II.c Double Integrals in Polar Coordinates (r, θ)

Let us suppose that the region boundary is now given in the form r = f(θ) or

θ = h(r), and/or the function being integrated is much simpler if polar coordinates

are used.

We begin with a brief review of polar coordinates.

We recall that a point P with coordinates (x, y) can also be specified by giving

its polar coordinates (r, θ) where

x = r cos θ r =√

x2 + y2

y = r sin θ tan θ =y

x.

r

x

y

( x , y )

θ

P

r

x

y

( x , y )

θ

P

Note that θ is always assumed given in radians. Clearly, and unlike the (x, y)

situation, a specific point has “many” (r, θ) coordinates obtained by increasing

(decreasing) θ by multiples of 2π. We shall allow r to be negative as follows.

Let r < 0. Then the point P = (r, θ) can be located by noting the diagram

95

θ

- r

P

y

x

I.e., this is the same point as given by (−r, θ+π). We note that in some books

r is not allowed to be negative. This can actually make a difference in a problem.

To conclude this brief review, we plot a couple of graphs.

Example 1. Sketch r = θ.

Answer. We mean the curve in the (x, y) plane, since in the (r, θ) plane r = θ is a

straight line! We start at θ = 0 and increase θ. When θ = 0 we have r = 0 and we

are at the origin. When θ = π/2, r = π/2 and as θ increases so does r. This means

that the curve spirals away from the origin.

(π , π) (2π , 2π)

(π / 2, π /2)

y

x

We must not forget what happens for θ < 0. We also get a spiral as follows:96

x

y

(−2π , −2π)

(−π / 2, −π / 2)

(−π , −π)

So the final picture is two spirals:

x

y

(−π , −π)

Example 2. Sketch r = cos θ.

Answer. We mean the curve in the (x, y) plane not in the (r, θ) plane. Again we

start at θ = 0 and note

θ = 0 r = 1

θ ↑ r ↓

θ =π

2r = 0.

Note that r ↓ means here we are moving towards the origin. So we obtain Fig a.97

0=θπ2

=θ r = 0 , r = 1,x

y

Fig. a

We now keep on increasing θ:

θ =π

2, r = 0

θ ↑, r ↓

θ = π, r = 1,

and obtain Fig. b.

=θ π , r = - 1

y

x

Fig. b

Observe that here r ↓ means we are moving away from the origin and the point is

in the fourth quadrant. By the properties of the cosine (i.e., periodicity), as θ keeps

on increasing, we shall retrace these curves over and over again, and the same will98

happen for θ negative. So the final total picture is

x

y

.

We can see that this is a circle as follows:

r = cos θ =⇒ r2 = r cos θ,

i.e., a point on r = cos θ must satisfy (in (x, y) coordinates)

x2 + y2 = x

or

(

x − 1

2

)2

+ y2 =1

4.

I.e., lie on a circle centered at ( 12, 0) of radius 1

2.

We shall practice sketching more graphs with the examples that follow.

Let us now suppose we have a “pie” shaped region R as shown.99

r

R

θ

Since the area of a circle is πr2, and there are 2π radians all the way around, then

Area of R = (πr2) · θ

2π=

θr2

2.

Consider now the next region as shown, with r, dr, dθ positive.

r

R

θ

We find

dA =(dθ)(r + dr)2

2− dθ

2r2

=dθ

2

[2rdr + (dr)2

]

= dθ

[

r dr +(dr)2

2

]

.

This formula is exact, but in cases of interest to us we shall let dr → 0. We can

neglect the term “(dr)2/2” in comparison with “dr” and thus shall state:

dA = r dr dθ .

An easy way to remember this is to note100

r

d

dA

dr

r d

θ

θ

and treat dA as for a “rectangle,” dA = r dθ dr. Let us return to our problem.

Consider a region D, bounded by r = h(θ), r = g(θ), θ = θ1, θ = θ2 as shown.

