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How to solve the following $$\iint\limits_{x^2+y^2\ge k}\frac{\exp(-(x^2+y^2)/2)}{2\pi}dxdy?$$

I think I should make the substitution $u=x^2+y^2$, but I don't know how the integral will look like.

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This is a Gaussian integral; a recommended method is to use polar coordinates. – Johannes Kloos Apr 19 '12 at 12:10
The word "solve", like one or two others, gets used as a catch-all by non-mathematicians. The right term here is "evaluate". On solves problems; one solves equations; one evaluates expressions. – Michael Hardy Apr 19 '12 at 12:35
In statistical language, if $X$ and $Y$ are independent standard normal random variables, then $X^2+Y^2$ has a $\chi^2$ distribution with 2 degrees of freedom, that is, an exponential distribution with mean 2. – Byron Schmuland Apr 19 '12 at 12:37

If you use polar coordinates, you have $dx\,dy = r\,dr\,d\theta$, so that: $$\iint\limits_{x^2+y^2\ge k}\frac{\exp(-(x^2+y^2)/2)}{2\pi}\,dx\,dy = \int\limits_{r\ge \sqrt{k}}\exp(-r^2/2)r\,dr$$ You can then use $u = r^2/2,~ du = r \,dr$ to get: $$\int\limits_{r\ge \sqrt{k}}\exp(-r^2/2)r\,dr = \int\limits_{u\ge k/2}\exp(-u)\,du = e^{-k/2}$$

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