Flux Through Sphere: Cylindrical Coordinates

[bowlbase]

Homework Statement

The trick to this problem is the E field is in cylindrical coordinates.
$E(\vec{r})=Cs^2\hat{s}$

Homework Equations

$\int E \cdot dA$

The Attempt at a Solution

I tried converting the E field into spherical coords and I can find the flux that way but it is a complicated answer. The problem suggests keeping the field in cylindrical and doing the integral of the circle in cylindrical instead of spherical. I'm sort of lost on how I would do that. Would I have the limits of s be 0→R and z -R→R and $\phi$ the same as hat it would normally be?

I doubt it is that simple but since I've never tried to use non-optimal coordinates for an object I'm not entirely sure how I would go about this.

 

[rude man]

And the problem is ... ?

 

[bowlbase]

The problem is to find the flux through a sphere where the E field is given in cylindrical coordinates. I can't convert the field into spherical as the question specifically asks that I do it the other way. And, I must also finally graph the divergence on the sz plane.

 

[bowlbase]

For example

$E=Cs^2\hat{s}$
$s=rsin(\theta)$ and $\hat{s}=sin(\theta)\hat{r}+cos(\theta)\hat{\theta}$
so $E=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})$
$\int E \cdot dA=E4\pi r^2=4\pi r^2(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})$

The next step it asks me to calculate the divergence of E and then graph it on the sz plane.

I can do this with the original equation but I now have answers in two different coordinate systems. Which I suppose sounds fair since they did gave me two also.

$∇\cdot E=\frac{1}{s}\frac{∂}{∂s}(sE_s)$
$=\frac{C}{s}(3s^2)=3Cs$

Finally, it asks that I now do the integral $\int (∇\cdot E) dV$ to show that the two methods are equivalent. At first glance I would say they are not. So I probably made a mistake somewhere.

 

[rude man]

I've never encountered "s" in any cylindrical coordinate system. The cylindrical coordinates are usually denoted r, theta, z or r, phi, z.

Also, your ## expressions are not being translated, at least not on my computer.

 

[bowlbase]

I get that a lot. It's something, I think, that is inherited from Griffiths since his books are popular on campus.

We normally do $(s,\phi, z)$

 

[rude man]

I get that a lot. It's something, I think, that is inherited from Griffiths since his books are popular on campus.

We normally do $(s,\phi, z)$

OK, that's fine. Griffith is very popular so maybe things have changed since my time ...

And I apologize for my comment about your ## expressions. I was looking at the "go advanced" window ...

Gotta think a bit.

Meanwhile, you might re-post this in the math section since it involves a dot-product in cylindrical coordinates. I would have to translate the components into cartesians before taking the dot product unless it's very simple.