<s>[INST] 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.
 [/INST] $$\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx\\
\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2-y^2-2y-1) \,dy\,dx\\
\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2)- (y+1)^2 \,dy\,dx$$
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.
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