Back in Calculus I we had the substitution rule that told us
that,
In essence this is taking an integral in terms of x’s and changing it into terms of u’s.
We want to do something similar for double and triple integrals. In fact we’ve already done this to a certain
extent when we converted double integrals to polar coordinates and when we
converted triple integrals to cylindrical or spherical coordinates. The main difference is that we didn’t
actually go through the details of where the formulas came from. If you recall, in each of those cases we
commented that we would justify the formulas for dA and dV
eventually. Now is the time to do that
justification.
While often the reason for changing variables is to get us
an integral that we can do with the new variables, another reason for changing
variables is to convert the region into a nicer region to work with. When we were converting the polar, cylindrical
or spherical coordinates we didn’t worry about this change since it was easy
enough to determine the new limits based on the given region. That is not always the case however. So, before we move into changing variables
with multiple integrals we first need to see how the region may change with a
change of variables.
First we need a little notation out of the way. We call the equations that define the change
of variables a transformation. Also we will typically start out with a
region, R, in xycoordinates and transform it into a region in uvcoordinates.
Example 1 Determine
the new region that we get by applying the given transformation to the region
R.
(a) R is the ellipse and the transformation is ,
. [Solution]
(b) R is the region bounded by ,
,
and and the
transformation is ,
. [Solution]
Solution
(a) R is the ellipse and the transformation is , .
There really isn’t too much to do with this one other than
to plug the transformation into the equation for the ellipse and see what we
get.
So, we started out with an ellipse and after the
transformation we had a disk of radius 2.
[Return to Problems]
(b) R is the region bounded by , , and and the transformation is , .
As with the first part we’ll need to plug the
transformation into the equation, however, in this case we will need to do it
three times, once for each equation.
Before we do that let’s sketch the graph of the region and see what
we’ve got.
So, we have a triangle.
Now, let’s go through the transformation. We will apply the transformation to each
edge of the triangle and see where we get.
Let’s do first.
Plugging in the transformation gives,
The first boundary transforms very nicely into a much
simpler equation.
Now let’s take a look at ,
Again, a much nicer equation that what we started with.
Finally, let’s transform .
So, again, we got a somewhat simpler equation, although
not quite as nice as the first two.
Let’s take a look at the new region that we get under the
transformation.
We still get a triangle, but a much nicer one.
[Return to Problems]

Note that we can’t always expect to transform a specific
type of region (a triangle for example) into the same kind of region. It is completely possible to have a triangle
transform into a region in which each of the edges are curved and in no way
resembles a triangle.
Notice that in each of the above examples we took a two
dimensional region that would have been somewhat difficult to integrate over
and converted it into a region that would be much nicer in integrate over. As we noted at the start of this set of
examples, that is often one of the points behind the transformation. In addition to converting the integrand into
something simpler it will often also transform the region into one that is much
easier to deal with.
Now that we’ve seen a couple of examples of transforming
regions we need to now talk about how we actually do change of variables in the
integral. We will start with double
integrals. In order to change variables
in a double integral we will need the Jacobian
of the transformation. Here is the definition
of the Jacobian.
Definition
The Jacobian is defined as a determinant of a 2x2 matrix, if
you are unfamiliar with this that is okay.
Here is how to compute the determinant.
Therefore, another formula for the determinant is,
Now that we have the Jacobian out of the way we can give the
formula for change of variables for a double integral.
Change of Variables
for a Double Integral
Note that we used du
dv instead of dA in the integral
to make it clear that we are now integrating with respect to u and v. Also note that we are
taking the absolute value of the Jacobian.
If we look just at the differentials in the above formula we
can also say that
Example 2 Show
that when changing to polar coordinates we have
Solution
So, what we are doing here is justifying the formula that
we used back when we were integrating with respect to polar coordinates. All that we need to do is use the formula
above for dA.
The transformation here is the standard conversion
formulas,
The Jacobian for this transformation is,
We then get,

So, the formula we used in the section on polar integrals
was correct.
Now, let’s do a couple of integrals.
Example 4 Evaluate
where R
is the ellipse given by and using the transformation ,
.
Solution
The first thing to do is to plug the transformation into
the equation for the ellipse to see what the region transforms into.
Or, upon dividing by 2 we see that the equation describing
R transforms into
or the unit circle.
Again, this will be much easier to integrate over than the original
region.
Note as well that we’ve shown that the function that we’re
integrating is
in terms of u
and v so we won’t have to redo that
work when the time to do the integral comes around.
Finally, we need to find the Jacobian.
The integral is then,
Before proceeding a word of caution is in order. Do not make the mistake of substituting or in for the integrands. These equations are only valid on the
boundary of the region and we are looking at all the points interior to the
boundary as well and for those points neither of these equations will be
true!
At this point we’ll note that this integral will be much
easier in terms of polar coordinates and so to finish the integral out will
convert to polar coordinates.

Let’s now briefly look at triple integrals. In this case we will again start with a
region R and use the transformation ,
,
and to transform the region into the new region S.
To do the integral we will need a Jacobian, just as we did with double
integrals. Here is the definition of the
Jacobian for this kind of transformation.
In this case the Jacobian is defined in terms of the
determinant of a 3x3 matrix. We saw how
to evaluate these when we looked at cross
products back in Calculus II. If you
need a refresher on how to compute them you should go back and review that
section.
The integral under this transformation is,
As with double integrals we can look at just the
differentials and note that we must have
We’re not going to do any integrals here, but let’s verify
the formula for dV for spherical
coordinates.
Example 5 Verify
that when using spherical coordinates.
Solution
Here the transformation is just the standard conversion
formulas.
The Jacobian is,
Finally, dV
becomes,
Recall that we restricted to the range for spherical coordinates and so we know
that and so we don’t need the absolute value bars
on the sine.

We will leave it to you to check the formula for dV for cylindrical coordinates if you’d
like to. It is a much easier formula to
check.