y2_{dy dx}

+ ˆ 3

1 ˆ √

9−x2

0

p

x2_+y2_{dy dx}

Double Integrals

Suppose we want to evaluate

¨

R

p

x

+

y

dA

where

R

is the

region given below:

1 2 3 1

2

3 _y=p 9−x2

y=p1−x2

R=R1∪R2 where

R1 =

n

(x, y) : 0≤x≤1, p1−x2_≤y≤p_9−x2o

R2 =

n

(x, y) : 1≤x≤3, 0≤y≤p9−x2o

Then

R p

x2_+y2 _{dA =} ˆ 1

0 ˆ √

9−x2

√ 1−x2

p

x2_+y2_{dy dx}

+ ˆ 3

1 ˆ √

9−x2

0

p

Double Integrals

Suppose we want to evaluate

¨

R

p

x

+

y

dA

where

R

is the

region given below:

1 2 3 1

2

3 _y=p 9−x2

y=p1−x2

R=R1∪R2 where

R1 =

n

(x, y) : 0≤x≤1, p1−x2_≤y≤p_9−x2o

R2 =

n

(x, y) : 1≤x≤3, 0≤y≤p9−x2o

Then

R p

x2_+y2 _{dA =} ˆ 1

0 ˆ √

9−x2

√ 1−x2

p

x2_+y2 _{dy dx}

+ ˆ 3

1 ˆ √

9−x2

0

p

x2_+y2_{dy dx}

Polar Coordinates

The polar coordinates (r, θ) are related to the rectangular

coordinates (x, y) by

(x, y)

θ r

r

=

x

+

y

x

=

r

cos

θ

y

=

r

sin

θ

Remark: In polar coordinates,

R the equation of a line through the pole isθ=k for some 0≤k <2π

R the equation of a circle centered at the pole of radiuskisr=k R the equation of a circle centered at (a, b) passing through the

Polar Coordinates

The polar coordinates (r, θ) are related to the rectangular

coordinates (x, y) by

(x, y)

θ r

r

=

x

+

y

x

=

r

cos

θ

y

=

r

sin

θ

Remark: In polar coordinates,

R the equation of a line through the pole isθ=k for some 0≤k <2π

R the equation of a circle centered at the pole of radiuskisr=k R the equation of a circle centered at (a, b) passing through the

origin is r= 2acosθ+ 2bsinθ

Polar Coordinates

The polar coordinates (r, θ) are related to the rectangular

coordinates (x, y) by

(x, y)

θ r

r

=

x

+

y

x

=

r

cos

θ

y

=

r

sin

θ

Remark: In polar coordinates,

R the equation of a line through the pole isθ=k for some 0≤k <2π

R the equation of a circle centered at the pole of radiuskisr=k

Polar Coordinates

The polar coordinates (r, θ) are related to the rectangular

coordinates (x, y) by

(x, y)

θ r

r

=

x

+

y

x

=

r

cos

θ

y

=

r

sin

θ

Remark: In polar coordinates,

R the equation of a line through the pole isθ=k for some 0≤k <2π

R the equation of a circle centered at the pole of radiuskisr=k R the equation of a circle centered at (a, b) passing through the

origin is r= 2acosθ+ 2bsinθ

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r

and [α, β] into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

f(rcosθ, rsinθ)r dr dθ

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ.

This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

f(rcosθ, rsinθ)r dr dθ

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

f(rcosθ, rsinθ)r dr dθ

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

Double Integral in Polar Coordinates

Define apolar rectangleas R={(r, θ) :a≤r≤b, α≤θ≤β}.

Subdivide [a, b] into m subintervals [ri−1, ri] of equal width ∆r and [α, β]

into n subintervals [θj−1, θj] of equal

width ∆θ. This divides R into polar subrectanglesRij.