θ1

θ 2

r = h ( )θ

r = g ( )θ

x

y

D

Our problem is to find

∫∫

D

f(x, y) dA.

We now decompose the region into “circular caps” of area dA = r dr dθ. These

replace dx dy. We then have

f(x, y) dA = f(r cos θ, r sin θ) r dr dθ.

Now assemble the caps into “spikes” (these replace the strips).101

θ1

θ 2

r = h ( )θ

r = g ( )θ

x

y

D

The spike between θ and θ + dθ contributes:

∫ r=f(θ) (spike ends)

r=g(θ) (spike starts)

f(r cos θ, r sin θ) r dr dθ.

Finally, assembling all the spikes gives:

∫∫

D

f(x, y) dA =

∫ θ=θ2

θ=θ1

∫ r=h(θ)

r=g(θ)

f(r cos θ, r sin θ) r dr dθ.

We shall illustrate this below. Before, note that if the boundary of D is given by

θ = h(r), θ = g(r), r = r1, r = r2, then we assemble the caps the other way around

and get

∫∫

D

f(x, y) dA =

∫ r=r2

r=r1

∫ θ=h(r)

θ=g(r)

f(r cos θ, r sin θ)r dr dθ.

102

= h ( r )θ

= g ( r )θ

r = r 1

r = r 2

x

y

One more remark: We found dA = r dr dθ under the assumption that r, dr,

dθ are all positive. Since r can be negative, this formula may not hold (if r < 0,

dr > 0, dθ > 0, then dA < 0!). We have several ways to deal with this problem and

some are as follows:

(1) If r = f(θ) changes sign, break the problem into r > 0 and r < 0 regions.

(2) Where possible, use symmetry. This is the easiest.

(3) If r = f(θ) is negative, rewrite the curve as r = g(θ) with g positive.

(4) If r = f(θ) is negative, then take both r and dr negative.

We illustrate the above four ways with the following example.

Example 3. Find the area enclosed by the curve r = cos θ.

Answer. As we saw in Example 2 this is the circle

(

x − 1

2

)2

+ y2 =1

4,

So we know the answer: π 14.

103

x

y

We however ignore this and use the example to practice the four ways indi-

cated. Note that r = cos θ changes sign at θ = π/2, so we split the problem by

considering first θ between 0 and π/2 and later between π/2 and π.

Now as θ varies between θ and π/2 we get a semicircle.

= 0θ= / 2θ π

r = cos θ

r = 0

The area is

∫ π/2

θ=0

∫ cos θ

r=0

r dr dθ =

∫ π/2

0

cos2 θ

2dθ =

1

4

∫ π/2

0

[1 + cos(2θ)] dθ

=1

4

[

θ +sin(2θ)

2

]π/2

0

=π

8.

A not unexpected result. Since the figure is symmetric (it’s a circle, or from the

properties of the cosine), we use this fact, and obtain that the total area is 2 ×

(π/8) = π/4 as we know. We now practice the other ways for the purposes of104

illustration. Keep in mind that we are trying to calculate the area enclosed by

r = cos θ, r = 0, θ =π

2, θ = π.

The next way (changing to positive r) gives as follows: A point P = (r, θ) is

on r = cos θ for some θ between π/2 and π iff P is also given by (−r, θ + π). I.e., if

we plot (− cos θ, θ + π) with π/2 < θ < π, we will get the same curve.

θP ( r , )

θ

θ - r+π

y

x

Put ϕ = θ+π. Then our curve is also given by r = − cos(ϕ−π) with 3π/2 < ϕ < 2π,

and now r is positive.

= 2φ

= 3 / 2φ πx

y

105

So we get

A =

∫ 2π

ϕ= 3π

2

∫ − cos(ϕ−π)

r=0

r dr dϕ =1

2

∫ 2π

3π

2

cos2(ϕ − π) dϕ

=1

2

∫ 2π

3π

2

1 + cos(2(ϕ − π))

2dϕ =

1

2

∫ 2π

3π

2

1 + cos(2ϕ)

2

=1

4

[

1 +sin(2ϕ)

2

]2π

3π

2

=1

4

[

2π − 3π

2

]

=π

8.

