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Question Number 20612 by ajfour last updated on 29/Aug/17

$$Evaluate{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dydx.$$

Answered by ajfour last updated on 29/Aug/17

$$let{x}^2+y^2=r^2$$$$thatisx=r\cos \theta ;y=r\sin \theta$$$$dx=-r\sin \theta d\theta ;dy=r\cos \theta d\theta$$$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dydx=I$$$$I={\int }_0^{\mathrm{\infty }}\left[{\int }_0^{\pi /2}{e}^{-r^2}rd\theta \right]dr$$$$=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}{e}^{-r^2}rdr=\frac{\pi }{4}{\int }_{\mathrm{\infty }}^0{e}^{-r^2}(-2rdr)$$$$I=\frac{\pi }{4}\left(e^{-r^2}\right){\mid }_{\mathrm{\infty }}^0=\frac{\pi }{4}.$$$$$$

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