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Double Integrals in Polar
Coordinates

Double Integrals in Polar Coordinates
To this point we’ve seen quite a few double integrals
...
In this section we want
to look at some regions that are much easier to describe in terms of polar coordinates
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In these cases
using Cartesian coordinates could be somewhat cumbersome
...

These would be,

−2 ≤ x ≤ 2

− 4 − x2 ≤ y ≤ 4 − x2
With these limits the integral would become,

∫∫
D

2

f ( x, y ) dA = ⌠
⎮ ∫
⌡ −2 −

4− x2
4− x

2

f ( x, y ) dy dx

Due to the limits on the inner integral this is liable to be an unpleasant integral to compute
...

So, if we could convert our double integral formula into one involving polar coordinates we
would be in pretty good shape
...
In computing double integrals to this point we have been using the fact that
dA = dx dy and this really does require Cartesian coordinates to use
...

So, let’s step back a little bit and start off with a general region in terms of polar coordinates and
see what we can do with that
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So, our general region will be defined by inequalities,

α ≤θ ≤ β
h1 (θ ) ≤ r ≤ h2 (θ )

Now, to find dA let’s redo the figure above as follows,

As shown, we’ll break up the region into a mesh of radial lines and arcs
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The area of this piece is ΔA
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Basic geometry then tells us that
the length of the inner edge is ri Δ θ while the length of the out edge is ro Δ θ where Δ θ is the
angle between the two radial lines that form the sides of this piece
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In order to arrive at this we had to make the assumption that the mesh was very small
...
Recall that the definition of a double integral is in terms of two
limits and as limits go to infinity the mesh size of the region will get smaller and smaller
...
This second way will not involve any assumptions either and so it maybe a little better
way of deriving this
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The actual formula for dA has
an r in it
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Now, if we’re going to be converting an integral in Cartesian coordinates into an integral in polar
coordinates we are going to have to make sure that we’ve also converted all the x’s and y’s into
polar coordinates as well
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∫∫
D

β

f ( x, y ) dA = ⌠
⎮
⌡α

∫

h 2 (θ )
h1 (θ )

f ( r cos θ , r sin θ ) r dr dθ

It is important to not forget the added r and don’t forget to convert the Cartesian coordinates in
the function over to polar coordinates
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Example 1 Evaluate the following integrals by converting them into polar coordinates
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[Solution]
(b)

∫∫ e

x2 + y 2

dA , D is the unit circle centered at the origin
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First let’s get D in terms of polar coordinates
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We want the region between them so we will have the
following inequality for r
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0 ≤θ ≤

π

2

Now that we’ve got these we can do the integral
...
Now, let’s simplify and make use of
the double angle formula for sine to make the integral a little easier
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D

In this case we can’t do this integral in terms of Cartesian coordinates
...
First, the region D is defined by,

0 ≤ θ ≤ 2π
0 ≤ r ≤1

In terms of polar coordinates the integral is then,

∫∫ e

x +y
2

2

D

2π

dA = ⌠
⌡0

∫

1
0

r e dr dθ
r2

Notice that the addition of the r gives us an integral that we can now do
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∫∫ e
D

x2 + y 2

2π

dA = ⌠
⌡0

∫

1
0

r e dr dθ

2π

r2

⌠ 1 r2
=⎮
e dθ
0
⌡0 2
1

2π

1
⌠
=⎮
( e − 1) dθ
⌡0 2
= π ( e − 1)
[Return to Problems]

Let’s not forget that we still have the two geometric interpretations for these integrals as well
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Solution
Here is a sketch of the region, D, that we want to determine the area of
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We
can determine these points by setting the two equations and solving
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Note as well that we’ve acknowledged that − π6 is another representation for the angle

11 π
6


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If we’d chosen to use 116π then as we increase from 76π to 116π we
would be tracing out the lower portion of the circle and that is not the region that we are after
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π

7π
− ≤θ ≤
6
6
2 ≤ r ≤ 3 + 2sin θ

To get the ranges for r the function that is closest to the origin is the lower bound and the function
that is farthest from the origin is the upper bound
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187
2
3

Example 3 Determine the volume of the region that lies under the sphere x 2 + y 2 + z 2 = 9 ,
above the plane z = 0 and inside the cylinder x 2 + y 2 = 5
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The function isn’t too bad
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We are looking at the region that lies under the sphere and above the plane

z = 0 (just the xy-plane right?) and so all we need to do is solve the equation for z and when
taking the square root we’ll take the positive one since we are wanting the region above the xyplane
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z = 9 − x2 − y 2
The region D isn’t too bad in this case either
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Since we only want the
portion of the sphere that actually lies inside the cylinder given by x 2 + y 2 = 5 this is also the
region D
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2

2

For reference purposes here is a sketch of the region that we are trying to find the volume of
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We are definitely going to want to do this integral in terms of polar coordinates so here are the
limits (in polar coordinates) for the region,

0 ≤ θ ≤ 2π

0≤r ≤ 5
and we’ll need to convert the function to polar coordinates as well
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Solution
Let’s start this example off with a quick sketch of the region
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The formula

V = ∫∫ f ( x, y ) dA
D

finds the volume under the function f ( x, y ) and we’re actually after the area that is above a
function
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First, notice that

V = ∫∫ 16 dA
D

will be the volume under z = 16 (of course we’ll need to determine D eventually) while

V = ∫∫ x 2 + y 2 dA
D

is the volume under z = x 2 + y 2 , using the same D
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Determining the region D in this case is not too bad
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This is because the top of
the region, where the elliptic paraboloid intersects the plane, is the widest part of the region
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Here are the inequalities for the region and the function we’ll be integrating in terms of polar
coordinates
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There is one more type of example that we need to look at before moving on to the next section
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We need to see an example of
how to do this kind of conversion
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1

⌠
⎮ ∫
⌡0 0

1− y 2

cos ( x 2 + y 2 ) dx dy

Solution
First, notice that we cannot do this integral in Cartesian coordinates and so converting to polar
coordinates may be the only option we have for actually doing the integral
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Let’s first determine the region that we’re integrating over and see if it’s a region that can be
easily converted into polar coordinates
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0 ≤ y ≤1

0 ≤ x ≤ 1− y2
Now, the upper limit for the x’s is,

x = 1− y2
and this looks like the right side of the circle of radius 1 centered at the origin
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The range for the y’s however, tells us that we are only going to have positive y’s
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So, we know that the inequalities that will define this region in terms of polar coordinates are
then,

0 ≤θ ≤

π

2
0 ≤ r ≤1

Finally, we just need to remember that,

dx dy = dA = r dr dθ

and so the integral becomes,
1

⌠
⎮ ∫
⌡0 0

1− y 2

cos ( x + y ) dx dy = ∫
2

2

π

2
0

∫

1

0

r cos ( r 2 ) dr dθ

Note that this is an integral that we can do
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π

1− y
⌠ 1
2
2
⌠
cos ( x + y ) dx dy = ⎮ sin ( r 2 ) dθ
⎮ ∫
⌡0 0
⌡0 2
0
1

2

2

π

1

2 1
⌠
= ⎮ sin (1) dθ
⌡0 2

=

π
4

sin (1)



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