Visualizing the Divergence Theorem for a Cylinder

[jammydav93]

Homework Statement

Prove the divergence theorem for the vector field A = p = (x,y) and taking the volume V to be the cylinder of radius a with its base centred at the origin, its axis
parallel to the z-direction and having height h.

I can find the dV side of the equation fine (I think) however am unable to do the dS side to show they are equal to each other

Homework Equations

∫∫∫div(A) dV = ∫∫ (A . n) dS

The Attempt at a Solution

∫∫∫div(A) dV = 2*∏*r^2*h
∫∫ (A . n) dS = ??


Many thanks for all hints and help

James

 

[HallsofIvy]

First, I have a problem with your "A = p = (x,y)" because that is a two dimensional vector and for this to make sense we must be working in three dimensions. I am going to assume that you mean (x, y, 0).

In that case, yes, your calculation of the volume integral is correct. Because the boundary is not "smooth" you will want to do it in three parts:
The base is the disc bounded by x^2+ y^2= r^2, z= 0. The unit vector normal to that surface, directed outward, is n= <0, 0, 1> and dS= dxdy.

The top is the disc bounded by x^2+ y^2= r^2, z= h. The unit vector normal to that surface, directed outward, is n= <0, 0, -1> and dS= dxdy. It would probably be simplest to do those two integrals in polar coordinates.

The curved surface is given by x^2+ y^2= r^2, with z any number between 0 and h. We can find the "vector differential of surface area" of any surface, given by the "position vector \vec{v}= x(s,t)\vec{i}+ y(s,t)\vec{j}+ z(s,t)\vec{k} is given by the cross product of \partial \vec{v}/\partial s and \partial \vec{v}/\partial t. Standard parametric equations for the circle, of radius r, are x= r cos(\theta) and y= r sin(\theta) so parametric equations for the curved surface of the cylinder, using \theta and z as parameters are x= r cos(\theta), y= r sin(\theta), z= z so a position vector for a point , \vec{r}(\theta, z)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ z\vec{k}.

Differentiate those with respect to \theta and z and take the cross product. That cross product, with signs chosen to give the "outward normal" (changing the order of multiplication in a cross product changes the sign so we may have to correct that to give the correct direction), times d\theta dz is \vec{n} dS.

(Although it is easy to get "dS" for the top and bottom by inspection, we could do the same thing there. A position vector for the top, the plane z= h, would be \vec{v}= x\vec{i}+ y\vec{j}+ h with x and y as parameters. The derivatives with respect to x and y would be just \vec{i} and \vec{j} so the cross product is \vec{k}. The "outward" normal would be upward so \vec{k}. For z= 0, the same calculation gives \vec{k} again, but now the outward normal is downward so -\vec{k}.)

 

[jammydav93]

Yes sorry I did mean (x,y,0)

For the top and bottom face integrals, are they both just trivial as (x,y,0).(0,0,1) = 0 and (x,y,0).(0,0,-1) = 0, or am I doing this wrong?

Thanks for the indepth explanation!

 

[Vargo]

After you have worked through the surface integral the way Halls of Ivy explained, try to redo the entire surface integration without parametrizing the surface or calculating dS in coordinates. You can work it out in your head, and you are spot on about the top and bottom.

 

[jammydav93]

\frac{dr}{d\vartheta}=(-rsin(\vartheta), rcos(\vartheta),0)

\frac{dr}{dz}=(0,0,1)

\frac{dr}{d\vartheta} x \frac{dr}{dz} = (rcos(\vartheta), -rsin(\vartheta),0)

A . (\frac{dr}{d\vartheta} x \frac{dr}{dz}) = (x,y,0) . (rcos(\vartheta), -rsin(\vartheta),0) = (rcos(\vartheta), rsin(\vartheta),0) . (rcos(\vartheta), -rsin(\vartheta),0)
= r^2*cos^2(\vartheta) - r^2*sin^2(\vartheta) + 0

is that right for A .n ?

Thanks!

 

[Vargo]

No, there is a sign error. It should come out to something simpler. Try to visualize A and n (drawing a good picture would help). Then you can see what their dot product should be.