Triple IntegrationDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/