Double integral with polar?

Question

I have the following integral :

$$\iint\limits_R \operatorname e^{-\frac{x^2+y^2}{2}} \operatorname d\!y \operatorname d\!x $$

Where R is: $$R=\{(x,y):x^2+y^2 \leq 1\}$$

I think I should convert to polar in order to solve it. Is this correct? Also how do you calculate the integration limits? Thanks

Answer 1

Set $x=r\cos(t)$ and $y=r\sin(t)$. We then have \begin{align} \iint_R e^{-(x^2+y^2)/2}dydx &= \int_{0}^{2\pi} \int_{r=0}^1 e^{-r^2/2}rdrdt = 2\pi \int_0^1 e^{-r^2/2} d(r^2/2)\\ & = 2\pi \int_0^{1/2}e^{-t}dt = 2\pi (1-e^{-1/2}) \end{align}

Answer 2

$R=\{(r,\theta):0\leq r\leq 1,0\leq \theta \leq 2\pi\}$ \begin{align*} \iint_Re^{-\frac{x^2+y^2}{2}}\mathrm dx\mathrm dy&=\int_{\theta = 0}^{\theta=2\pi}\left[\int_{r=0}^{r=1}e^{-\frac{1}{2}r^2}r\mathrm dr\right]\mathrm d\theta \end{align*}