So I find myself stuck on this integral where $R^2$ is the real plane:
$$\int\int_{R^2}x^2e^{-\sqrt{x^2+y^2}} $$
I know I'm suposed to perform some variable substitution but I just don't see it. Polar coordinates got me nothing but trouble. When I try that path I end up with something looking like this:
$$\frac{1}{2}\int_0^{2\pi}(\cos{2\theta}+1)d\theta\int_0^\infty \frac{r^3}{e^r}dr$$
In which the second integral gives me grief.
It seems it's been a long week but thanks to Winther I realised that it is possible to continue with my integral after polar coordinate substitution using partial integration thrice:
$$ \pi\int_0^\infty \frac{r^3}{e^r}dr=\pi\Big(\big[-e^{-r}r^3\big]^\infty_0 +3\int_0^\infty\frac{r^2}{e^r}dr\Big)=\\ =\pi\Big(\big[-e^{-r}r^3\big]^\infty_0+\big[-3e^{-r}r^2\big]^\infty_0+6\int_0^\infty\frac{r}{e^r}dr\Big)=\\ =\pi\Big(\big[-e^{-r}r^3\big]^\infty_0+\big[-3e^{-r}r^2\big]^\infty_0+\big[-6e^{-r}r\big]^\infty_0+6\int_0^\infty e^{-r}dr\Big)=$$
$$=\pi\Big(-0+0+3\times0-3\times0-6\times0+6\times0-6\times0+6\Big)=6\pi. $$