Question about joint probability density function Given the joint probability density function of $X$ and $Y$ is
$$
f(x,y)=\cases{xy \exp\bigl(-\textstyle{1\over2}(x^2+y^2)\bigr),&$x>0,y>0$\cr 0\phantom{\Bigl[},& elsewhere.}
$$
Find $P(X^2+Y^2>4)$.
What approach should I use to solve this question?
Transformation?
Hope someone can help me !_!
 A: Hint: Transformation to polar coordinates suggests itself here. Let $x=r\cos\theta$, $y=r\sin\theta$. Then $dx\,dy=r\,dr\,d\theta$. You will end up with 
$$\int_{\theta=0}^{\pi/2}\int_{r=2}^\infty (\cos\theta\sin\theta) r^3e^{-r^2/2}\,dr\,d\theta.$$
For the integration with respect to $r$, I suggest a substitution, then integration by parts, although direct integration by parts also works. For the trigonometric part, one can let $t=\sin\theta$.
A: This looks a lot like homework, but since an answer has already been posted, here
are alternatives that do not use a transformation to polar coordinates
and needing to integrate trigonometric functions.
Since $\frac{\mathrm d}{\mathrm dx} \exp(-x^2/2) = -x\exp(-x^2/2)$, it is possible
to compute
$$P\{X^2+Y^2 \leq 4\} = \int_{x=0}^2 x\exp(-x^2/2)\int_{y=0}^{\sqrt{4-x^2}}
y\exp(-y^2/2)\,\mathrm dy\,\mathrm dx$$
without resorting to a transformation to polar coordinates.
The inner integral should work out to something like 
$1 - \exp(-(4-x^2)/2) = 1 - e^{-2}\exp(x^2/2)$ which will change the
integrand of the outer integral to something like $x\exp(-x^2/2)-e^{-2}x$
which can also be easily
integrated.
As yet another approach, note that $X$ and $Y$ are independent
Rayleigh random variables and so $X = \sqrt{Z_1^2+Z_2^2},~ Y = \sqrt{Z_3^2+Z_4^2}$ where $Z_1, Z_2, Z_3, Z_4$ are independent
standard normal random variables. Thus
$P\{X^2+Y^2 > 4\}$ is the probability that a $\chi^2$ random
variable $W = X^2+Y^2 = Z_1^2+Z_2^2+Z_3^2+Z_4^2$ exceeds $4$.
