# Surface Integral of cylinder.

1. Jan 8, 2010

### hhhmortal

1. The problem statement, all variables and given/known data

I’m trying to integrate the surface of a cylinder.
I know when integrating the surface of a cylinder the surface element is:
ρdØdz

Where ρ² + z² = r²

And for a sphere it is:
r²sinθdθdØ

In a sphere r=ρ

But in a cylinder when I’m integrating its surface, could it be written as:

(r² - z² )½ . dθ.dz

For example ∫∫zdS over a cylinder from 0<z<5 and 0<Ø<2π

Would it be:

∫∫z. (r² - z² )½ . dθ.dz ?

2. Jan 9, 2010

### HallsofIvy

You seem to be confusing "$\rho$" and "r". In spherical coordinates, $\rho$ is the straight line distance from the origin to a point. In cylindrical coordinates, "r" is the same as in polar coordinates- the straight line distance from the origin to a point in the xy-plane. In three dimensional cylindrical coordinates r is the distance from the origin to t he point (x,y,0) directly "below" the point (x,y,z). The "differential of area", on the surface of a sphere of (fixed) radius r, is $r d\theta dz$ where r is a constant. I can see no reason to introduce "$(\rho^2- z^2)^{1/2}$".

Last edited by a moderator: Jan 9, 2010