Here we consider integrating the function
*f*(*x*,*y*,*z*) = *xz* over the volume
inside
*x*^{2} +
*y*^{2} = 9,
in the first octant, and under the plane
*z* = *y* + 2.
This region is shown to the right, below.

To set up the integral, we first need to decide on a coordinate system
in which to work. Because this volume is defined on part of a
cylindrical domain, and because it has straight (in *z*) sides,
*cylindrical* coordinates are a good choice. The fact that the
sides are vertical in *z* makes *spherical* coordinates
less desirable. We could do it in *rectangular* coordinates,
but in general with a region like this we expect the choice of
cylindrical coordinates to be the best.

Next, we figure out the limits of integration. We'll do this two ways: first by thinking about the outer two limits of integration as a block, then by thinking about the inner two together.

**Thinking about the outer two limits**:

For the first of these, we think about the outer two variables first.
To do this, we choose which they'll be, and project the volume down
into the plane defined by those two variables. It makes sense here to
consider the *r*-*theta* plane, which is the same as the
*xy*-plane. This projection is shown in figure 1 to the
right.

We see that the projection is just a quarter circle of radius 3, so
the limits on *r* and *theta* are
0 <= *r* <= 3 and
0 <= *theta* <= pi/2. Because both of
these sets of limits are constant, it doesn't matter what order we put
them in.

Finally, we need the limits for the inner variable, *z*. To find
these, we think about any point in the projection and mentally extend
lines in the *z* direction, seeing what the first and last values
of *z* that are of interest are.
To see this visually,
click here a couple of times
(click here to clear the
arrows). Looking at these,
we can see that the limits are
0 <= *z* <= *y*+2, or
0 <= *z* <= *r*sin(*theta*)+2. Thus,
noting that
*f*(*x*,*y*,*z*) = *xz* is the same as
*f*(*x*,*y*,*z*) =
*r* cos(*theta*) *z*,
the final integral is

,
or

**Thinking about fixing the outer variable**:

For the second method, we're thinking about just one outer variable
first, and then find what "slices" of the volume look like when that
variable is held fixed. Here, either *r* or *theta* are the
best outer variables to consider, because with *z* fixed for
*z* <2 the slices have a different shape than when we
look at *z* >2. (Be sure you can see this by looking at
the figure.)

Let's think about *theta* as the outermost variable. Then a
slice with *theta* fixed is shown in figure 2. The possible
values for *theta* are any values in the first octant, so we need
0 <= *theta* <= pi/2.

Now in each slice, we want to think about forming a double integral in
the remaining variables (*r* and *z*). We'll do this the
same way we normally do: pick one for the outer variable, find the
limits of integration for that, and then at each value of that outer
variable figure out what the range of values for the inner variable is.

Let's think about *r* as the outer variable in the slice. Notice
that no matter where we make the slice we always have
0 <= *r* <= 3,
so this looks like a good choice for the outer slice variable. Then
for each value of *r*, what does *z* do? As before, we
think about taking any value of *r* in the slice (which can be at
any of the possible values for *theta*, of course), and draw
vertical lines to see what the values of *z* are
click here a couple of times to
see this visually
(click here to clear the
arrows). Looking at these,
we can (again) see that the limits are
0 <= *z* <= *r*sin(*theta*)+2. Thus,
the final integral becomes

,

which, because the outer two limits are constant, can be rewritten as

**Evaluating the integral**:

A quick integration. It's easiest to evaluate this
*dz* *dtheta* *dr*:

(Note that to do the integration in *theta*, we use the
substitution *u* = (*r* sin(*theta*) + 2).

*(How were the figures here generated? In Maple, with this
maple worksheet.)*

3d integration: review problem

Last modified: Fri Mar 19 12:55:14 EST 2004

Comments to:glarose@umich.edu

©2004 Gavin LaRose, UM Math Dept.