Computations of the expressions for the gradient, divergence, curl and Laplacian in Cartesian, cylindrical and spherical coordinates

[Win_odd Dhamnekar]

Suppose I want to compute curl of the vector-valued function \(\displaystyle f = f_1 \hat{i} + f_2 \hat{j} + f_3 \hat{k} \) in spherical coordinates.

How can I comute it given that curl in cartesian coordinates is \(\displaystyle \nabla \times f = \left( \frac{\partial f_3}{\partial y} -\frac{\partial f_2}{\partial z}\right)\hat{i} + \left(\frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x}\right) \hat{j} + \left(\frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y}\right)\hat{k} \)

Author computed gradient, divergence, curl and laplacian in spherical coordinates in the following table:




These above given computations are difficult to understand.

 

[Dr.Peterson]

Suppose I want to compute curl of the vector-valued function \(\displaystyle f = f_1 \hat{i} + f_2 \hat{j} + f_3 \hat{k} \) in spherical coordinates.

How can I comute it given that curl in cartesian coordinates is \(\displaystyle \nabla \times f = \left( \frac{\partial f_3}{\partial y} -\frac{\partial f_2}{\partial z}\right)\hat{i} + \left(\frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x}\right) \hat{j} + \left(\frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y}\right)\hat{k} \)

Author computed gradient, divergence, curl and laplacian in spherical coordinates in the following table:



View attachment 34660
These above given computations are difficult to understand.

I assume that by "compute" you don't just mean to use the formulas you show, but to derive them. Is that right?

Please show what you have tried. Did you look for a textbook or other source that shows the derivation? I found some, though they might give different forms of the formula. And did you try carrying out the work yourself and get stuck? Then show us.

I found, for example, this textbook that shows your formula but says

The derivation of the above formulas for cylindrical and spherical coordinates is straightforward but extremely tedious. The basic idea is to take the Cartesian equivalent of the quantity in question and to substitute into that formula using the appropriate coordinate transformation. As an example, we will derive the formula for the gradient in spherical coordinates.​

Reading the derivation they do show may help you.

Wikipedia has a different approach to the derivation. It also uses a different convention for the names of coordinates.