$P(X^2+Y^2<1)$ 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.

Box-Muller transform suggests that if $X, Y$ are two independent standard normal random variables, then if we let $X = R cos\Theta, Y = R sin\Theta, R \in (0, \infty), \Theta \in [0, 2\pi)$ (polar coordinate), then this bivariate transform is 1-to-1 and onto. Further, $R, \Theta$ are also independent. And $\frac{1}{2}R^2 \sim Exponential(1), \Theta \sim Uniform(0, 2\pi)$, i.e., $\frac{1}{2}R^2$ follows an Exponential distribution with $\lambda = 1$ and $\Theta$ follows a Uniform distribution on $(0, 2\pi)$.
Hence, $P(X^2 + Y^2 < 1) = P(R^2cos^2\Theta + R^2sin^2\Theta < 1) = P(R^2 < 1) = P(\frac{1}{2}R^2 < \frac{1}{2})$. And since $\frac{1}{2}R^2 \sim Exponential(1)$, we can denote $Z = \frac{1}{2}R^2$. Using the pdf of an $Exponential(1)$ distribution, $P(\frac{1}{2}R^2 < \frac{1}{2}) = P(Z < \frac{1}{2}) = \int_{-\infty}^{\frac{1}{2}} e^{-z} \, dz = 1 - e^{-\frac{1}{2}}$.
