Volume of paraboloid using divergence theorem (gives zero)

[unscientific]

Homework Statement

A surface S in three dimensional space may be specified by the equation
f(x, y, z) = 0, where f(x, y, z) is a real function. Show that a unit vector nˆ normal to
the surface at point (x0, y0, z0) is given by

Homework Equations

The Attempt at a Solution

r = (x, y, z)

∇f = 2(x, y, -z)

n = 1/r (x, y, -z)

dS = dx dy √[1 + (x/z)² + (y/z)²] (1/r) (x, y, -z)

When i take r (dot) dS it gives x² + y² - z² which = 0..



Method 2 (Cylindrical Coordinates)

dS = n (r dr d∅)

but this still gives 0 when n (dot) r

 

[haruspex]

When i take r (dot) dS it gives x² + y² - z² which = 0..

... which is correct for that part of the surface. But there's another part, governed by z = h.

 

[unscientific]

... which is correct for that part of the surface. But there's another part, governed by z = h.

Ah i see, I missed out that surface entirely! (Which happens to be simply a disc)

Since divergence theorem only works for closed volumes.