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To this point we’ve seen quite a few double integrals. However, in every case in 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
that,
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
this.
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
integrals.
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Example 1 Evaluate
the following integrals by converting them into polar coordinates.
(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. [Solution]
(b) ,
D is the unit circle centered at
the origin. [Solution]
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
r.
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
defined by,
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]
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Let’s not forget that we still have the two geometric
interpretations for these integrals as well.
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Example 2 Determine
the area of the region that lies inside and outside .
Solution
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
is then,
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