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. 

Draw the set $M = \{(x,y)\in\mathbb R^2 \mid x<2y\}$. What you should get is the half-plane above the line $y=\frac{x}{2}$. Since the given pdf $f$ is radially symmetric, you get the result
$$
P(X<2Y) = \int_M f(x,y) \, \mathrm dx\,  \mathrm dy = \frac{1}{2}
$$
without any calculations.
