To this point we’ve seen quite a few double integrals. However, in every case we've seen to this point the region D could be easily described in terms of
simple functions in Cartesian coordinates.
In this section we want to look at some regions that are much easier to
describe in terms of polar coordinates.
For instance, we might have a region that is a disk, ring, or a portion
of a disk or ring. In these cases using
Cartesian coordinates could be somewhat cumbersome. For instance let’s suppose we wanted to do
the following integral,
To this we would have to determine a set of inequalities for
x and y that describe this region.
These would be,
With these limits the integral would become,
Due to the limits on the inner integral this is liable to be
an unpleasant integral to compute.
However, a disk of radius 2 can be defined in polar
coordinates by the following inequalities,
These are very simple limits and, in fact, are constant
limits of integration which almost always makes integrals somewhat easier.
So, if we could convert our double integral formula into one
involving polar coordinates we would be in pretty good shape. The problem is that we can’t just convert the
dx and the dy into a dr and a . In computing double integrals to this point
we have been using the fact that and this really does require Cartesian
coordinates to use. Once we’ve moved
into polar coordinates and so we’re going to need to determine just
what dA is under polar coordinates.
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. Here is a sketch of some region
using polar coordinates.
So, our general region will be defined by inequalities,
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. Now, if we pull one of
the pieces of the mesh out as shown we have something that is almost, but not
quite a rectangle. The area of this
piece is . The two sides of this piece both have length where is the radius of the outer arc and is the radius of the inner arc. Basic geometry then tells us that the length
of the inner edge is while the length of the out edge is where is the angle between the two radial lines that
form the sides of this piece.
Now, let’s assume that we’ve taken the mesh so small that we
can assume that and with this assumption we can also assume that
our piece is close enough to a rectangle that we can also then assume that,
Also, if we assume that the mesh is small enough then we can
also assume that,
With these assumptions we then get .
In order to arrive at this we had to make the assumption
that the mesh was very small. This is
not an unreasonable assumption. 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. In fact, as the mesh size gets smaller and
smaller the formula above becomes more and more accurate and so we can say
We’ll see another way of deriving this once we reach the Change of Variables section later in this
chapter. This second way will not
involve any assumptions either and so it maybe a little better way of deriving
Before moving on it is again important to note that . The actual formula for dA has an r in it. It will be easy to forget this r on occasion, but as you’ll see without
it some integrals will not be possible to do.
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. To do this we’ll
need to remember the following conversion formulas,
We are now ready to write down a formula for the double integral
in terms of polar coordinates.
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.
Let’s look at a couple of examples of these kinds of
Example 1 Evaluate
the following integrals by converting them into polar coordinates.
D is the portion of the region
between the circles of radius 2
and radius 5 centered at the origin
that lies in the first quadrant. [Solution]
D is the unit circle centered at
the origin. [Solution]
(a) , D
is the portion of the region between the circles of radius 2 and radius 5
centered at the origin that lies in the first quadrant.
First let’s get D
in terms of polar coordinates. The
circle of radius 2 is given by and the circle of radius 5 is given by . We want the region between them so we will
have the following inequality for r.
Also, since we only want the portion that is in the first
quadrant we get the following range of ’s.
Now that we’ve got these we can do the integral.
Don’t forget to do the conversions and to add in the extra
Now, let’s simplify and make use of the double angle formula for sine
to make the integral a little easier.
[Return to Problems]
(b) , D
is the unit circle centered at the origin.
In this case we can’t do this integral in terms of
Cartesian coordinates. We will however
be able to do it in polar coordinates.
First, the region D is
In terms of polar coordinates the integral is then,
Notice that the addition of the r gives us an integral that we can now do. Here is the work for this integral.
[Return to Problems]
Let’s not forget that we still have the two geometric
interpretations for these integrals as well.
Example 2 Determine
the area of the region that lies inside and outside .
Here is a sketch of the region, D, that we want to determine the area of.
To determine this area we’ll need to know that value of θ for which the two curves intersect. We can determine these points by setting
the two equations and solving.
Here is a sketch of the figure with these angles added.
Note as well that we’ve acknowledged that is another representation for the angle . This is important since we need the range
of θ to actually enclose the regions as we
increase from the lower limit to the upper limit. If we’d chosen to use then as we increase from to we would be tracing out the lower portion of
the circle and that is not the region that we are after.
So, here are the ranges that will define the region.
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.
The area of the region D
Example 3 Determine
the volume of the region that lies under the sphere ,
above the plane and inside the cylinder .
We know that the formula for finding the volume of a
In order to make use of this formula we’re going to need
to determine the function that we should be integrating and the region D that we’re going to be integrating
The function isn’t too bad. It’s just the sphere, however, we do need
it to be in the form . We are looking at the region that lies
under the sphere and above the plane
(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 xy-plane. Here is the function.
The region D
isn’t too bad in this case either. As
we take points, ,
from the region we need to completely graph the portion of the sphere that we
are working with. Since we only want
the portion of the sphere that actually lies inside the cylinder given by this is also the region D. The region D is the disk in the xy-plane.
For reference purposes here is a sketch of the region that
we are trying to find the volume of.
So, the region that we want the volume for is really a
cylinder with a cap that comes from the sphere.
We are definitely going to want to do this integral in
terms of polar coordinates so here are the limits (in polar coordinates) for
and we’ll need to convert the function to polar coordinates
The volume is then,
Example 4 Find
the volume of the region that lies inside and
below the plane .
Let’s start this example off with a quick sketch of the
Now, in this case the standard formula is not going to
work. The formula
finds the volume under the function and we’re actually after the area that is
above a function. This isn’t the
problem that it might appear to be however.
First, notice that
will be the volume under (of course we’ll need to determine D eventually) while
is the volume under ,
using the same D.
The volume that we’re after is really the difference
between these two or,
Now all that we need to do is to determine the region D and then convert everything over to
Determining the region D
in this case is not too bad. If we
were to look straight down the z-axis
onto the region we would see a circle of radius 4 centered at the
origin. This is because the top of the
region, where the elliptic paraboloid intersects the plane, is the widest
part of the region. We know the z
coordinate at the intersection so, setting in the equation of the paraboloid gives,
which is the equation of a circle of radius 4 centered at
Here are the inequalities for the region and the function
we’ll be integrating in terms of polar coordinates.
The volume is then,
In both of the previous volume problems we would have not
been able to easily compute the volume without first converting to polar
coordinates so, as these examples show, it is a good idea to always remember
There is one more type of example that we need to look at
before moving on to the next section.
Sometimes we are given an iterated integral that is already in terms of x and y and we need to convert this over to polar so that we can actually
do the integral. We need to see an
example of how to do this kind of conversion.
Example 5 Evaluate
the following integral by first converting to polar coordinates.
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.
Notice that the function will convert to polar coordinates nicely and
so shouldn’t be a problem.
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. Here are the inequalities
that define the region in terms of Cartesian coordinates.
Now, the upper limit for the x’s is,
and this looks like the right side of the circle of radius
1 centered at the origin. Since the
lower limit for the x’s is it looks like we are going to have a portion
(or all) of the right side of the disk of radius 1 centered at the origin.
The range for the y’s
however, tells us that we are only going to have positive y’s.
This means that we are only going to have the portion of the disk of
radius 1 centered at the origin that is in the first quadrant.
So, we know that the inequalities that will define this
region in terms of polar coordinates are then,
Finally, we just need to remember that,
and so the integral
Note that this is an integral that we can do. So, here is the rest of the work for this