X and Y are two independent random variables, and are standard normal. Determine $P(X\lt Y \lt 2X)$ X and Y are two independent  random variables, and are standard normal. Determine $P(X\lt Y \lt 2X)$.
I'm having difficulties understanding how to solve this and would appreciate any hints or steps.
 A: Hint: The set $\{(x,y):x<y<2x\}$ is a radial wedge (make a drawing). Do you know the probability density $f_{X,Y}(x,y) $ ? 
If so write down an integral and change to polar coordinates. This is probably recommendable as a good exercise in integration.
But, even simpler is to notice that the wedge is angular symmetric and so is the PDF. The probability is therefore proportional to the angle spanned by the two radial half-lines bounding the domain. So  the result, $\frac{\tan^{-1}(2)-\tan^{-1}(1)}{2\pi}=\frac{\tan^{-1}(2)}{2\pi}-\frac18$ may be obtained essentially without calculations.
A: $$\mathbb{P}(X<Y<2X)=\frac{1}{2\pi}\int_{0}^{\infty}\int_{x}^{2x}\exp\left(-\frac{x^2+y^2}{2}\right)dydx$$
set
$$x=r\cos\theta\\
y=r\sin\theta$$
hence
$$1\le \frac{y}{x}\le 2$$
thus
$$\frac{\pi}{4}\le\theta\le \tan^{-1}(2)$$
finally, we have
$$\mathbb{P}(X<Y<2X)=\frac{1}{2\pi}\int_{\frac{\pi}{4}}^{\tan^{-1}(2)}\int_{0}^{\infty}r\exp\left(-\frac{r^2}{2}\right)drd\theta$$
$$\mathbb{P}(X<Y<2X)=\frac{1}{2\pi}\int_{\frac{\pi}{4}}^{\tan^{-1}(2)}d\theta=\frac{1}{2\pi}\left(\tan^{-1}(2)-\frac{\pi}{4}\right)$$
