Solved

# Outward flux of surface

Posted on 2008-10-31

Consider the parametric curve: (sinh(t), cosh(t)), -1 <= t <= 1.

Revolve it about the x-axis to form a 3D surface, S.

I'm asked to use the Divergence Theorem to find the volume of the shape.

Obviously, if we choose f=(x,y,z), then div f = 3, and so:

(Volume Integral over V) div f = 3 * Volume of shape

and so, by the Divergence Theorem:

Volume of shape = 1/3 * (Surface Integral over S, D+, and D-) f dot N

where D+ and D- are the flat discs/surfaces either side of the shape, and N is the surface normal.

If I calculate and sum the surface integrals for D+ and D-, I get: 2*Pi*Sinh(1)*Cosh^2(1). This part seems trivial.

However, I don't seem to get an integrable function when I attempt to integrate f dot N over S.

Here's the basic outline of my attempt:

Expressing the surface parametrically in terms of (u,t):

x = t, y = Sqrt(1 + t^2)*Cos(u), z = Sqrt(1 + t^2)*Sin(u).

So, f = (x,y,z) = (t, Sqrt(1+t^2)*Cos(u), Sqrt(1+t^2)*Sin(u)).

Then I calculate the surface normal to be:

N = [t, -Sqrt(1+t^2)*Cos^2(u), -Sqrt(1+t^2)*Sin(u)] / Sqrt(2+Cos(2u)).

And then f dot N seems to simplify to:

f dot N = -1 / Sqrt(2 + Cos(2u)).

However, first of all, this doesn't seem easily integrable (elliptic function, blah blah?), but secondly, doesn't seem to depend on t (which I would expect of it).

So I'm guessing I've gone wrong somewhere. Does anyone get the surface normal to be different to me? Or, f dot N? Perhaps there's a more suitable parameterization?

Any help would be much appreciated, thanks.