# Volume of paraboloid using divergence theorem (gives zero)

1. May 28, 2013

### unscientific

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

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

2. Relevant equations

3. 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

2. May 28, 2013

### haruspex

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

3. May 29, 2013

### unscientific

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

Since divergence theorem only works for closed volumes.

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