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Question Number 128262 by Ahmed1hamouda last updated on 06/Jan/21

Answered by mr W last updated on 06/Jan/21

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{dxdy}{{(x^2+y^2+a^2)}^2}$$$$={\int }_0^{\mathrm{\infty }}\left[{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2+y^2+a^2)}^2}\right]dy$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{2(y^2+a^2)}{[\frac{{\tan }^{-1}\frac{x}{\sqrt{y^2+a^2}}}{\sqrt{y^2+a^2}}+\frac{x}{x^2+y^2+a^2}]}_0^{\mathrm{\infty }}dy$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{2(y^2+a^2)}\times \frac{\pi }{2\sqrt{y^2+a^2}}dy$$$$=\frac{\pi }{4}{\int }_0^{\mathrm{\infty }}\frac{dy}{(y^2+a^2)\sqrt{y^2+a^2}}$$$$=\frac{\pi }{4a^2}{\left[\frac{y}{\sqrt{y^2+a^2}}\right]}_0^{\mathrm{\infty }}$$$$=\frac{\pi }{4a^2}\times \underset{y\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt{1+\frac{a^2}{y^2}}}$$$$=\frac{\pi }{4a^2}$$

Answered by mnjuly1970 last updated on 06/Jan/21

$$x=rcos\left(\theta \right)$$$$\Rightarrow \mid j(r,\theta )=\mid \frac{\partial (x,y)}{\partial (r,\theta )}\mid =r$$$$y=rsin\left(\theta \right)$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{r}{{(r^2+a^2)}^2}drd\theta$$$$={\int }_0^{\frac{\pi }{2}}{\left[\frac{-1}{2(r^2+a^2)}\right]}_0^{\mathrm{\infty }}d\theta ={\int }_0^{\frac{\pi }{2}}\frac{1}{2a^2}d\theta =\frac{\pi }{4a^2}$$$$\mathrm{\Omega }=\frac{\pi }{4a^2}✓$$

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