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. 
 A: 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 $\tfrac12$.
A: 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.
A: if we consider the domain D* symmetrical of D with respect to O, the integral on D * is equal to that on D by parity of the function: f(x,y)=f(-x,-y) and the change (x, y) in (-x, -y)
and as D and D* are disjoint and DUD*= RxR  (to see on the drawing!)
the integral sought is half of that on RXR
A: So this is my version that I have understood from @pre-kidney's answer please note that this is just an elaboration of his answer.
$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2+y^2}} \ ;\  -\infty<\ x \ <\infty\ ; -\infty < y < \infty$
$X=rcosθ,Y=rsinθ$
$f(r, \theta) = f(\theta).f(r) = (\frac {1}{2\pi} ).(re^{-r})\ \ ;\ \ 0 < r < \infty\ ;\ 0 <  \theta < 2\pi$
Now, $P(X < 2Y) = P(rcos\theta < 2rsin\theta) = P(tan\theta > \frac {1}{2}) = P(\theta > tan^-\frac {1}{2}) = \frac {1}{2\pi} \int_{tan^-\frac {1}{2}}^{2 \pi}d\theta = \frac {1}{2\pi} [\theta]_{tan^-\frac {1}{2}}^{\pi + tan^-\frac {1}{2}} = \frac {1}{2\pi}[\pi + tan^-\frac {1}{2} - -tan^-\frac {1}{2}]) = \frac {1}{2\pi} . \pi = 0.5$
