<s>[INST] 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. 
 [/INST] 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, $P(X < 2Y) = P(rcos\theta < 2rsin\theta) = P(tan\theta > \frac {1}{2}) = P(\theta > tan^-\frac {1}{2}) = \frac {1}{2\pi} \int_{tan^-\frac {1}{2}}^{2 \pi}d\theta = \frac {1}{2\pi} [\theta]_{tan^-\frac {1}{2}}^{\pi + tan^-\frac {1}{2}} = \frac {1}{2\pi}[\pi + tan^-\frac {1}{2} - -tan^-\frac {1}{2}]) = \frac {1}{2\pi} . \pi = 0.5$
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