Question Number 4389 by Yozzii last updated on 17/Jan/16
$$\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{exp}\left(\frac{−\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right){dydx}=? \\ $$
Commented by prakash jain last updated on 19/Jan/16
$${transform}\:{to}\:{polar} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} {r}^{\mathrm{2}} {e}^{−{r}^{\mathrm{2}} /\mathrm{2}} {drd}\theta \\ $$
Commented by Yozzii last updated on 19/Jan/16
$${In}\:{letting}\:{x}={rcos}\theta\:{and}\:{y}={rsin}\theta, \\ $$$${what}\:{is}\:{the}\:{justification}\:{used}\:{in}\: \\ $$$${determining}\:{that}\:\mathrm{0}\leqslant{r}<\infty\:{when}\:{the} \\ $$$${integral}\:{initial}\:{integral}\:{had}\:{negative} \\ $$$${x}\:{and}\:{y}\:{values}\:{as}\:{limits}?\:{Is}\:{it}\:{because} \\ $$$${r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{so}\:{that}\:\mathrm{0}\leqslant{r}<\infty\:{as}\: \\ $$$${x},{y}\in\left(−\infty,\infty\right)\:? \\ $$
Commented by prakash jain last updated on 22/Jan/16
$${x}=\left(−\infty,\infty\right)\:\mathrm{and}\:{y}\left(−\infty,\infty\right)\:\mathrm{covers}\:\mathrm{entire} \\ $$$$\mathrm{2D}\:\mathrm{space}.{i}.{e}\:{r}\mathrm{0},\infty\:\mathrm{and}\:\theta\:−\mathrm{0}−\mathrm{2}\pi \\ $$$$\mathrm{For}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{such}\:\mathrm{as}\:\mathrm{a}\:\left({x},{y}\right)\:\mathrm{over}\:\mathrm{an}\:\mathrm{ellipse} \\ $$$$\mathrm{limits}\:\mathrm{on}\:{r}\:\mathrm{may}\:\mathrm{depend}\:\mathrm{on}\theta. \\ $$