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Question Number 119647 by IE last updated on 26/Oct/20

Answered by Ar Brandon last updated on 26/Oct/20

$$\mathrm{En}\mathrm{coordonn}\acute{\mathrm{e}es}\mathrm{sph}\acute{\mathrm{e}riques}\mathrm{nous}\mathrm{avons}$$$$\{\begin{array}{ll}\mathrm{x}=\mathrm{rsin}\phi \cos \theta & \mathrm{r}\in [0,\mathrm{R}]\\ \mathrm{y}=\mathrm{rsin}\phi \sin \theta & \theta \in [0,\pi ]\\ \mathrm{z}=\mathrm{rcos}\phi & \phi \in [0,\alpha ]\end{array}$$$$\Rightarrow \mathrm{\Omega }={\int }_0^{\pi }{\int }_0^{\alpha }{\int }_0^{\mathrm{R}}\frac{\mid \mid \mathrm{J}\mid \mid }{\mathrm{rcos}\phi }\mathrm{drd}\phi \mathrm{d}\theta ,\mid \mid \mathrm{J}\mid \mid =\mathrm{la}\mathrm{matrice}\mathrm{Jacobien}$$$$={\int }_0^{\pi }{\int }_0^{\alpha }{\int }_0^{\mathrm{R}}\frac{{\mathrm{r}}^2\sin \phi }{\mathrm{rcos}\phi }\mathrm{drd}\phi \mathrm{d}\theta ={\int }_0^{\pi }{\int }_0^{\alpha }{\int }_0^{\mathrm{R}}\mathrm{rtan}\phi \mathrm{drd}\phi \mathrm{d}\theta$$$$={\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\mathrm{R}}{\left[\frac{\ln \mid \sec \phi \mid }{1}\right]}_0^{\alpha }{\left[\frac{\theta }{1}\right]}_0^{\pi }=\frac{\pi {\mathrm{R}}^2\ln \mid \sec \alpha \mid }{2}$$

Commented by Ar Brandon last updated on 26/Oct/20

$$\mid \mathrm{J}\mid =\left|\begin{array}{ccc}\frac{\partial \mathrm{x}}{\partial \mathrm{r}}& \frac{\partial \mathrm{x}}{\partial \phi }& \frac{\partial \mathrm{x}}{\partial \theta }\\ \frac{\partial \mathrm{y}}{\partial \mathrm{r}}& \frac{\partial \mathrm{y}}{\partial \phi }& \frac{\partial \mathrm{y}}{\partial \theta }\\ \frac{\partial \mathrm{z}}{\partial \mathrm{r}}& \frac{\partial \mathrm{z}}{\partial \phi }& \frac{\partial \mathrm{z}}{\partial \theta }\end{array}\right|=\left|\begin{array}{ccc}\sin \phi \cos \theta & \mathrm{rcos}\phi \cos \theta & -\mathrm{rsin}\phi \sin \theta \\ \sin \phi \sin \theta & \mathrm{rcos}\phi \sin \theta & \mathrm{rsin}\phi \cos \theta \\ \cos \phi & -\mathrm{rsin}\phi & 0\end{array}\right|$$$$=\sin \phi \cos \theta \left({\mathrm{r}}^2{\sin }^2\phi \cos \theta \right)+\mathrm{rcos}\phi \cos \theta \left(\mathrm{rsin}\phi \cos \phi \cos \theta \right)$$$$+\mathrm{rsin}\phi \sin \theta ({\mathrm{rsin}}^2\phi \sin \theta +{\mathrm{rcos}}^2\phi \sin \theta )$$$$={\mathrm{r}}^2{\sin }^3\phi {\cos }^2\theta +{\mathrm{r}}^2\sin \phi {\cos }^2\phi {\cos }^2\theta +{\mathrm{r}}^2\sin \phi {\sin }^2\theta \underset{1}{({\sin }^2\phi +{\cos }^2\phi )}$$$$={\mathrm{r}}^2\sin \phi {\cos }^2\theta \underset{1}{({\sin }^2\phi +{\cos }^2\phi )}+{\mathrm{r}}^2\sin \phi {\sin }^2\theta$$$$={\mathrm{r}}^2\sin \phi \underset{1}{({\cos }^2\theta +{\sin }^2\theta )}={\mathrm{r}}^2\sin \phi$$

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