Is Polar Conversion the Best Approach for This Double Integral Problem?

[CINA]

Homework Statement

http://img23.imageshack.us/img23/3118/intx.th.jpg

Homework Equations

I'm guessing polar conversion?

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

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 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.

 

[Hootenanny]

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

[tex]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}[/tex]

[tex]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)[/tex]

[tex]I = \int_{-1}^1dz\left(2\sqrt{1-z^2}\right)[/tex]

 

[CINA]

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

[tex]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}[/tex]

[tex]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)[/tex]

[tex]I = \int_{-1}^1dz\left(2\sqrt{1-z^2}\right)[/tex]

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.

 

[Hootenanny]

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.

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.