Question

# Following the derivation of gradient and divergence from rectangular (x, y, z) to cylindrical (p, 0, z) coordinates,derive the similarly by transforming from rectangular (x, y, z) to spherical (r,

0, \$) coordinates. That is show,following the same method as discussed in lectures, the relationships for gradient and divergence in spherical coordinates. \text { 1. } \nabla \psi=\hat{\mathbf{r}} \frac{\partial \psi}{\partial r}+\hat{\phi} \frac{1}{r} \frac{\partial \psi}{\partial \phi}+\hat{\theta} \frac{1}{r \sin \phi} \frac{\partial \psi}{\partial \theta} \text { where we have } \psi(x, y, z) \Longrightarrow \psi(r, \theta, \phi) \text {. } \text { If the vector } \overrightarrow{\mathbf{A}}=\hat{\mathbf{r}} \mathrm{A}_{r}(r, \theta, \phi)+\hat{\theta} \mathrm{A}_{\theta}(r, \theta, \phi)+\hat{\phi} \mathrm{A}_{\phi}(r, \theta, \phi) \text { as is expressed in spherical system of coordinates, } then we can obtain the divergence in spherical system of coordinates as: then we can obtain the divergence in spherical system of coordinates as: \nabla \cdot \overrightarrow{\mathbf{A}}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \mathrm{~A}_{r}\right)+\frac{1}{r \sin \phi} \frac{\partial}{\partial \phi}\left(\mathrm{A}_{\phi} \sin \phi\right)+\frac{1}{r \sin \phi} \frac{\partial \mathrm{A}_{\theta}}{\partial \theta} \nabla \cdot \overrightarrow{\mathbf{A}}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \mathrm{~A}_{r}\right)+\frac{1}{r \sin \phi} \frac{\partial}{\partial \phi}\left(\mathrm{A}_{\phi} \sin \phi\right)+\frac{1}{r \sin \phi} \frac{\partial \mathrm{A}_{\theta}}{\partial \theta}

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