A number of good review-type problems involving setting up integrals and what-have-you.

**1. Problem**:
Set up double and triple integrals for the volume inside
both of the cylinders
*x*^{2} +
*y*^{2} = 1 and
*y*^{2} +
*z*^{2} = 4.

**Answer**: Note that we don't actually have to be
able to see the three dimensional volume to set this up. For
the triple integral, we want
,
where *E* is the volume we're considering. To do this,
think about projecting it into one plane and looking at
.

What do the cylinders look like? A quick sketch is shown in figures 1 and 2 to the right.

We'll project into the *xy*-plane, where the projection is just
*x*^{2} +
*y*^{2} = 1. So the integral
over the projection in the *xy*-plane is easily written in polar
coordinates:
.

Then the limits on the innermost integral are determined by the range
of *z*-values taken on in the volume. These are given by the
second cylinder,
*y*^{2} +
*z*^{2} = 4.
On the bottom of this,
*z* = -sqrt(4 - *y*^{2})
and on the top,
*z* = sqrt(4 - *y*^{2}).

Thus the volume integral is
Volume = (after
converting *y* into polar coordinates).

The double integral giving volume is directly related to this: we want
, where (height) is the vertical height
of the volume at any point in the projected region *R*. This is
the same as simply evaluating the innermost integral in the
cylindrical coordinates triple integral that we found above. Thus we
get
Volume = .

**2. Problem**:
If *delta* = 3*x* + *y* + *z* is
the mass of an "ice cream cone", set up an integral in spherical,
cylindrical and rectangular coordinates to find the mass.
The side of the cone is
*z* = 2 sqrt(*x*^{2} +
*y*^{2})
and the top is a spherical cap with radius 3.

**Answer**:
The ice cream cone is shown in figure 3 to the right. Because we
are told that the boundaries are a *spherical cap* and a
*cone*, both of which are given by constant values of
*rho* and *phi*, respectively, we would normally set this up
in spherical coordinates. In spherical coordinates, we can
immediately see that
0 <= *rho* <= 3 and
0 <= *theta* <= 2*pi*.

To find the range for *phi*, look at the figure from the side:
the smallest value of *phi* that we want is when
*phi* = 0 (that is, along the *z*-axis), and the
largest value gives the edge of the cone. Because the figure looks
the same from all sides, we can consider any view we like. Look at
the *yz*-plane. There *x* = 0, so *z* =
2*y*. This says that the ratio of the two legs in the triangle
drawn on the right side of the cone is 1:2 (*y*:*z*), and
the maximum angle is *phi* = arctan(1/2). (Be sure that you
see why this is.)

Thus, the integral in spherical coordinates (after converting the
density *delta* into spherical coordinates) is

What about cylindrical coordinates? In this case we project down into
the *xy*-plane to get a circle. The radius of the circle is
where the cone intersects the sphere of radius 3, so we have both that
*z* = 2 sqrt(*x*^{2} +
*y*^{2})
and
*x*^{2} +
*y*^{2} +
*z*^{2} = 9. Thus
*x*^{2} +
*y*^{2} +
4 (*x*^{2} +
*y*^{2}) = 9, or
5 (*x*^{2} +
*y*^{2}) = 9,
so that
*r* = 3/sqrt(5).

Thus the outer two limits for the cylindrical coordinates integral are
0 <= *r* <= 3/sqrt(5) and
0 <= *theta* <= 2*pi*.

For the *z* limits, notice that at any point in the circle the
bottom value of *z* we want is given by the cone, and the top is
given by the sphere. Thus we must have
2 sqrt(*x*^{2} +
*y*^{2}) < =
*z* <=
sqrt(9 - (*x*^{2} +
*y*^{2})), or
2 *r* <= *z* <=
sqrt(9 - *r*^{2}).

Thus the final integral in cylindrical coordinates is

Finally, in rectangular coordinates, we get the integral

The spherical coordinates integral is nicer.

**3. Problem**:
It the top of the ice cream cone is flat, say, at *z* = 3,
how does your answer change?

**Answer**:
In this case it doesn't make sense to use spherical coordinates.
The
flat top (*z* = 3) is *not* well described in
spherical coordinates, so we'd default to cylindrical coordinates.
The integral will be similar to that shown above, but because we've
said that *z* = 3, *not* *rho* = 3 the
radius of the circle changes and we get

Check that you understand how all of the limits were arrived at for this problem.

**4. Problem**:
Sketch and convert to cylindrical coordinates the integral

**Answer**:
If we didn't have to sketch this, we could do the conversion almost
routinely. The domain in the *xy*-plane we can see from the
outer two limits in the integral above is a quarter circle with radius
2. Thus we can convert the integral without thinking too hard: the
outer two limits are
0 <= *r* <= 2 and
0 <= *theta* <= *pi*/2.
Finally, we just have to change the *z* limits into polar
coordinates, and we're done:

What does the region look like, though? The domain in the
*xy*-plane is a quarter circle, and the top of the figure is
determined by the plane
*z* = 2sqrt(2) - *x* - *y*. A crude
sketch is shown to the right above.

review

Last modified: Wed Nov 17 18:06:03 EST 2004

Comments to:glarose(at)umich(dot)edu

©2004 Gavin LaRose, UM Math Dept.