## Surface integral (Flux) with cylinder and plane intersections

 Quote by Dick Then the second term in your integrand picked up extra R someplace.
What I said was $$F(θ,z) \cdot (R\cos\theta i+ R\sin\theta j) = R\cos\theta z i + R^2\cos\theta \sin\theta j + R\sin\theta z k \cdot (R\cos\theta i + R\sin\theta j) = R^2 cos^2\theta + R^3 cos\theta sin^2 \theta$$

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

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 Quote by CAF123 What I said was $$F(θ,z) \cdot (R\cos\theta i+ R\sin\theta j) = R\cos\theta z i + R^2\cos\theta \sin\theta j + R\sin\theta z k \cdot (R\cos\theta i + R\sin\theta j) = R^2 cos^2\theta + R^3 cos\theta sin^2 \theta$$
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.

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 Quote by CAF123 Edit: thanks for clarifying things. Ok, I will check my work
Yours is still the wrong one. Use what you got but drop the extra R from the dS!

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

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 Quote by CAF123 Ok, I get your answer now. Thanks a lot for staying with me and for your time.
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.
 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.

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 Quote by 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.
$$\int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA.$$

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.

 Quote by Dick $$\int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA.$$ 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.
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.

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 Quote by 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.
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.
 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

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

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 Quote by 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.
Yes, r is the Jacobian. The cross product factor also represents the Jacobian.
 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##?

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 Quote by CAF123 I see. So they are essentially using the form $$dS = \frac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA = \hat{n} r\,dr\,d\theta$$
Yes, that's it.
 Did you catch my edit in the last post?

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 Quote by 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##?