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Question Number 131245 by Ahmed1hamouda last updated on 02/Feb/21

$$\mathrm{Find}$$$${\int }_{-\mathrm{b}}^{\mathrm{b}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{\mathrm{d}xdy}{{(x^2+y^2+h^2)}^{3/2}}$$

Answered by Ar Brandon last updated on 03/Feb/21

$$\mathcal{I}={\int }_{-\mathrm{b}}^{\mathrm{b}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{\mathrm{dxdy}}{{({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{h}}^2)}^{3/2}}=4{\int }_0^{\mathrm{b}}{\int }_0^{\mathrm{a}}\frac{\mathrm{dxdy}}{{({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{h}}^2)}^{3/2}}$$$$\{\begin{array}{ll}\mathrm{x}=\mathrm{rcos}\theta & \mathrm{r}\in [0,\mathrm{\infty })\\ \mathrm{y}=\mathrm{rsin}\theta & \theta \in [0,2\pi ]\end{array},\mathrm{r}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2},\theta ={\tan }^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$$$$\mathcal{I}=4{\int }_0^{{\tan }^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}{\int }_0^{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\frac{\mathrm{rdrd}\theta }{{({\mathrm{r}}^2+{\mathrm{h}}^2)}^{3/2}}$$$$=-4{\left[\frac{1}{{({\mathrm{r}}^2+{\mathrm{h}}^2)}^{1/2}}\right]}_0^{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\times {\left[\frac{\theta }{1}\right]}_0^{{\tan }^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$$$$=4(\frac{1}{\mid \mathrm{h}\mid }-\frac{1}{{({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{h}}^2)}^{1/2}})\times {\tan }^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$$

Answered by mathmax by abdo last updated on 03/Feb/21

$$\mathrm{I}={\int }_{-\mathrm{b}}^{\mathrm{b}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{\mathrm{dxdy}}{{({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{h}}^2)}^{\frac{3}{2}}}\Rightarrow \mathrm{I}=4{\int }_0^{\mathrm{b}}.{\int }_0^{\mathrm{a}}\frac{\mathrm{dxdy}}{{({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{h}}^2)}^{\frac{3}{2}}}$$$$\mathrm{we}\mathrm{consider}\mathrm{the}\mathrm{diffeomorphism}\{\begin{array}{l}\mathrm{x}=\mathrm{rcos}\theta \\ \mathrm{y}=\mathrm{rsin}\theta \end{array}$$$$0⩽\theta ⩽2\pi \mathrm{we}\mathrm{have}{\mathrm{x}}^2+{\mathrm{y}}^2⩽{\mathrm{a}}^2+{\mathrm{b}}^2\Rightarrow 0⩽\mathrm{r}⩽\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\Rightarrow$$$$\mathrm{I}=4{\int }_0^{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\int }_0^{2\pi }\frac{\mathrm{rdrd}\theta }{{({\mathrm{r}}^2+{\mathrm{h}}^2)}^{\frac{3}{2}}}=8\pi {\int }_0^{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{r}{({\mathrm{r}}^2+{\mathrm{h}}^2)}^{-\frac{3}{2}}\mathrm{dr}$$$$=-8\pi {\left[{({\mathrm{r}}^2+{\mathrm{h}}^2)}^{-\frac{1}{2}}\right]}_0^{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=-8\pi \{{({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{h}}^2)}^{-\frac{1}{2}}-{\mathrm{h}}^{-1}\}$$$$\mathrm{I}=8\pi \{\frac{1}{\mathrm{h}}-\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{h}}^2}}\}$$

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