Compute $\iint_D (x^4-y^4) \,dx\,dy$ How should I proceed? the question is :
$$\iint_D (x^4-y^4) \,dx\,dy$$
$$D= \left\{(x,y):1<x^2-y^2<4, \quad \sqrt{17}<x^2+y^2<5,\quad x<0,\ \  y>0\right\}$$
I've tried to solve it with change of variable:
$$u=x^2-y^2, \quad v=x^2+y^2$$ $$|J|=\frac{1}{8xy}, \quad \frac{1}{8}\iint uv\frac{1}{xy} \,du\,dv$$
How should I proceed?
Any suggestion would be great, thanks

Oddly enough polar coordinates will do the trick and does it well as an alternative. The region is one integral if done angular first (which is easy to see because the region is between two circular arcs)
$$1<x^2-y^2<4 \implies \pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)<\theta<\pi-\frac{1}{2}\cos^{-1}\left(\frac{4}{r^2}\right)$$
in the second quadrant. Which makes the integral
$$\int_{\sqrt[4]{17}}^{\sqrt{5}}\int_{\pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)}^{\pi-\frac{1}{2}\cos^{-1}\left(\frac{4}{r^2}\right)}r^5\cos2\theta\:d\theta dr = \int_{\sqrt[4]{17}}^{\sqrt{5}} \frac{r^3}{2}\left(\sqrt{r^2-1}-\sqrt{r^2-4}\right)dr$$
$$ = \int_{\sqrt{17}}^{5} \frac{t}{4} \left(\sqrt{t-1}-\sqrt{t-4}\right)dt$$
Integration by parts takes care of the rest.
$$ = \boxed{\frac{113}{15}-\frac{\sqrt{17}}{6}\left[(\sqrt{17}-1)^{\frac{3}{2}}-(\sqrt{17}-4)^{\frac{3}{2}}\right]+\frac{1}{15}\left[(\sqrt{17}-1)^{\frac{5}{2}}-(\sqrt{17}-4)^{\frac{5}{2}}\right]}$$
As a side note, this answer would have come out a lot more cleanly had the lower bound been $4$ instead of $\sqrt{17}$ giving
$$\frac{113}{15}-\frac{7\sqrt{3}}{5}$$
