You are about to erase your work on this activity. Are you sure you want to do this?

Updated Version Available

There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed?

Mathematical Expression Editor

We study integrals over general regions by integrating .

Now we will integrate over regions that are more complex than rectangles and boxes.
In particular we will find areas of regions bounded by curves in , and volumes of
regions bounded by surfaces in . To do this we will use integrals whose integrand is .
The integrand is the “thing” you are integrating: Think about this: The integral
computes the length between and . This is because the area between the
constant curve of height one and the -axis is just the length of the curve! In a
similar way: Here you imagine a surface of constant height above a region :

The volume under the plane of constant height above the region is numerically
equal to the area of . In an entirely similar way, if is a subset of , then In this
section we will focus on integrals with a trivial integrand, meaning it is , and hence
the section is called “Integrals with trivial integrands.”

Double integrals and area

We start with a third version of Fubini’s Theorem.

Fubini’s Theorem Let be a closed, bounded region in the -plane and let be a
continuous function on .

If

where and are continuous functions on , then

If

where and are continuous functions on , then

It is important to note that when using Fubini’s Theorem, we must always have
numbers as the limits of the outer-most integral and curves (note constant curves are
numbers) as the limits of the inner-most integral: Note, if we set and (or and ) to
be constant functions, we recover the version of Fubini’s Theorem from the previous
section.

Whenever you learn a new technique, you should always “try it out” on a
computation where you know the answer through a different method. So let’s used an
integral to find the area of a triangle.

Set-up and evaluate an iterated integral that will compute the area of the region
below via a double integral:

Our region above can be defined by: The area of this region is given by, write with me,

Set-up and evaluate an iterated integral that will compute the area of the region
below via a double integral:

Our region above can be defined by: The area of this region is given by, write with me,

Changing the order of integration

We can change the order of integration. This can make a big difference in the
difficulty of the integral in question. This can be somewhat challenging. If you have a
graph of the region in question, this can help quite a bit. If not, then a strategy is to

Find absolute bounds for each variable.

Solve inequalities that will allow you to integrate:

Let’s see an example.

Consider: Set-up an iterated integral that integrates with respect to and then
integrates with respect to .

Start by finding absolute bounds for our variables. In this
case:

Now, write in terms of . Write with me:

We may now write our desired integral:

Let’s see another example.

Consider: Set-up an iterated integral that computes the area of that integrates with
respect to and then with respect to .

Start by finding absolute bounds for our
variables. In this case:

Now, write in terms of . Write with me:

We may now write our desired integral:

Sometimes when you switch the order of integration, you will need to write the sum
of iterated integrals. You can recognize this by the fact that you won’t be able to find
a single curve to bound your inner integral. This will be more clear with an
example.

Consider the region:

Set-up iterated integrals that compute the area of .

First we’ll integrate with respect
to and then . Here and so our integral is:

Now let’s integrate with respect to and then . If we give absolute bounds to our inequalites
above, we see something strange: This is actually two inequalities (note we can see this
from the graph):

Solving for in these inequalities we find:

Here is bounded by two different curves! When this happens, it is probably best to
look at the graph of the situation:

Thus we see that the area can be found by the sum:

Why change the order of integration?

You may ask yourself, “Why change the order of integration?” Sometimes changing
the region can make a difficult (impossible) antiderivative easier (not impossible).
Let’s see an example.

Compute:

Here you don’t stand a chance if you try to antidifferentiate with
respect to . So the trick is to immediately swap the order of antidifferentation.
Starting by writing absolute bounds for and : Now we see that:

This author does not know how to solve the problem above without changing the
order of integration.

Triple integrals and volume

We start by introducing a fourth version of Fubini’s Theorem.

Fubini Let be a closed, bounded region in and let be a continuous function on . If

where and are continuous functions on , and and are continuous on the
region then There are six reorderings total with three variables. We will
spare the young mathematician the details, and trust that you will sort it
out.

Again note that when using Fubini’s Theorem, we must always have numbers as the
limits of the outer-most integral, curves as the limits of the middle integral, and
surfaces as the limits of the inner-most integral:

Now with Fubini’s help, we will use triple integrals to compute volumes.

Set-up and evaluate an iterated integral that will compute the volume of the region
bounded by:

The plane .

The plane .

The plane .

The plane .

Our region above produces a tetrahedron, a triangular-based pyramid. It intersects
the -axis at , the -axis at , and the -axis at . The region can be defined by:

The volume of this region is given by, write with me,

In the previous example, we integrated with respect to , then , then . Set-up
an integral that computes the volume of that integrates with respect to ,
then , then . Start by finding overall bounds for our variables. In this case:

At this point we see that our bounds for are to . Now we will find our
bounds for . We must find an expression for in terms of and . Write with me:

However, we know that is nonnegative, so , and is bounded below by . So runs from
to .

Finally we must write in terms of . Unfortunately, from our inequalities above, there
is no direct way to get this. We must think about what our solid looks like. Recall
that the plane bounding the solid is . If , then our plane is the line . Hence We may
now write our desired integral:

Set-up an iterated integral that will compute the volume of the region bounded by
the cone below:

First note that

Hence, the volume of this region is given by

In the previous example, we integrated with respect to , then , then . Set-up an
iterated integral that computes the volume of that integrates with respect to , then ,
then .