# Help with a Double Integral

1. Dec 1, 2009

### CINA

1. The problem statement, all variables and given/known data
http://img23.imageshack.us/img23/3118/intx.th.jpg [Broken]

2. Relevant equations

I'm guessing polar conversion?

http://en.wikipedia.org/wiki/Polar_...rting_between_polar_and_Cartesian_coordinates

3. The attempt at a solution

I'm having trouble tackling this problem, on one hand Cartesian coordinates seem like a hassle, but turning this into a polar problem will give http://img269.imageshack.us/img269/8738/int2o.th.jpg [Broken] right? This seems like an equally unpleasant integral, what with the cos^3 and all. Am I setting this problem up incorrectly? Can someone tell me the first few steps in doing this problem? I'm fine understanding the problem which the integral applies, just the mechanics on it is getting me.

Last edited by a moderator: May 4, 2017
2. Dec 1, 2009

### Hootenanny

Staff Emeritus
Why are you bothering with polar coordinates? I think that this problem is simpler than you're making out

$$I = \int_{-1}^1dz\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \left(1-2x^3\right) dx = \int_{-1}^1 dz \left[x-\frac{1}{2}x^4\right]_{-\sqrt{1-z^2}}^\sqrt{1-z^2}$$

$$I = \int_{-1}^1dz\left(2\sqrt{1-z^2} - \frac{1}{2}\left(1-z^2}\right)^2 + \frac{1}{2}\left(1-z^2}\right)^2\right)$$

$$I = \int_{-1}^1dz\left(2\sqrt{1-z^2}\right)$$

3. Dec 1, 2009

### CINA

Doh! That was a lot easier than I thought it was(!), that's the perils of late-night mathematics I suppose. Thanks a lot! You've saved me many needlessly lost hairs.

4. Dec 1, 2009

### Hootenanny

Staff Emeritus
Not a problem.

A handy rule to remember is that if you integrate an odd function (such as x3) over a symmetric interval about the origin (as we have here), the result is always zero.