Integration and Laplacian in polar coordinates

[chilge]

Homework Statement

I have a function y that is axisymmetric, so that y=y(r).

I want to solve for r such that ∇²y(r) = Z.

Can anyone tell me if I'm following the right procedure? I'm not sure since there are two "∂/∂r"s present...

Homework Equations

∇² = (1/r)(∂/∂r)(r*(∂/∂r)) + (1/r²)*(∂²/∂θ²)
--> The terms involving theta become zero since y is a function of r only

The Attempt at a Solution

∇²y = (1/r)(∂/∂r)(r*(∂y/∂r)) = Z
(∂/∂r)(r*(∂y/∂r)) = Zr
integrate with respect to r: r(∂y/∂r) = (1/2)Z*r² + C
(∂y/∂r) = (1/2)Zr + C/r
integrate with respect to r again: y = (1/4)Z*r² + Cln(r) + K

 

[Dick]

Homework Statement

I have a function y that is axisymmetric, so that y=y(r).

I want to solve for r such that ∇²y(r) = Z.

Can anyone tell me if I'm following the right procedure? I'm not sure since there are two "∂/∂r"s present...

Homework Equations

∇² = (1/r)(∂/∂r)(r*(∂/∂r)) + (1/r²)*(∂²/∂θ²)
--> The terms involving theta become zero since y is a function of r only

The Attempt at a Solution

∇²y = (1/r)(∂/∂r)(r*(∂y/∂r)) = Z
(∂/∂r)(r*(∂y/∂r)) = Zr
integrate with respect to r: r(∂y/∂r) = (1/2)Z*r² + C
(∂y/∂r) = (1/2)Zr + C/r
integrate with respect to r again: y = (1/4)Z*r² + Cln(r) + K

Looks fine to me. If you substitute that back into the original equation you do get Z, right? That's a good way to check.