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Question Number 60717 by peter frank last updated on 24/May/19

Commented by Prithwish sen last updated on 25/May/19

$$=\left|\begin{array}{c}\frac{\partial \mathrm{x}}{\partial \mathrm{r}}\frac{\partial \mathrm{x}}{\partial \theta }\frac{\partial \mathrm{x}}{\partial \mathrm{\varnothing }}\\ \frac{\partial \mathrm{y}}{\partial \mathrm{r}}\frac{\partial \mathrm{y}}{\partial \theta }\frac{\partial \mathrm{y}}{\partial \varphi }\end{array}\right|$$$$\left|\begin{array}{c}\frac{\partial \mathrm{z}}{\partial \mathrm{r}}\frac{\partial \mathrm{z}}{\partial \theta }\frac{\partial \mathrm{z}}{\partial \varphi }\end{array}\right|$$$$=\sin \theta \cos \varphi \left({\mathrm{r}}^2{\sin }^2\theta \cos \varphi \right)+\sin \theta \sin \varphi \left({\mathrm{r}}^2{\sin }^2\theta \sin \varphi \right)$$$$+\cos \theta ({\mathrm{r}}^2\sin \theta \cos \theta {\cos }^2\varphi +r^{2}cos\theta \sin \theta {\sin }^2\theta )$$$$={\mathrm{r}}^2{\sin }^3\theta ({\cos }^2\varphi +{\sin }^2\varphi )+{\mathrm{r}}^2\sin \theta {\cos }^2\theta$$$$={\mathrm{r}}^2\sin \theta$$

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