Finding $P(X<2Y)$ given joint pdf $f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}$ for $x,y\in\mathbb R$ 
The joint pdf of $X$ and $Y$ is $$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}\,; \quad x,y\in \mathbb{R}$$ Find $P(X<2Y)$.

I have tried this:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{2y} f(x, y)\, dx\,dy$$
@StubbornAtom suggested polar transformations. 
So here it goes, 
Let $X = r \cos\theta, \, Y = r \sin\theta$. 
Then The integral gets transformed as $\int \int re^{-r}d\theta$.
Could some one help find the new limits. 

The joint distribution of $(X,Y)$ is rotationally symmetric, since the density depends only on the distance from the origin. This means that the conditional distribution of $(X,Y)$ given $X^2+Y^2$ is uniformly distributed on the circle of radius $r=\sqrt{X^2+Y^2}$. Writing $X=r\cos\Theta$ and $Y=r\sin\Theta$ with $\Theta\in [0,2\pi)$ uniformly random, we have
$$
\mathbb P(X<2Y\mid X^2+Y^2)=\mathbb P(\cos\Theta<2\sin \Theta)$$$$=\mathbb P\Bigl(\tan\Theta>\frac{1}{2}, \cos\theta > 0\Bigr)+\mathbb P\Bigl(\tan\Theta<\frac{1}{2}, \cos\theta < 0\Bigr),$$
which can be seen to equal
$$\mathbb P\Bigl(\tan^{-1}\frac{1}{2}<\Theta<\pi+\tan^{-1}\frac{1}{2}\Bigr)=\frac12.
$$
Since the conditional probability is $\tfrac12$ independent of $X^2+Y^2$, it follows that the unconditional probability is also $\tfrac12$.
