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$

Question

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.

Answer 1

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

$$I=2\sqrt2 \int_0^{2\pi}\int_0^1 (1-r^2)rdr d\theta= \sqrt2\pi$$

Answer 2

Integrate each element in turn and find:

$$2+\pi$$

$$-\pi$$

$$-\frac{20 + 9 \pi}{6}$$

$$4 + 2 \pi$$

Final result: $\frac{8}{3}+\frac{\pi }{2}$

Answer 3

$$\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx\\ \int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2-y^2-2y-1) \,dy\,dx\\ \int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2)- (y+1)^2 \,dy\,dx$$

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.