Now, simply add up the -values found in the rectangles that are within our region and multiply by . Compute .

We study integrals over basic regions.

### Double integrals

Suppose you have a function . A graph of this function is a surface in . For example:

We are interested in the “signed volume” or “net volume” between this surface and the -plane. This means that the space above the plane under will have a positive volume. Space above and under the -plane will have a “negative” volume. This is similar to the notion of “signed area” used before. If we want to compute the signed volume of a surface defined by over a rectangular region, say the rectangle defined by we break into slices parallel to the -axis, and slices parallel to the -axis. This allows us to consider boxes of dimension where is a point in -rectangle:

Computing the volume of each of these boxes approximates and summing them together approximates the signed volume enclosed by the surface:

Letting the number of rectangles in the -direction and -direction go to infinity, we
will have that goes to zero, and we will find the exact volume enclosed by our
surface when bounded by the region . This leads to our definition of a *double
integral*:

How do we compute a double integral with calculus? We use an *iterated integral*. At
this point we will introduce something called *Fubini’s Theorem*.

#### Fubini’s Theorem

Fubini’s Theorem gives us a recipe for computing double integrals. In this class, we
are going to have many different versions of Fubini’s Theorem. The common factor
between all of these theorems is that with each, the “punch-line” will be: where an
*iterated integral* is nothing more than two applications of our familiar friend/foe: the
single integral.

**two**things at once:

- Double integrals can be computed via iterated integrals.
- The order of integration can be changed, as long as the bounds are changed as well.

Now let’s work some examples:

Now integrate with respect to , treating as a constant:

Now let’s compute this integral using a **different order of integration.** Write with me,

Note our answers are **the same** regardless of the order of integration.

In our next example, we will see that it is sometimes easier to apply Fubini’s Theorem and integrate with respect to one variable instead of the other.

Now we notice that this is an improper integral! We must therefore compute Now we must integrate with respect to . This is also tricky. Our second trick is to rewrite our current integrand as and now “see” that this results from the quotient rule being applied to So now we must compute:

Now, for our third trick, we’ll use L’Hôpital’s rule. Note that the numerator and denominator both go to zero and are differentiable, so this is OK.

So putting this all together, we have Whew. This was hard!

Now use Fubini’s Theorem and integrate with respect to first! Fubini’s theorem says:
provided you switch the order of integration. Note, in this case the limits of integration
for both and are the same, but we *did* switch them. Let’s get on with it! Write with me:

Man alive! That was easy! Note, we get the same final answer regardless of the order of integration. Thank-you Fubini!

### Triple integrals

Using a similar technique to how we made boxes to define double integrals, we can
make four-dimensional *boxes* to define a triple integral that computes the signed
hypervolume bounded by a hypersurface and a three-dimensional region. How do we
compute a triple integral with calculus? We use our second version of Fubini’s
Theorem. This time, it will say something like:

*six*combinations of orders of integration will work, and be equal, provided that the bounds are changed appropriately.

Let’s do an example.

We’ve just begun our journey with multiple integrals. Next, we’ll think about more complex regions! For some interesting extra reading check out: