integration review problems

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 x² + y² = 1 and y² + z² = 4.

figure 1: the first cylinder

figure 2: the second cylinder

figure 3: the "ice cream cone"

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 x² + y² = 1. So the integral over the projection in the xy-plane is easily written in polar coordinates: $int_0^{2pi} int_0^1 dA$ .

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, y² + z² = 4. On the bottom of this, z = -sqrt(4 - y²) and on the top, z = sqrt(4 - y²).

Thus the volume integral is Volume = $int_0^{2pi} int_0^1 int_{-sqrt(4-(r sin(t))^2)}^{sqrt(4-(r sin(t))^2)} r dz dr dt$ (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 = $int_0^{2pi} int_0^1 2 sqrt(4 - (r sin(t))^2) r dr dt$ .

2. Problem: If delta = 3x + 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(x² + y²) 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 <= 2pi.

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 = 2y. 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

$int_0^{2pi} int_0^3 int_0^arctan(1/2) (3 rho sin(phi) cos(theta) + rho sin(phi) sin(theta) + rho cos(phi)) (rho^2 sin(phi)) dphi drho dtheta$

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(x² + y²) and x² + y² + z² = 9. Thus x² + y² + 4 (x² + y²) = 9, or 5 (x² + y²) = 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 <= 2pi.

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(x² + y²) < = z <= sqrt(9 - (x² + y²)), or 2 r <= z <= sqrt(9 - r²).

Thus the final integral in cylindrical coordinates is

$int_0^{2pi} int_0^{3/sqrt(5)} int_{2r}^{sqrt(9-r^2)} (3 r cos(theta) + r sin(theta) + z) (r) dz dr dtheta$

Finally, in rectangular coordinates, we get the integral

$int_{-3/sqrt(5)}^{3/sqrt(5)} int_{-sqrt((9/5)-x^2)}^{sqrt((9/5)-x^2)} int_{2sqrt(x^2+y^2)}^{sqrt(9-(x^2+y^2))} (3x + y + z) dz dy dx$

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?

figure 4: the last region

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

$int_0^{2pi} int_0^{3/2} int_{2r}^{3} (3 r cos(theta) + r sin(theta) + z) (r) dz dr dtheta$

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

$int_0^2 int_0^{sqrt(4-y^2)} int_0^{2sqrt(2)-x-y} 5xy dz dx dy$

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:

$int_0^{pi/2} int_0^2 int_0^{2sqrt(2)-r cos(theta)-r sin(theta)} 5r^2 cos(theta) sin(theta) r dz dr dtheta$

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.