Choose a point (r∗_i, θ∗_j) ∈ R and ap-proximate the area ofRij by

∆Aij =ri∗∆θ∆r

Therefore, we have_¨

R

f(x, y)dA = lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)∆Aij

= lim

m,n→∞

m

X

i=1 n

X

j=1

f(r∗icosθ

∗

j, r

∗

isinθ

∗

j)r

∗

i∆r∆θ

= ˆ β

α

ˆ b

a

f(rcosθ, rsinθ)r dr dθ

Double Integral in Polar Coordinates

Change to Polar Coordinates in a Double Integral

If

f

is continuous on a polar rectangle

R

=

{

(r, θ) :

a

≤

r

≤

b, α

≤

θ

≤

β

}

, where 0

≤

α

−

β

≤

2π,

then

¨

R

f

(

x, y

)

dA

=

ˆ

α

ˆ

a

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 y=p9−x2

y=p1−x2

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

=13π 3

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 y=p9−x2

y=p1−x2

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

y=p1−x2

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

=13π 3

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

=13π 3

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ

= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

=13π 3

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ

= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

=13π 3

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26θ π

2

Double Integral in Polar Coordinates

Example

Evaluate

¨

R

p

x

+

y

dA

where

R

is the region given below.

Solution.

1 2 3 1

2

3 r= 3

r= 1

In polar coordinates,

R=n(r, θ) : 1≤r≤3, 0≤θ≤π

2 o

.

Hence, ¨

R

p

x2_+y2 _{dA =}

ˆ π

2

0

ˆ 3

1

r·r dr dθ= ˆ π

2

0

ˆ 3

1

r2 dr dθ

= ˆ π

2

0

r3

3

r=3

r=1

dθ= ˆ π

2

0

26 3dθ

= 26 3θ

π

2

0

=13π 3

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

=π

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

=π

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

=π

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

(3−1)dθ

= 2θ π

2

0

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

0

(3−1)dθ= 2θ π

2

0

=π

Double Integral in Polar Coordinates

Example

Evaluate ˆ 2√2

0 ˆ √

8−x2

0

1 p

x2_+y2_{+ 1}dy dx.

Solution:

R=n(x, y) : 0≤x≤2√2,0≤y≤p8−x2o

1 2 3 1

2 3

0

In polar coordinates,

R=n(r, θ) : 0≤r≤2√2,0≤θ≤ π

2 o

.

ˆ 2√2

0

ˆ √8−x2

0

1 p

x2_+y2_{+ 1}dy dx =

ˆ π

2

0

ˆ 2√2

0

1

√

r2_{+ 1} r dr dθ

= ˆ π

2

0

p r2_{+ 1}

r=2√2

r=0

dθ

= ˆ π

2

(3−1)dθ= 2θ π

2

Double Integral in Polar Coordinates

Double Integral over General Polar Region

If

f

is continuous on a polar region of the form

D

=

{

(r, θ) :

α

≤

θ

≤

β, h

(θ)

≤

r

≤

h

(θ)

}

then

¨

D

f

(

x, y

)

dA

=

ˆ

α

ˆ

h1(θ)

f

(

r

cos

θ, r

sin

θ

)

r dr dθ

.

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

R=

(r, θ) : 0≤θ≤π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

R=

(r, θ) : 0≤θ≤π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

= 54

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

R=

(r, θ) : 0≤θ≤π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

R=

(r, θ) : 0≤θ≤π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ

r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

= 54

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

R=

(r, θ) : 0≤θ≤π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

= 54

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

= 54

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

= 54

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

4

π

2

Double Integral over General Polar Region

Example

Evaluate ¨

R

x dA, whereRis the region in the first quadrant between

x2_{+ (y−3)}2_{= 9 andx}2_+y2_{= 36.}

Solution:

(r, θ) : 0≤θ≤ π

2,6 sinθ≤r≤6 . Hence,

1 2 3 4 5 6 1

2 3 4 5 6

r= 6 sinθ r= 6

¨

R

x dA = ˆ π

2

0

ˆ 6

6 sinθ

(rcosθ)r dr dθ

= ˆ π

2

0

ˆ 6

6 sinθ

r2cosθ dr dθ

= ˆ π

2

0

r3

3 cosθ

r=6

r=6 sinθ

dθ

= ˆ π

2

0

72 cosθ−72 sin3θcosθ dθ

= 72 sinθ−18 sin4θ π

2

0

= 54

Exercises

1 _{Evaluate the following double integrals by converting them to polar}

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2+y2)dx dy

b.

ˆ 2

0 ˆ 0

−√4−y2

x2y dx dy

c.

