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Question Number 203349 by Mathspace last updated on 17/Jan/24

$$calculate\int {\int }_{{[0,a]}^2}{e}^{-x^2-y^2}dxdy$$$$canyoufind{\int }_0^a{e}^{-x^2}dx?$$$$a>0$$

Answered by witcher3 last updated on 17/Jan/24

$$\mathrm{the}\mathrm{idee}\int {\int }_{{[0,\mathrm{\infty }]}^2}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{dxdy}={\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\mathrm{f}(\mathrm{rcos}\left(\mathrm{a}\right),\mathrm{rsin}\left(\mathrm{a}\right))\mathrm{rdrda}$$$${\mathrm{can}}^{\prime }\mathrm{t}\mathrm{hellp}\mathrm{ther}$$$$\mathrm{for}\mathrm{a}>0{[0,\mathrm{a}]}^2\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{if}\mathrm{you}\mathrm{try}\mathrm{to}\mathrm{fill}\mathrm{it}\mathrm{withe}\mathrm{disc}$$$$\mathrm{is}\mathrm{not}\mathrm{easy}$$$$0⩽\mathrm{rcos}\left(\mathrm{z}\right)⩽\mathrm{a}$$$$0⩽\mathrm{rsin}\left(\mathrm{z}\right)⩽\mathrm{a}$$$$0<\mathrm{r}<\frac{\mathrm{a}}{\cos \left(\mathrm{z}\right)},0⩽\mathrm{r}<\frac{\mathrm{a}}{\sin \left(\mathrm{z}\right)}$$$$\Rightarrow \mathrm{r}<\min (\frac{\mathrm{a}}{\cos \left(\mathrm{z}\right)},\frac{\mathrm{a}}{\sin \left(\mathrm{z}\right)})$$$$={\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{\mathrm{a}}{\cos \left(\mathrm{z}\right)}}{\mathrm{re}}^{-{\mathrm{r}}^2}\mathrm{drdz}+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\int }_0^{\frac{\mathrm{a}}{\sin \left(\mathrm{z}\right)}}{\mathrm{re}}^{-{\mathrm{r}}^2}\mathrm{drdz}$$$$={\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{\mathrm{a}}{\cos \left(\mathrm{z}\right)}}{2\mathrm{re}}^{-{\mathrm{r}}^2}\mathrm{drdz}$$$$={\int }_0^{\frac{\pi }{4}}{(-{\mathrm{e}}^{-{\mathrm{r}}^2}]}_0^{\frac{\mathrm{a}}{\cos \left(\mathrm{z}\right)}}\mathrm{dz}$$$$={\int }_0^{\frac{\pi }{4}}1-{\mathrm{e}}^{-\frac{{\mathrm{a}}^2}{{\cos }^2\left(\mathrm{z}\right)}}\mathrm{dz}$$$$\mathrm{tg}\left(\mathrm{z}\right)=\mathrm{t}$$$$={\int }_0^1(1-{\mathrm{e}}^{-{\mathrm{a}}^2(1+{\mathrm{t}}^2)})\frac{\mathrm{dt}}{1+{\mathrm{t}}^2},\mathrm{not}\mathrm{easy}\mathrm{from}\mathrm{heree}$$$$\mathrm{we}\mathrm{can}\mathrm{use}\mathrm{feyneman}\mathrm{and}\mathrm{givre}\mathrm{answer}\mathrm{withe}‘‘{\mathrm{erf}}^{″}$$$${\int }_0^{\mathrm{a}}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}....\mathrm{should}\mathrm{bee}\mathrm{first}\mathrm{Quation}=\frac{\sqrt{\pi }}{2}\mathrm{erf}\left(\mathrm{a}\right)$$$$\mathrm{erf}\left(\mathrm{z}\right)=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{z}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}$$$$\mathrm{Than}\int \int {\mathrm{e}}^{-{\mathrm{x}}^2-{\mathrm{y}}^2}\mathrm{dxdy}={\int }_0^{\mathrm{a}}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}.{\int }_0^{\mathrm{a}}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}$$$$={\left({\int }_0^{\mathrm{a}}{\mathrm{e}}^{-{\mathrm{x}}^2}\right)}^2\mathrm{dx}=\frac{\pi }{4}.{\mathrm{erf}}^2\left(\mathrm{a}\right)$$$$$$

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