TITLE: $P(X^2+Y^2<1)$ of two independent n(0,1) random variables
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: Since $X$  and $Y$ are independent, and because $\mu=0$ and $\sigma=1$, then we know that $$f_{X,Y}(x,y)=f_X(x)f_Y(y)=\frac{1}{2\pi}\exp(\frac{-(x^2+y^2}{2}).$$
Of course, since our domain over which we are integrating is a circle with a radius of 1,  we convert $x^2+y^2=r^2$, and $\text{d}x \text{d}y=r\text{d}r\text{d}\theta$, using 
$x=r\cos\theta$ and $y=r\sin \theta$ to get the standard polar coordinate transformation.  Notice, we are converting the joint pdf $f_{X,Y}(x,y)$ to make our lives easier, and the circle defined by the question, ie $X^2+Y^2<1$, merely comes into play when looking at what we integrate over.  
\begin{align*}
P(X^2+Y^2<1)&=\frac{1}{2\pi}\int_{r=0}^{1}\int_{\theta=0}^{2\pi}e^{(\frac{-r^2}{2})}r \text{d}r \text{d}\theta\\
&=2\pi \frac{1}{2\pi}\int_{r=0}^{1}e^{(\frac{-r^2}{2})}r \text{d}r\\
&=\int_{u=1}^{e^{-1/2}}-\text{d}\underbrace{(e^{-r^2/2})}_u=-u\bigg|_1^{e^{-1/2}}\\
&=-e^{1/2}--1=1-e^{-1/2}
\end{align*}
The u-substitution may add some confusion, especially with changing the limits.  For this question, it may be more obvious to just skip the explicit substitution, and just leave $-\text{d}e^{-r^2/2}$ with $e^{-r^2/2}$ the variable in the differential term, as that may be clearer and more obvious, as this is a well known derivative.