# Question-133905

Question Number 133905 by mohammad17 last updated on 25/Feb/21
Answered by Dwaipayan Shikari last updated on 25/Feb/21
$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}} {dxdy} \\$$$$=\sqrt{\mathrm{3}\pi}\:.\mathrm{2}\sqrt{\pi}\:=\mathrm{2}\sqrt{\mathrm{3}}\pi \\$$
Answered by mathmax by abdo last updated on 25/Feb/21
$$\mathrm{I}=\int\int_{\mathrm{R}^{+^{\mathrm{2}} } } \:\:\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{4}}} \:\mathrm{dxdy}\:\mathrm{with}\:\mathrm{diffeomorphism}\:\:\begin{cases}{\mathrm{x}=\sqrt{\mathrm{3}}\mathrm{rcos}\theta}\\{\mathrm{y}=\mathrm{2rsin}\theta\:\:\:\mathrm{we}\:\mathrm{get}}\end{cases} \\$$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \left(\mathrm{2}\sqrt{\mathrm{3}}\right)\mathrm{r}\:\mathrm{drd}\theta \\$$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{3}}\right)\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{r}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{dr}\:=\pi\sqrt{\mathrm{3}}\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\infty} \:=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}} \\$$$$\\$$