In the previous section we gave the definition of the double
integral. However, just like with the
definition of a single integral the definition is very difficult to use in
practice and so we need to start looking into how we actually compute double
integrals. We will continue to assume
that we are integrating over the rectangle
We will look at more general regions in the next section.
The following theorem tells us how to compute a double
integral over a rectangle.
Fubini’s Theorem
Note that there are in fact two ways of computing a double
integral and also notice that the inner differential matches up with the limits
on the inner integral and similarly for the outer differential and limits. In other words, if the inner differential is dy then the limits on the inner integral
must be y limits of integration and
if the outer differential is dy then
the limits on the outer integral must be y
limits of integration.
Now, on some level this is just notation and doesn’t really
tell us how to compute the double integral.
Let’s just take the first possibility above and change the notation a
little.
We will compute the double integral by first computing
and we compute this by holding x constant and integrating with respect to y as if this were a single integral. This will give a function involving only x’s which we can in turn integrate.
We’ve done a similar process with partial derivatives. To take the derivative of a function with
respect to y we treated the x’s as constants and differentiated with
respect to y as if it was a function
of a single variable.
Double integrals work in the same manner. We think of all the x’s as constants and integrate with respect to y or we think of all y’s
as constants and integrate with respect to x.
Let’s take a look at some examples.
Example 1 Compute
each of the following double integrals over the indicated rectangles.
(a) , [Solution]
(b) , [Solution]
(c) , [Solution]
(d) , [Solution]
(e) , [Solution]
Solution
(a) ,
It doesn’t matter which variable we integrate with respect
to first, we will get the same answer regardless of the order of
integration. To prove that let’s work
this one with each order to make sure that we do get the same answer.
Solution 1
In this case we will integrate with respect to y first. So, the iterated integral that we need to
compute is,
When setting these up make sure the limits match up to the
differentials. Since the dy is the inner differential (i.e. we are integrating with respect
to y first) the inner integral
needs to have y limits.
To compute this we will do the inner integral first and we
typically keep the outer integral around as follows,
Remember that we treat the x as a constant when doing the first integral and we don’t do any
integration with it yet. Now, we have
a normal single integral so let’s finish the integral by computing this.
Solution 2
In this case we’ll integrate with respect to x first and then y. Here is the work for
this solution.
Sure enough the same answer as the first solution.
So, remember that we can do the integration in any order.
[Return to Problems]
(b) ,
For this integral we’ll integrate with respect to y first.
Remember that when integrating with respect to y all x’s are treated as constants and so as far as the inner integral
is concerned the 2x is a constant
and we know that when we integrate constants with respect to y we just tack on a y and so we get 2xy from the first term.
[Return to Problems]
(c) ,
In this case we’ll integrate with respect to x first.
Don’t forget your basic Calculus I substitutions!
[Return to Problems]
(d) ,
In this case because the limits for x are kind of nice (i.e.
they are zero and one which are often nice for evaluation) let’s integrate
with respect to x first. We’ll also rewrite the integrand to help
with the first integration.
[Return to Problems]
(e) ,
Now, while we can technically integrate with respect to
either variable first sometimes one way is significantly easier than the
other way. In this case it will be
significantly easier to integrate with respect to y first as we will see.
The y
integration can be done with the quick substitution,
which gives
So, not too bad of an integral there provided you get the
substitution. Now let’s see what would
happen if we had integrated with respect to x first.
In order to do this we would have to use integration by
parts as follows,
The integral is then,
We’re not even going to continue here as these are very
difficult integrals to do.
[Return to Problems]

As we saw in the previous set of examples we can do the
integral in either direction. However,
sometimes one direction of integration is significantly easier than the other
so make sure that you think about which one you should do first before actually
doing the integral.
The next topic of this section is a quick fact that can be
used to make some iterated integrals somewhat easier to compute on occasion.
Fact
So, if we can break up the function into a function only of x times a function of y then we can do the two integrals
individually and multiply them together.
Let’s do a quick example using this integral.
Example 2 Evaluate
,
.
Solution
Since the integrand is a function of x times a function of y
we can use the fact.

We have one more topic to discuss in this section. This topic really doesn’t have anything to do
with iterated integrals, but this is as good a place as any to put it and there
are liable to be some questions about it at this point as well so this is as
good a place as any.
What we want to do is discuss single indefinite integrals of
a function of two variables. In other
words we want to look at integrals like the following.
From Calculus I we know that these integrals are asking what
function that we differentiated to get the integrand. However, in this case we need to pay
attention to the differential (dy or dx) in the integral, because that will
change things a little.
In the case of the first integral we are asking what
function we differentiated with respect to y
to get the integrand while in the second integral we’re asking what function
differentiated with respect to x to
get the integrand. For the most part answering
these questions isn’t that difficult.
The important issue is how we deal with the constant of integration.
Here are the integrals.
Notice that the “constants” of integration are now functions
of the opposite variable. In the first
integral we are differentiating with respect to y and we know that any function involving only x’s will differentiate to zero and so when integrating with respect
to y we need to acknowledge that
there may have been a function of only x’s
in the function and so the “constant” of integration is a function of x.
Likewise, in the second integral, the “constant” of
integration must be a function of y
since we are integrating with respect to x. Again, remember if we differentiate the
answer with respect to x then any
function of only y’s will
differentiate to zero.