We need to find the integral $$\iiiint\limits_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}dxdydudv$$ I was only able to get to this point ... $$\iiiint\limits_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}dxdydudv=\iint\limits_{x^2+y^2\leq 1}e^{x^2+y^2}\left ( \iint\limits_{u^2+v^2\leq 1-x^2-y^2}\frac{dudv}{e^{u^2+v^2}} \right )dxdy$$ I don’t know how to solve it further ...
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1 Answer
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Use polar coordinates (twice):
In the inner integral change variables
$u\rightarrow r\cos{\theta}$, $v\rightarrow r\sin{\theta}$, to get
$$\iint\limits_{u^2+v^2\leq 1-x^2-y^2}\frac{dudv}{e^{u^2+v^2}}= \int_{0}^{2\pi}\int_{0}^{\sqrt{1-x^2-y^2}}r e^{-r^2} dr\,d\theta =\pi \left(1-e^{-1}e^{x^2+y^2}\right)$$
Then use polar coordinates once more to compute
$$\int_{0\leq x^2+y^2\leq 1} e^{x^2+y^2} dx dy=\int_{0}^{2\pi} \int_{0}^{1} r e^{r^2} dr d\theta=\pi (e-1)$$
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1$\begingroup$ I just want to add that you had already solved the tricky part right... $\endgroup$– MedoCommented Apr 20, 2021 at 22:43
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1$\begingroup$ Oh, everything was so simple ... I did not see the transition to polar coordinates. Thank you! $\endgroup$– DmitryCommented Apr 20, 2021 at 22:51