For the final way: We now keep θ between π/2 and π. Suppose we increase θ

by dθ and decrease r (remember r is negative!!) by dr.

θd

- r

dA

x- dr

θd

y

We then obtain in the fourth quadrant a dA given by

dA = (−r dθ)(−dr) = r dθ dr.

This looks the same as before, except now both r and dr are negative! We calculate

the area of a spike, from the biggest (least negative) r to the smallest (most negative)

r—since our dr are negative—and get

∫ cos θ

r=0

r dr dθ.

106

So that, since dθ is positive, we get

A =

∫ π

π

2

[∫ cos θ

r=0

r dr

]

dθ =

∫ π

π

2

cos2 θ

2dθ =

1

4

∫ π

π

2

(1 + cos(2θ))dθ =π

8.

It is obvious that symmetry (when it exists) yields the easiest approach! We

now consider more examples.

Example 4. Evaluate∫∫

D(x + 3y) dA if D is the circle centered at the origin of

radius 2.

Answer. Note that the boundary of D is given by x2 + y2 = 4. This is the

motivation for changing to polar coordinates, since we can describe D in the much

simpler form r = 2!

x

y

We then have:

∫ 2π

θ=0

∫ 2

r=0

(cos θ + 3 sin θ)r2 dr dθ =

∫ 2π

θ=0

(cos θ + 3 sin θ)r3

3

∣∣∣∣

2

0

dθ

=

∫ 2π

0

(cos θ + 3 sin θ)8

3dθ

=8

3[sin θ − 3 cos θ

∣∣∣∣

2π

0

= 0.

107

Imagine we tried to do this problem using (x, y).

4 - x y = 2

y = - 4 - x 2

- 2 2x

y

We would have:

∫ 2

x=−2

∫ √4−x2

y=−√

4−x2

(x + 3y) dy dx.

This can be done, but the integrals are more complicated.

Example 5. Find∫∫

D(x2 + y2) dA where D is the cardioid r = 1 + cos θ.

Answer. Since the boundary of D is given in polar coordinates, we use this system.

Can you imagine how complicated this would become if we tried to switch to (x, y)?

We first need to sketch the cardioid.

θ = 0 r = 2

θ ↑ r ↓

θ =π

2r = 1

θ ↑ r ↓

θ = π r = 0108

θ ↑ r ↑

θ =3π

2r = 1

θ = 2π r = 2.

We only plot the graph for 0 ≤ θ ≤ 2π since cos θ is periodic.

= / 2 , r = 1πθ

= , r = 0πθ

θ = 0 , r = 2

y

x

We now write the integral in polar coordinates:

x = r cos θ, y = r sin θ

and so

x2 + y2 = r2.

We then have

(x2 + y2)dA = (r2)(r dr dθ)

and so

∫∫

D

(x2 + y2) dA =

∫ 2π

θ=0

∫ 1+cos θ

r=0

r2r dr dθ

109

=

∫ 2π

0

r4

4

∣∣∣∣

1+cos θ

0

dθ =1

4

∫ 2π

0

(1 + cos θ)4 dθ.

Note that

(1 + cos θ)4 =((1 + cos θ)2

)2= (1 + 2 cos θ + cos2 θ)2

= 1 + 4 cos θ + 6 cos2 θ + 4 cos3 θ + cos4 θ

= 1 + 4 cos θ + 6

[1 + cos 2θ

2

]

+ 4 cos3 θ +

[1 + cos 2θ

2

]2

= 1 + 4 cos θ + 3(1 + cos 2θ) + 4 cos3 θ +1

4[1 + 2 cos 2θ + cos2 2θ]

= 41

4+ 4 cos θ + 3

1

2cos 2θ + 4 cos3 θ +

1

4cos2 2θ

= 41

4+ 4 cos θ + 3

1

2cos 2θ + 4 cos3 θ +

1

4

[1 + cos 4θ

2

]

= 43

8+ 4 cos θ + 3

1

2cos 2θ + 4 cos3 θ +

1

8cos 4θ.

