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Question Number 133905 by mohammad17 last updated on 25/Feb/21

Answered by Dwaipayan Shikari last updated on 25/Feb/21

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x^2}{3}-\frac{y^2}{4}}dxdy$$$$=\sqrt{3\pi }.2\sqrt{\pi }=2\sqrt{3}\pi$$

Answered by mathmax by abdo last updated on 25/Feb/21

$$\mathrm{I}=\int {\int }_{{\mathrm{R}}^{{+}^2}}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{3}-\frac{{\mathrm{y}}^2}{4}}\mathrm{dxdy}\mathrm{with}\mathrm{diffeomorphism}\{\begin{array}{l}\mathrm{x}=\sqrt{3}\mathrm{rcos}\theta \\ \mathrm{y}=2\mathrm{rsin}\theta \mathrm{we}\mathrm{get}\end{array}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{-{\mathrm{r}}^2}\left(2\sqrt{3}\right)\mathrm{r}\mathrm{drd}\theta$$$$=\frac{\pi }{2}\left(2\sqrt{3}\right){\int }_0^{\mathrm{\infty }}\mathrm{r}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{dr}=\pi \sqrt{3}{[-\frac{1}{2}{\mathrm{e}}^{-{\mathrm{r}}^2}]}_0^{\mathrm{\infty }}=\frac{\pi \sqrt{3}}{2}$$$$$$

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