HELP Setting up triple integral in spherical coordinate

kyva1929
Messages
16
Reaction score
0
HELP! Setting up triple integral in spherical coordinate

Homework Statement



http://img517.imageshack.us/img517/9139/83291277.jpg

Homework Equations



I set up the bound for this problem as following:

r=0..2/cos(phi), phi=pi/2..3pi/4, theta=0..2pi, but maple always return an error in calculating such integral. I've included the Jcobian transformation in the integral, and convert x^2+y^2 to rho^2 sin(phi). I found that the problem was due to the denominator cos(phi) become zero at the lower bound. But the volume representing the respective volume integral makes sense, so how to correct for this? Am I setting up the integral wrong?

second problem:

http://img695.imageshack.us/img695/3585/38264414.jpg
The same thing happens, please help!

Thank you!
 
Last edited by a moderator:
Physics news on Phys.org


Use cylindrical coordinates instead.
 


Thank you in advance. But is there a way to do it in spherical coordinate?

And for question 2, I did the command int(function,[x=..,y..,z..]) and it returns error

Error, (in assuming) when calling '`property/ConvertProperty`'. Received: 'real is an invalid property'

but I can evaluate it either by hand or by evaluating the integrals one by one in maple and that gives me [2*Pi*(3^21-1)]/21.

What's wrong?

Thank you
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top