Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$ Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$
The original problem is : 
$$\text{Evaluate} \iint _R ye^{-x^2-y^2}dxdy$$
Where $R=\left\{ (x,y) \vert x\geq0,y\geq0\right\}$
I sub-ed $x=r\cos\theta, y=r\sin\theta$ and this changed to
$\left(\int_0^{\frac{\pi}{2}}\sin\theta d\theta \right)\left(\int_0^{\infty} r^2 e^{-r^2}dx\right)$
So I must calculate  $\int_0^{\infty} r^2 e^{-r^2}dr$, is there an easy way?

Let $z=\sqrt{2}r$ so that $dz=\sqrt{2}dr$ and $r^2=\frac{1}{2}z^2$. Then:
$$
\int_0^\infty r^2e^{-r^2}dr=\int_0^\infty\frac{1}{2}z^2e^{-\frac{1}{2}z^2}\frac{dz}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\int_0^{\infty}z^2e^{-z^2/2}dz=\frac{1}{4\sqrt{2}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz.
$$
The last equality above uses symmetry. Next, we note:
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz=1
$$
because it is the second-moment of a standard normal distribution. And so:
$$
\int_0^\infty r^2e^{-r^2}dr=\frac{1}{4\sqrt{2}}\sqrt{2\pi}=\frac{\sqrt{\pi}}{4}.
$$
