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 easiest way to do this is transforming the integral in the $(x,y)$ space to an integral using polar coordinates.
First, we take care of the variances by transforming $X$ and $Y$ into standard normals, $S$ and $T$ (both with variance equal to 1): let $S=X/\sqrt{a}$ and $T=Y/\sqrt(b)$. The desired probability is rewritten as:
\begin{align*}
\Pr\left(\{X+Y>0\}\cap \{Y>0\}\right) = \Pr\left(\{\sqrt{a}S+\sqrt{b}T>0\}\cap \{T>0\}\right).
\end{align*}
Next, we write the probability as an integral of the joint probability distribution of two standard normals, $S$ and $T$:
\begin{align*}
\Pr\left(\{\sqrt{a}S+\sqrt{b}T>0\}\cap \{T>0\}\right) = 
\frac{1}{2 \pi } \iint_{s\sqrt{a}+t\sqrt{b}>0, t>0}e^{-\frac{1}{2} \left(s^2 + t^2\right)} dsdt.
\end{align*}
Then, we convert $(s,t)$ coordinates into polar coordinates using $r^2 = s^2 + t^2$, $\tan(\theta)= t/s$. The two things to take care are: (i) $dxdy$ changes to $rdrd\theta$, and (ii) we need to convert the area we are integrating to polar coordinates. 
The area is a cone from the origin (that's why using polar coordinates is going to be useful), so we are integrating over rays from the origin of any (positive) length, that is over $r\in[0,\infty]$. We now need the angles of the rays on the edge of the cone: (i) the x-axis (between the quadrants I and IV), which is a ray of angle $\theta=0$, and (ii) the line $t=-\sqrt{\frac{a}{b}}s$ (in quadrant II, where $t>0$ and $s<0$). The angle of this ray is $\theta = \pi +\arctan(-\sqrt{a/b}) = \pi -\arctan(\sqrt{a/b})$.
The resulting integral is the following:
\begin{align*}
\frac{1}{2 \pi} \int_{0}^{\pi - \arctan\left(\sqrt{ \frac{a}{b}} \right)}\int_0^{\infty} e^{-\frac{r^2}{2}} rdrd\theta.
\end{align*}
This integral is easy to evaluate. We can integrate with respect to $\theta$ to get:
\begin{align*}
\frac{1}{2 \pi} \left(\pi - \arctan\left(\sqrt{ \frac{a}{b}} \right)\right)\int_0^{\infty} e^{-\frac{r^2}{2}} rdr. 
\end{align*}
For the remaining integral, if we make the change of variable $z=r^2$ with $dz = 2 r dr$, we realize that the integral is equal to one, and we are done!
\begin{align*}
\Pr\left(\{X+Y>0\}\cap \{Y>0\}\right) &= 
\frac{\left( \pi - \arctan \left( \sqrt{\frac{a}{b}}\right) \right)}{2\pi}\\
&= \frac{1}{2}- \frac{1}{2\pi}\arctan \left(- \sqrt{\frac{a}{b}}\right).
\end{align*}
