Is this a Cauchy random variable? Suppose $X$ and $Y$ are independent $n(0,1)$ random variables.
How do I find $P(X^2+Y^2\leq 1)$ and $P(X^2\leq 1)$, after verifying that $X^2$ is distributed $\chi_{1}^{2}$? 
My attempt:
$$P(X^2+Y^2\leq 1)= \int_{-\infty}^{\infty}\int_{-\infty}^\sqrt{1-y^2} \frac{1}{2\pi} e^\frac{-x^2+y^2}{2} dxdy $$
How do I proceed from it?
 A: Since $X,Y\sim\ N(0,1)$, then $X^2$ and $Y^2$ are chi-squared with 1 degree of freedom each.  
Since $X^2,Y^2\sim\chi^2(1)$ and they are independent, then $X^2+Y^2\sim\chi^2(2)$.  
So to compute those probabilities, use a chi-squared density (or R).
A: The crux of your question appears to be: "How do I show that if $X \sim N(0, 1)$, then $X^2 \sim \chi^2(1)$?"
Here's the gist: Let $f(x), F(x)$ be the density and cdf of a standard normal variable, respectively. (Note that $f(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, but that $F(x)$ doesn't have a clean closed-form expression.) We want to know more about the variable $U = X^2$, so call its density and cdf by $g(x)$ and $G(x)$, respectively.
First, note that for $x > 0$,
\begin{align*}G(x) &= \mathbb P(U \leq x) \\ &= \mathbb P(X^2 \leq x) \\ &= \mathbb P(-\sqrt x \leq X \leq \sqrt x) \\  &= F(\sqrt x) - F(-\sqrt x).
\end{align*}
It follows that
\begin{align*}
  g(x) &= G'(x) \\ &= \frac{\textrm d}{\textrm d x} [F(\sqrt x) - F(-\sqrt x)] \\
&= f(\sqrt x) \cdot \frac{1}{2 \sqrt x} + f(-\sqrt x) \cdot \frac{1}{2 \sqrt x}
\end{align*}
for all $x > 0$, and if you evaluate that, it happens to be exactly the density of a $\chi^2$ variable with 1 df.
There are lots of details in the steps above, but I wasn't sure about your background in the material, so let me know if there are steps that need clarification.
