Surface integral (Flux) with cylinder and plane intersections

CAF123

What I said was [tex] F(θ,z) \cdot (R\cos\theta i+ R\sin\theta j) = R\cos\theta z i + R^2\cos\theta \sin\theta j + R\sin\theta z k \cdot (R\cos\theta i + R\sin\theta j) = R^2 cos^2\theta + R^3 cos\theta sin^2 \theta[/tex]

Edit: thanks for clarifying things. Ok, I will check my work

Dick

That's correct. See my last post. You don't need extra r in the volume element. I was using n=(cos(theta),sin(theta),0) the unit normal.

Dick

Yours is still the wrong one. Use what you got but drop the extra R from the dS!

CAF123

Ok, I get your answer now. Thanks a lot for staying with me and for your time.

Dick

No problem. BTW checking dimensions on terms can be handy. R and H are both lengths. So all of your terms should have the same number of 'lengths' in them. So I could tell that your answer in post 13 MUST be wrong, even if I hadn't done the calculation.

CAF123

Just a quick question about one of the things here. I said ##dA = dz d\theta##. Why is this the case? In polars, we have ##dA = r\,dr\,d\theta## which makes sense as an area since ##\theta## is dimensionless.

Dick

[tex] \int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA. [/tex]

If you look at the above the dS has the dimensions of area. In the cross product form the extra r is coming from the cross product.

CAF123

In general, is ##dS = ( r_u \times r_v)\,dA, ##where ##dA = du\,dv##. So in polars, I could say ##dS = r\,dr\,d\theta##, where ##dA= dr\,d\theta## if I am correct.

Dick

That's it. Work out a simple problem like finding the area of the unit disk. You'll that that the cross product term is r.

CAF123

I was looking at some other site (referenced below) and they take dA=r dr dθ. Why is this?

http://calculus7.com/sitebuildercont...ntesdblint.pdf

Dick

Because they aren't using the form with the cross product in it.

CAF123

What's the other form you mentioned? When I thought about it later, I see that they were just converting from Cartesian to polar and so the r was the Jacobian.

Dick

Yes, r is the Jacobian. The cross product factor also represents the Jacobian.

CAF123

I see. So they are essentially using the form $$ dS = \frac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA = \hat{n} r\,dr\,d\theta$$
EDIT: if this is the case, then shouldn't ##|r_u\times r_v| dA = r\,dr\,d\theta, ##instead of just ##dA##?

Dick

Yes, that's it.

CAF123

Did you catch my edit in the last post?

Dick

Assuming we are still on polar coordinates ## \vec{r}=(r cos \theta, r sin \theta, 0)## , ##|\vec{r}_r\times \vec{r}_\theta|=r##, ##dA= dr d\theta##. So ##|\vec{r}_r\times \vec{r}_\theta| dr d\theta## IS ##r dr d\theta##.