Originally Posted by

**Ackbeet** No, I don't think you've quite got it yet. Just plugging in gives:

$\displaystyle \delta=10z^4+6(x^2-y^2)^2=10z^4+6((r\cos(\theta))^2-(r\sin(\theta))^2)^2=10z^{4}+6(r^{2}\cos^{2}(\thet a)-r^{2}\sin^{2}(\theta))^{2}$

$\displaystyle =10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=

10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s in^{2}(2\theta).$

You see how I work here: step-by-step. I don't often skip steps.

The other thing you have to worry about is that the volume differential in cylindrical components is $\displaystyle r\,dr\,d\theta\,dz,$ not just $\displaystyle dr\,d\theta\,dz.$

So, putting this all together, what does your integral look like now?