Calculate $ \int\limits_{-1}^{1} \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx$ Is there a nice way to substitute something in this double integral:
$$ \int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx$$
Can calculate it easier?
Wolfram Alpha gives me $\sqrt{2}\pi$ as solution.
 A: Examine the integral limits and recognize that the integration is over the elliptical region given by
$$x^2 + \frac{(y+1)^2}2 =1$$
Rescale the valuables with $u=x$ and $v=\frac{(y+1)}{\sqrt2}$ to transform the region to the unit circle $u^2+v^2=1$. As a result, $dxdy = \sqrt2 dudv$ and the integral simplifies to
$$I=\sqrt2 \int_{u^2+v^2\le 1} 2(1-u^2-v^2)dudv$$
Then, integrate in its polar coordinates to obtain
$$I=2\sqrt2 \int_0^{2\pi}\int_0^1 (1-r^2)rdr d\theta= \sqrt2\pi$$
A: Integrate each element in turn and find:
$$2+\pi$$
$$-\pi$$
$$-\frac{20 + 9 \pi}{6}$$
$$4 + 2 \pi$$
Final result:  $\frac{8}{3}+\frac{\pi }{2}$
A: $$\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx\\
\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2-y^2-2y-1) \,dy\,dx\\
\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2)- (y+1)^2 \,dy\,dx$$
This now integrates more elegantly
$$\int\limits_{-1}^1 (2-2x^2)- \frac 13 (2-2x^2)^{\frac 32} \,dx$$
We should probably break this up into two integrals at this point, but each is pretty straightforward.
