Probability of Independent Normal Random Variables There are two Random Variables X and Y. Both of them are Normally distributed with 0 mean and variance a and b respectively. X and Y are independent of each other.
What is the probability of 
What should be the easiest way to do this?

The joint probability distribution of $X$ and $Y$ is $\frac{1}{2\pi \sqrt{ab}}e^{\frac{-1}{2} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \right)}$.
Hence, we are looking for the value of the following integral.
\begin{align*}
\frac{1}{2 \pi \sqrt{ab}} \iint_{x+y>0, y>0}e^{\frac{-1}{2} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \right)} dxdy
\end{align*}
We make the substitution $s = \frac{x}{a}$ and $t = \frac{y}{b}$, and this becomes
\begin{align*}
\frac{\sqrt{ab}}{2 \pi} \iint_{as+bt>0, t>0}e^{\frac{-1}{2} \left(s^2 + t^2 \right)} dsdt
\end{align*}
Now, we want to switch to polar coordinates, but we have to be careful about the domain over which we integrate. You might want to draw a picture to find that it is the sector $0 < \theta < \pi + \arctan\left( \frac{-a}{b} \right)$. This gives the following expression.
\begin{align*}
\frac{\sqrt{ab}}{2 \pi} \int_{0}^{\pi + \arctan\left( \frac{-a}{b} \right)}\int_0^{\infty} e^{\frac{-r^2}{2}} rdrd\theta =& \frac{\sqrt{ab}\left( \pi + \arctan \left( \frac{-a}{b}\right) \right)}{2\pi} \int_0^{\infty}e^{\frac{-r^2}{2}}d\left( \frac{r^2}{2} \right) \\
=& \frac{\sqrt{ab}\left( \pi + \arctan \left( \frac{-a}{b}\right) \right)}{2\pi}
\end{align*}
which is your solution.
