Triple Integration

Date: 05/09/2003 at 10:49:40
From: Mike
Subject: Triple integration

I have to solve this triple integral.

Consider a frustum with a lower circular base radius equal to R, and 
an upper circular base radius and the height both equal to R/2. If 
the center of the lower base is located at the origin (so the center 
of the upper base is on the z-axis), evaluate the following triple 
integral: S = integral sign

SSS(z(x^2+y^2))dV, and E is the frustum that is described above. E

I know that I need to start with rectangular coordinates and then 
convert to polar coordinates using spherical coordinates. I know 
that the z-axis is not needed, as it does not change, and theta is 
between 0 and 2Pi.

Date: 05/09/2003 at 13:53:26
From: Doctor Luis
Subject: Re: Triple integration

Hi Mike,

To solve this problem, I suggest cylindrical coordinates, not 
spherical coordinates, where

  r = sqrt(x^2 + y^2), x = r cos(theta), y = r sin(theta)

This form is useful because our figure is symmetrical with respect to 
theta, and also, the integrand zr^2 is independent of theta: therefore 
the theta integral from 0 to 2pi just contributes a constant factor of 
2pi to the triple integral.

Look at the following diagram:



You can clearly see that as z changes from from 0 to height R/2, the 
radius r of the circle changes too, from radius R down to radius R/2. 
You can see that as z changes, r(z) = R-z.

So,

  / / /                  / 2pi / R/2 / r(z)
  | | |                  |     |     |
  | | | z*(x^2+y^2) dV = |     |     |  z*r^2 dr dz d(theta)
  | | |                  |     |     |
 / / /                  / 0   / 0   / 0
frustum

First integrate zr^2 with respect to r, and evaluate from r = 0 to 
r = R-z. Then integrate that answer with respect to z, and then
evaluate from z = 0 to z = R/2. Then integrate all that with respect
to theta. Since there's no theta dependence, this last step is easy: 
it's just a constant factor of 2pi. Does this make sense?

Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/