So, finally,

1

4

∫ 2π

0

(1 + cos θ)4 =1

4

∫ 2π

0

(

43

8+ 4 cos θ + 3

1

2cos 2θ +

1

8cos 4θ + 4 cos3 θ

)

dθ

=1

4

[

43

8· 2π + 4

∫ 2π

0

cos3 θ dθ

]

=1

4

[

43

8· 2π + 4

∫ 2π

0

cos2 θ cos θ dθ

]

=1

4

[

43

8· 2π + 4

∫ 2π

0

(1 − sin2 θ) cos θ dθ︸ ︷︷ ︸

put u=sin θ, du=cos θdθ

]

=1

4

[

43

8· 2π + 4

(

sin θ − sin3 θ

3

) ∣∣∣∣

2π

0

]

=1

4

(

43

8

)

2π.

110

Example 6. Find the area of one loop of r = cos 2θ.

Answer. Again we start with a sketch.

θ = 0 r = 1

θ ↑ r ↓

θ =π

4r = 0

θ ↑ r ↓

θ =π

2r = −1

θ ↑ r ↑

θ =3π

4r = 0

θ ↑ r ↑

θ = π r = 1

θ ↑ r ↓

θ =5π

4r = 0

θ ↑ r ↓

θ =3π

2r = −1

θ =7π

4r = 0

θ = 2π r = 1.

111

x

y

= / 4

= / 2

= , r = 1

= 3 / 4 = 3 / 2 , r = -1

= 0

θ

θ

θ

π

π

, r = - 1

θ

θ π

πθ

π

Note that the graph has four equal and symmetric loops, and that r changes from

positive to negative and back again. What I feel is the easiest way is to calculate

the area for θ between 0 and π/4 and multiply by 2. We get

Area of a loop = 2

∫ π/4

θ=0

∫ cos 2θ

r=0

r dr dθ

=

∫ π/4

θ=0

[cos2 2θ

]dθ

=

∫ π/4

0

1 + cos 4θ

2dθ =

1

2

(

θ +sin 4θ

4

)∣∣∣∣

π/4

0

=1

2

(π

4

)

=π

8.

This seems the best way. To practice, let us calculate the area for θ between π/4

and π/2 and again multiply by 2.112

θ

θ

= π / 2

= π / 4

- r

dr

Here r is negative and we choose dr < 0. Then

Area = 2

∫ π/2

π/4

∫ cos 2θ

0

r dr dθ =

∫ π/2

π/4

cos2 2θ dθ

=1

2

∫ π/2

π/4

[1 + cos 4θ] dθ =1

2

(

θ +sin 4θ

4

)∣∣∣∣

π/2

π/4

=π

8

as before.

Example 7. Find the volume outside the cone z =√

x2 + y2 and inside the

sphere x2 + y2 + z2 = 1.

Answer. Note that it is easiest to figure out just the volume common to the cone

and sphere, and subtract this from the volume of the sphere.

113

z = r

2 2z + r = 1

We can set up the problem using (x, y) and then switch to (r, θ) or start right away

with (r, θ). Let us start with (r, θ) directly. If we look down on our solid we see a

circle coming from the intersection of the sphere and cone, i.e.,

z2 + r2 = 1, z = r

or 2r2 = 1 or r = 1/√

2.

2r = 1 /

y

x

114

We now cut the disc into regions of size dA = rdr dθ and obtain that the volume of

a column over dA = (height)(cross-sectional area)

= (√

1 − r2

︸ ︷︷ ︸

from thesphere

− r︸︷︷︸

from thecone

)r dr dθ.

