polar coordinate subtitution 
$$\int\int_{(x^2+y^2) \le1} \left(\frac{x^4}{a^4}+ \frac{y^4}{b^4} \right)dxdy$$ 

I try to solve this multivariable integration, by substitute $x=ar \sin \theta$ and $y=br \sin \theta$
to  $(\frac{x^4}{a^4}+ \frac{y^4}{b^4} )$ so it will produce $r^4 $ with Jacobin $abr$ but when I substitute to $x^2+y^2 \le1$
it won't produce $r^2$ 
$a^2r^2\sin^2 \theta+b^2r^2 \cos^2 \theta \le1$, so I cannot change it to polar coordinate.
Is my approach wrong? Can someone give me hint too!  
 A: Just use plain old polar coordinates. Let $x = r\cos \theta$ and $y = r\sin \theta$, then everything boils down to
\begin{align}
\int^{2\pi}_0 \int^1_0\ \left( \frac{r^4 \cos^4\theta}{a^4}+\frac{r^4\sin^4\theta}{b^4}\right)\ rdrd\theta = \frac{1}{6} \int^{2\pi}_0 \frac{\cos^4 \theta}{a^4}+\frac{\sin^4\theta}{b^4}\ d\theta.
\end{align}
Additional:
\begin{align}
\int^{2\pi}_0 \cos^{2n}\theta\ d\theta =&\ \frac{1}{2^{2n}}\int^{2\pi}_0 \left( e^{i\theta}+e^{-i\theta}\right)^{2n} d\theta = \frac{1}{2^{2n}} \int^{2\pi}_0 \sum^{2n}_{k=0} \binom{2n}{k}e^{ik\theta} e^{-i(2n-k)\theta}\ d\theta\\
 =&\ \frac{1}{2^{2n}}\sum^{2n}_{k=0}\binom{2n}{k} \int^{2\pi}_0 e^{i(2k-2n)\theta}\ d\theta = \binom{2n}{n}\frac{\pi}{2^{2n-1}} = \frac{(2n)! \pi}{(n!)^2 2^{2n-1}}
\end{align}
and
\begin{align}
\int^{2\pi}_0 \sin^{2n}\theta\ d\theta =&\ \frac{(-1)^n}{2^{2n}} \int^{2\pi}_0 (e^{i\theta}-e^{-i\theta})^{2n}\ d\theta = \frac{(-1)^n}{2^{2n}} \int^{2\pi}_0 \sum^{2n}_{k=0} \binom{2n}{k}(-1)^{2n-k}e^{i(2k-2n)\theta}\ d\theta\\
=&\  \frac{(-1)^n}{2^{2n}} \sum^{2n}_{k=0} \binom{2n}{k}(-1)^{2n-k} \int^{2\pi}_0 e^{i(2k-2n)\theta}\ d\theta = \frac{(2n)!\pi}{(n!)^22^{2n-1}}.
\end{align}