ˆ √ 2 2

0

ˆ √1−y2

y

(x+y)dx dy

d.

ˆ 2

0 ˆ √

2x−x2

0

p

x2_+y2 _{dy dx}

2 _Evaluate

¨

R

y3dA, whereR is the region enclosed by the liney=x,

the circle (x−1)2_+y2_{= 1 and thex-axis.}

3 _{Find the volume of the solid that lies inside the sphere}

Exercises

1 _{Evaluate the following double integrals by converting them to polar}

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2+y2)dx dy

b.

ˆ 2

0 ˆ 0

−√4−y2

x2y dx dy

c.

ˆ √ 2 2

0

ˆ √1−y2

y

(x+y)dx dy

d.

ˆ 2

0 ˆ √

2x−x2

0

p

x2_+y2 _{dy dx}

2 _Evaluate

¨

R

y3dA, whereR is the region enclosed by the liney=x,

the circle (x−1)2_+y2_{= 1 and thex-axis.}

3 _{Find the volume of the solid that lies inside the sphere}

x2_+y2_+z2_{= 16 and outside the cylinderx}2_+y2_{= 4.}

Exercises

1 _{Evaluate the following double integrals by converting them to polar}

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2+y2)dx dy

b.

ˆ 2

0 ˆ 0

−√4−y2

x2y dx dy

c.

ˆ √ 2 2

0

ˆ √1−y2

y

(x+y)dx dy

d.

ˆ 2

0 ˆ √

2x−x2

0

p

x2_+y2 _{dy dx}

2 _Evaluate

¨

R

y3dA, whereR is the region enclosed by the liney=x,

the circle (x−1)2_+y2_{= 1 and thex-axis.}

3 _{Find the volume of the solid that lies inside the sphere}

Exercises

1 _{Evaluate the following double integrals by converting them to polar}

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2+y2)dx dy

b.

ˆ 2

0 ˆ 0

−√4−y2

x2y dx dy

c.

ˆ √ 2 2

0

ˆ √1−y2

y

(x+y)dx dy

d.

ˆ 2

0 ˆ √

2x−x2

0

p

x2_+y2 _{dy dx}

2 _Evaluate

¨

R

y3dA, whereR is the region enclosed by the liney=x,

the circle (x−1)2_+y2_{= 1 and thex-axis.}

3 _{Find the volume of the solid that lies inside the sphere}

x2_+y2_+z2_{= 16 and outside the cylinderx}2_+y2_{= 4.}

Exercises

1 _{Evaluate the following double integrals by converting them to polar}

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2+y2)dx dy

b.

ˆ 2

0 ˆ 0

−√4−y2

x2y dx dy

c.

ˆ √ 2 2

0

ˆ √1−y2

y

(x+y)dx dy

d.

ˆ 2

0 ˆ √

2x−x2

0

p

x2_+y2 _{dy dx}

2 _Evaluate

¨

R

y3dA, whereR is the region enclosed by the liney=x,

the circle (x−1)2_+y2_{= 1 and thex-axis.}

3 _{Find the volume of the solid that lies inside the sphere}

Exercises

1 _{Evaluate the following double integrals by converting them to polar}

coordinates.

a.

ˆ 1

−1

ˆ √1−y2

0

cos(x2+y2)dx dy

b.

ˆ 2

0 ˆ 0

−√4−y2

x2y dx dy

c.

ˆ √ 2 2

0

ˆ √1−y2

y

(x+y)dx dy

d.

ˆ 2

0 ˆ √

2x−x2

0

p

x2_+y2 _{dy dx}

2 _Evaluate

¨

R

y3dA, whereR is the region enclosed by the liney=x,

the circle (x−1)2_+y2_{= 1 and thex-axis.}

3 _{Find the volume of the solid that lies inside the sphere}

x2_+y2_+z2_{= 16 and outside the cylinderx}2_+y2_{= 4.}

References

1 _{Stewart, J., Calculus, Early Transcendentals, 6 ed., Thomson}

Brooks/Cole, 2008

2 _{Leithold, L.,The Calculus 7, Harper Collins College Div., 1995}

3 _{Dawkins, P.,Calculus 3, online notes available at}