Hence

volume over a spike =

∫ 1/√

2

r=0

(√

1 − r2 − r)r dr dθ

and volume V , common to sphere and cone, is:

V =

∫ 2π

0

∫ 1/√

2

r=0

(√

1 − r2 − r)r dr dθ

=

∫ 2π

0

{[∫ 1/√

2

r=0

√

1 − r2r dr

]

−∫ 1/

√2

r=0

r2 dr

}

dθ

=

∫ 2π

0

[

− 1

3(1 − r2)

3

2 − r3

3

]1/√

2

r=0

dθ

=

∫ 2π

0

(1

3

)(

1 −[(

1 − 1

2

)3/2

+1

23

2

])

dθ

=

∫ 2π

0

(1

3

)(

1 − 2

23

2

)

dθ =1

3

(

1 − 1√2

)

· 2π.

So the final answer is

Volume of sphere − V =4

3π − 2π

3

(

1 − 1√2

)

=4

3π

(1

2+

1

2√

2

)

.

Example 8. Find the volume bounded above by the cone z = r and below by

y = 0, y = x, x = 1 and the xy plane.

Answer. Since the altitude is z = r (simple), we do this problem using (r, θ), even

though the boundary is given by y = f(x). We rewrite the boundary of the base115

triangle as:

y = 0 −→ θ = 0

y = x −→ r sin θ = r cos θ −→ tan θ = 1 −→ θ =π

4

x = 1 −→ r cos θ = 1 −→ r = sec θ.

r = sec θ

x

y

z

θ = π / 4

θ = 0

y

x

So,

Volume over dA = altitude × cross sectional area

= (r)(r dr)dθ.

And we get

V =

∫ π/4

θ=0

∫ sec θ

r=0

r2 dr dθ =1

3

∫ π/4

0

sec3 θ dθ.

Now to evaluate:

∫ π/4

0

sec3 θ dθ =

∫ π/4

0

sec2 θ︸ ︷︷ ︸

u′

(sec θ︸︷︷︸

v

)dθ

(u = tan θ, v′ = sec θ tan θ)

= tan θ sec θ

∣∣∣∣

π/4

0

−∫ π/4

0

tan2 θ sec θ dθ

= tan θ sec θ

∣∣∣∣

π/4

0

−∫ π/4

0

(sec2 θ − 1) sec θ dθ

116

= tan θ sec θ

∣∣∣∣

π/4

0

−∫ π/4

0

sec3 θ dθ +

∫ π/4

0

sec θ dθ,

or

2

∫ π/4

0

sec3 θ dθ = tan θ sec θ

∣∣∣∣

π/4

0

+ ln | sec θ + tan θ|∣∣∣∣

π/4

0

= tanπ

4sec

π

4+ ln

∣∣∣sec

π

4+ tan

π

4

∣∣∣

=2√2

+ ln

(2√2

+ 1

)

.

And, finally,

V =1

3

[1√2

+1

2ln

(2√2

+ 1

)]

.

Example 9. Find the volume under the plane 4x + 6y + z = 12 and above the

region in the (x, y) plane with boundary x2 + y2 = x.

Answer. Note that x2 + y2 = x (or (x− 12)2 + y2 = 1

4 ) is a circle. We change it to

polar coordinates as follows: r2 cos2 θ + r2 sin2 θ = r cos θ or r = cos θ. Observe the

values of θ which are needed to trace the circle, and the fact that r goes negative!

x

θ = 0

θ = − π / 2

θ = π / 2

y

We could split the problem up into regions where r is positive and where it’s neg-117

ative, but the best way is probably to instead have θ range over −π/2 ≤ θ ≤ π/2.

This also covers the circle. Next note that the plane does not cross the circle. In-

deed, in the circle x ≤ 1 and y ≤ 1/2 and so z = 12 − 4x− 6y ≥ 12 − 4 − 6 · 12

> 0.

