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Surface integral (Flux) with cylinder and plane intersections

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Dick
#19
Nov4-12, 01:32 PM
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Quote Quote by CAF123 View Post
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]
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
#20
Nov4-12, 01:35 PM
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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!
CAF123
#21
Nov4-12, 01:36 PM
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Quote Quote by Dick View Post
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.
Dick
#22
Nov4-12, 02:08 PM
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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.
CAF123
#23
Nov26-12, 04:54 PM
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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
#24
Nov26-12, 05:19 PM
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Quote Quote by CAF123 View Post
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.
[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
#25
Nov27-12, 02:38 AM
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Quote Quote by Dick View Post
[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.
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
#26
Nov27-12, 06:23 AM
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Quote Quote by CAF123 View Post
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.
CAF123
#27
Nov27-12, 08:10 AM
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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
#28
Nov27-12, 10:01 AM
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Quote Quote by CAF123 View Post
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.
CAF123
#29
Nov27-12, 10:06 AM
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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
#30
Nov27-12, 10:14 AM
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Quote Quote by CAF123 View Post
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.
CAF123
#31
Nov27-12, 10:24 AM
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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
#32
Nov27-12, 10:28 AM
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Quote Quote by CAF123 View Post
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.
CAF123
#33
Nov27-12, 10:31 AM
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Did you catch my edit in the last post?
Dick
#34
Nov27-12, 10:58 AM
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Quote Quote by CAF123 View Post
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##?
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##.
CAF123
#35
Nov27-12, 11:07 AM
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Quote Quote by Dick View Post
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##.
I see that this implies that dS = r dr dθ. But in the example from the website they also have dA = r dr dθ. How does this follow if dA = dr dθ? How can it equal two different things?
Dick
#36
Nov27-12, 11:16 AM
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Quote Quote by CAF123 View Post
I see that this implies that dS = r dr dθ. But in the example from the website they also have dA = r dr dθ. How does this follow if dA = dr dθ? How can it equal two different things?
They are two different uses of the symbol 'dA'. They are two different things.


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