Surface integral (Flux) with cylinder and plane intersections

[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]

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]

Did you catch my edit in the last post?

 

[Dick]

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]

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]

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.

 

[CAF123]

They are two different uses of the symbol 'dA'. They are two different things.

Ok, thanks. I think this also answers one of my other questions: The form dA=dr dθ is not really an area element because the dimensions don't check out. What is the difference between the two forms?

 

[Dick]

Ok, thanks. I think this also answers one of my other questions: The form dA=dr dθ is not really an area element because the dimensions don't check out. What is the difference between the two forms?

We've been over that several times already. dr dθ is the area element without the jacobian part 'r'. You use that in the cross product form of the surface integral because the cross product contains the jacobian part, 'r'. If you are just integrating a function in polar coordinates you want the full area element, i.e. with jacobian r dr dθ.

 

[CAF123]

We've been over that several times already. dr dθ is the area element without the jacobian part 'r'. You use that in the cross product form of the surface integral because the cross product contains the jacobian part, 'r'. If you are just integrating a function in polar coordinates you want the full area element, i.e. with jacobian r dr dθ.

Ok, everything makes sense. Thanks so much!