By changing variables to spherical coordinates \(r, \theta, \phi\), such that \(x=r \sin \theta \cos \phi\), \(y=r \sin \theta \sin \phi, z=r \cos \theta\), show that the laplacian of \(p\) becomes $$ \nabla^{2} p=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial p}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial p}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2} p}{\partial \phi^{2}} $$

Question: Show that the Laplacian of a scalar function p, defined in Cartesian coordinates as \(\nabla^{2} p = \frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2} + \frac{\partial^2 p}{\partial z^2}\), transforms to the following expression in spherical coordinates \((r, \theta, \phi)\): \(\nabla^2 p = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial p}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial p}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 p}{\partial\phi^2}\). Answer: We can show that the Laplacian transforms to the given expression in spherical coordinates by calculating the partial derivatives of the function p with respect to the spherical coordinates, substituting them into the formula of the Laplacian in Cartesian coordinates, and simplifying the expression. The final expression for the Laplacian in spherical coordinates is: \(\nabla^2 p = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial p}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial p}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 p}{\partial\phi^2}\).