$P(X^2+Y^2<1)$ of two independent n(0,1) random variables Suppose that X and Y are independent n(0,1) random variables.
a) Find $P(X^2+Y^2<1)$
Attempt: 
a) Let $U = X^2 + Y^2$, $V = Y$.
Then $X = \sqrt{V^2 -U}$, $Y = V$.
$J  = \left| \begin{array}{ccc}
\frac{-1}{\sqrt{V^2-U}} & \frac{V}{V^2-U} \\
0 & 1\\
\end{array} \right| $
Then the joint distribution of $f_{u,v}(u,v)$ is:
$$f_{u,v}(u,v)= \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}$$
Then $P(X^2 +Y^2 <1)$ is:
$$\int_0^\infty \int_0^{v^2-u} \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}dudv$$
However, at this point I simply do not know how any tricks to complete this integration.

Since $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is the product of the individual densities, so it is
$$\frac{1}{2\pi}e^{-(x^2+y^2)/2}$$
on the entire $x$-$y$ plane. We want to find
$$\int_D \frac{x^2+y^2}{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx,$$
where $D$ is the unit disk. Switch to polar coordinates. We want to find
$$\int_{r=0}^{1}\int_{\theta=0}^{2\pi} \frac{r^3}{2\pi}e^{-r^2/2}\,d\theta\,dr.$$
The integration with respect to $\theta$ gets rid of the $2\pi$ in the denominator.
So we want
$$\int_0^1 r^3 e^{-r^2/2}\,dr.$$
This can be done by integration by parts. To make things more familiar, you may wish to first make the substitution $t=r^2/2$. Our integral becomes
$$\int_0^{1/2} 2te^{-t}\,dt.$$
