Question Number 128262 by Ahmed1hamouda last updated on 06/Jan/21 | ||
Answered by mr W last updated on 06/Jan/21 | ||
$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{{dxdy}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left[\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}\left({y}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\left[\frac{\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\:\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }}}{\:\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }}+\frac{{x}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} {dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}\left({y}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}×\frac{\pi}{\:\mathrm{2}\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{dy} \\ $$$$=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{dy}}{\:\left({y}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$$=\frac{\pi}{\mathrm{4}{a}^{\mathrm{2}} }\left[\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\pi}{\mathrm{4}{a}^{\mathrm{2}} }×\underset{{y}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}} \\ $$$$=\frac{\pi}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$ | ||
Answered by mnjuly1970 last updated on 06/Jan/21 | ||
$${x}={rcos}\left(\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mid{j}\left({r},\theta\right)=\mid\frac{\partial\left({x},{y}\right)}{\partial\left({r},\theta\right)}\mid={r} \\ $$$${y}={rsin}\left(\theta\right) \\ $$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\infty} \frac{{r}}{\left({r}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{drd}\theta \\ $$$$=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left[\frac{−\mathrm{1}}{\mathrm{2}\left({r}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\right]_{\mathrm{0}} ^{\infty} {d}\theta=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }{d}\theta=\frac{\pi}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\frac{\pi}{\mathrm{4}{a}^{\mathrm{2}} }\:\:\checkmark \\ $$ | ||