Question

Prove that the offset of cartesian --> spherical is ρ^2sinφ

Answer 1

not sure what you mean by 'offset' but this is from the Jacobian matrix of the change of variables. Recall that going from spherical to Cartesian we use the following:$$x=\rho\sin \phi\cos\theta\\y=\rho\sin\phi\sin\theta\\z=\rho\cos\phi$$Computing derivatives, we observe:$$\frac{\partial x}{\partial \rho}=\sin\phi\cos\theta,\frac{\partial x}{\partial \phi}=\rho\cos\phi\cos\theta,\frac{\partial x}{\partial \theta}=-\rho\sin\phi\sin\theta\\\frac{\partial y}{\partial \rho}=\sin\phi\sin\theta,\frac{\partial y}{\partial \phi}=\rho\cos\phi\sin\theta,\frac{\partial y}{\partial \theta}=\rho\sin\phi\cos\theta\\\frac{\partial z}{\partial\rho}=\cos\phi,\frac{\partial z}{\partial\phi}=-\rho\sin\phi,\frac{\partial z}{\partial\theta}=0$$ hence our Jacobian matrix is given by:$$\mathbf{J}=\begin{bmatrix}\sin\phi\cos\theta&\rho\cos\phi\cos\theta&-\rho\sin\phi\sin\theta\\\sin\phi\sin\theta&\rho\cos\phi\sin\theta&\rho\sin\phi\cos\theta\\\cos\phi&-\rho\sin\phi&0\end{bmatrix}$$now we merely compute its derivative:$$\det\mathbf{J}=\rho^2\cos^2\phi\sin\phi\cos^2\theta+\rho^2\sin^3\phi\sin^2\theta+\rho^2\cos^2\phi\sin\phi\sin^2\theta+\rho^2\sin^3\phi\cos^2\theta\\\ \ \ \ \ \ \ =\rho^2\sin\phi\ (\cos^2\theta\cos^2\phi+\sin^2\phi\sin^2\theta+\cos^2\phi\sin^2\theta+\sin^2\phi\cos^2\theta)\\\ \ \ \ \ \ \ =\rho^2\sin\phi\ (\cos^2\phi+\sin^2\phi)(\cos^2\theta+\sin^2\theta)\\\ \ \ \ \ \ \ =\rho^2\sin\phi$$

Answer 2

You really know how to use that equation tool. Well said!