Flux through rotating cylinder using divergence theorem

Question

A vector field $V$ has the divergence $div(V)=4$.

What is the total flux in through the surface $\partial C$ of a massive rotating cylinder $C$ that has the height $h=2$ and the radius $r=5$?

My parameterization is:

$$r(u,v)=(5\cos v,5\sin v,u)$$

The normal vector to the cylinder I've calculated is:

$$\vec{n}=(-5\cos v,-5\sin v,0)$$

I don't know where to go from here. I'd appreciate your help.

zoli · Answer 1 · 2017-03-29T11:41:22.483

0

According to the Divergence Theorem, the surface integral of a vector field over a "simple" and closed surface equals the volume integral of the divergence of the same over the solid defined by the surface.

Since the divergence given is constant, the integral of its divergence equals the divergence times the volume of the solid. $V=5^2\cdot2\cdot \pi=50\pi$. So, the surface integral has to equal $4\cdot 50\pi=200\pi.$

I don't think that the rotating nature of the cylinder counts. (If ibeing massive or is rotating about its vertical axis of symmetry. If it behaves another way then the problem is not defined.) The cylinder determines the same body even if rotating or if hollow.

edited Mar 29 '17 at 11:41

answered Mar 29 '17 at 10:26

zoli

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The MapleTA I'm using says it's wrong, sorry. What about the height? – Steve Mar 29 '17 at 10:40
In $V=5^2\cdot2\cdot \pi=50\pi$: $2$ is the height. – zoli Mar 29 '17 at 11:08
It's for a cylinder though. Shouldn't it be $2\pi$ then? – Steve Mar 29 '17 at 11:23
The area of the bottom is $r^2\pi$ theheight is $h $.. – zoli Mar 29 '17 at 11:34
Oh sorry, I remembered the formula wrong. You're right, but it's still not the solution to my problem. – Steve Mar 29 '17 at 11:39
What is the alleged solution to your problem? – zoli Mar 29 '17 at 13:12

Flux through rotating cylinder using divergence theorem

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