See attached file.
Before addressing the problem, Let's first derive the gradient operator (denoted as nabla), divergence and Laplacian in spherical coordinates. You could, of course, skip this and just use standard formulas, but this is not recommended since you have difficulties with this problem. After you've read the derivations (the problem itself is trivial), you should derive the expressions for the gradient, divergence and Laplacian in cylindrical coordinates.
By r we denote the position vector and by |r| it's length. If f(r) is a function of the position vector, then changing r by an infinitesimal amount dr will lead to a change of the function. The gradient operator is defined such that change of the function, denoted by df is given by:
df = (nabla f) dot dr (1)
I.e. the infinitesimal change of the function, df, is given by the inner product of the gradient of the function and the infinitesimal change in the position vector, dr.
To derive the expression for the gradient operator, we can use the fact that for any differentiable function of n variables x1, x2, ...xn we always have:
df = (df/dx1) dx1 + (df/dx2) dx2 + (df/dx3) dx3 + ... + (df/dxn) dxn (2)
Here the derivatives are partial derivatives. To find the expression for the gradient operator you just compare (1) to (2). In case of spherical coordinates you have three variables |r|, theta and phi, where |r| is the magnitude of the position vector, theta the angle with the z-axis and phi is the angle the projected vector on the x-y plane makes with the x-axis (loosely speaking the angle we use to rotate the vector around the z-axis). The components of the infinitesimal displacement vector dr are given by:
dr = r-hat d|r| + theta-hat |r| d theta + phi-hat |r| sin(theta) d phi (3)
Here r-hat, theta-hat and phi-hat are the (mutually orthogonal) unit vectors in the directions ...
A detailed solution is given.