If the plane crossed, we would have to split the problem, since the altitude would

sometimes be z and sometime −z. So we get

(12 − 4x − 6y)dA = (12 − 4r cos θ − 6r sin θ)r dθ dθ

and

Volume =

∫ π/2

θ=−π/2

∫ cos θ

r=0

(12 − 4r cos θ − 6r sin θ)r dr dθ

=

∫ π/2

−π/2

(

6r2 − 4r3

3cos θ − 2r3 sin θ

) ∣∣∣∣

cos θ

0

dθ

=

∫ π/2

−π/2

(

6 cos2 θ − 4

3cos4 θ − 2 cos3 θ sin θ

)

dθ.

Now

∫ π/2

−π/2

6 cos2 θ dθ =

∫ π/2

−π/2

3(1 + cos 2θ)dθ = 3

(

θ +sin 2θ

2

) ∣∣∣∣

π/2

−π/2

= 3π

∫ π/2

−π/2

−4

3cos4 θ dθ =

∫ π/2

−π/2

(

−4

3

)[1 + cos 2θ

2

]2

dθ

=

∫ π/2

−π/2

(

−1

3

)

[1 + 2 cos 2θ + cos2 2θ] dθ

=

∫ π/2

−π/2

(

−1

3

)[

1 + 2 cos 2θ +1

2(1 + cos 4θ)

]

dθ

= −1

3

[

θ + sin 2θ +1

2

(

θ +sin 4θ

4

)]π/2

−π/2

= −1

3· 3

2· π = −π

2.

118

Finally,

∫ π/2

−π/2

−2 cos3 θ sin θ dθ =cos4 θ

2

∣∣∣∣

π/2

−π/2

= 0.

(u = cos θ, du = − sin θ dθ)

So

V = 3π − π

2=

5π

2.

Example 10. Convert to polar coordinates and evaluate

∫ a

−a

∫ √a2−x2

0

(x2 + y2)1/2dy dx.

Answer. We begin by sketching the region D. Note that x varies from −a to a, and

for a given x between −a and a, y goes from 0 to√

a2 − x2. But if y =√

a2 − x2,

then y2 + x2 = a2 (i.e., on the circle) and y ≥ 0. So that D is:

x + y = a 222

D

x

y

- a a

Now

(x2 + y2)1/2 = r and x2 + y2 = a2 −→ r = a.119

So we get

∫ θ=π

θ=0

∫ a

r=0

(r)(r dr dθ) =

∫ π

0

a3

3dθ =

πa3

3.

120

Further Exercises:

1) Evaluate∫∫

Ay2 dA if A is the area enclosed by r = 4(1 − cos θ).

2) Change to polar coordinates and evaluate

∫ 2

−2

∫√

4−y2

−√

4−y2

e−(x2+y2) dx dy.

3) Find the total area enclosed by r2 = 2 cos(2θ).

4) Find by means of polar coordinates the area in the circular cap bounded above

by x2 + y2 = 1 and below by y = 1/√

2.

5) Find the area of the region common to the cardioid r = 1 + cos θ and the

circle r = sin θ.

6) Change to polar coordinates and evaluate

∫ 1

z=0

∫ √1−z2

y=0

sin(y2 + z2) dy dz.

7) Calculate the area bounded by the curves:

θ = 0, r =π

4, r =

π

2, θ =

(

r − 3π

8

)2

.

8) Evaluate∫∫

B

√

x2 + y2 dx dy by means of polar coordinates if B is the region

bounded by the circle r = − sin θ.

Note: the answer is positive.

9) Change to (x, y) coordinates and evaluate:

∫ π

2

θ= π

4

∫ csc θ

r=0

r5 cos2 θ sin2 θ dr dθ.

121

10) Evaluate∫∫

D(1+x2+y2)10 dx dy where D is the region bounded by the curves

y = 0, y = x, x2 + y2 = 1.

11) Evaluate∫∫

D

√

x2 + y2 dy dx where D is the triangle with vertices (0, 0),

(1, 0), (1, 1).

122