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Question Number 141219 by mathmax by abdo last updated on 16/May/21

$$\mathrm{find}\int {\int }_{{[0,1]}^2}{\mathrm{e}}^{-({\mathrm{x}}^2+{\mathrm{y}}^2)}\sqrt{{\mathrm{x}}^4+{\mathrm{y}}^4}\mathrm{dxdy}$$

Answered by Ar Brandon last updated on 16/May/21

$$\mathcal{I}={\int }_0^1{\int }_0^1{\mathrm{e}}^{-({\mathrm{x}}^2+{\mathrm{y}}^2)}\sqrt{{\mathrm{x}}^4+{\mathrm{y}}^4}\mathrm{dxdy}$$$$={\int }_0^{\frac{\pi }{2}}{\int }_0^1{\mathrm{re}}^{-{\mathrm{r}}^2}\sqrt{{\mathrm{r}}^4-{2\mathrm{r}}^4{\sin }^2\theta {\cos }^2\theta }\mathrm{drd}\theta$$$$={\int }_0^{\frac{\pi }{2}}{\int }_0^1{\mathrm{r}}^3{\mathrm{e}}^{-{\mathrm{r}}^2}\sqrt{1-\frac{1}{2}{\sin }^{2}2\theta }\mathrm{drd}\theta$$$$=-\frac{1}{2}{\int }_0^1{\mathrm{r}}^2\cdot (-{2\mathrm{re}}^{-{\mathrm{r}}^2})\mathrm{dr}\times {\int }_0^{\frac{\pi }{2}}\sqrt{1-\frac{1}{2}{\sin }^{2}2\theta }\mathrm{d}\theta$$$$=-\frac{1}{2}{[{\mathrm{r}}^2\cdot \frac{{\mathrm{e}}^{-{\mathrm{r}}^2}}{2}-\int {\mathrm{re}}^{-{\mathrm{r}}^2}\mathrm{dr}]}_0^1\times {\int }_0^{\frac{\pi }{2}}\sqrt{1-\frac{1}{2}{\sin }^{2}2\theta }\mathrm{d}\theta$$$$=-\frac{1}{2}{[{\mathrm{r}}^2\cdot \frac{{\mathrm{e}}^{-{\mathrm{r}}^2}}{2}+\frac{{\mathrm{e}}^{-{\mathrm{r}}^2}}{2}]}_0^1\times {\int }_0^{\frac{\pi }{2}}\sqrt{1-\frac{1}{2}{\sin }^{2}2\theta }\mathrm{d}\theta$$

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