# Stokes's theorem in spherical coordinates

1. Sep 12, 2013

### wifi

Problem:

Say we have a vector function $\vec{F} (\vec{r})=\hat{\phi}$.

a. Calculate $\oint_C \vec{F} \cdot d\vec{\ell}$, where C is the circle of radius R in the xy plane centered at the origin

b. Calculate $\int_H \nabla \times \vec{F} \cdot d\vec{a}$, where H is the hemisphere above the xy plane with boundary curve C.

c. Calculate $\int_D \nabla \times \vec{F} \cdot d\vec{a}$, where D is the disk in the xy plane bounded by C.

(Verify Stokes's theorem in both cases.)

Solution:

First off, isn't it unnecessary to say verify Stokes's theorem in both cases? If we calculate a line integral, and then a surface integral bounded by the same curve, Stokes's theorem states that they're the same, yes?

Moving on...

a. We have $$\vec{F} \cdot d\vec{\ell}=(0,0,1)(dr, d\theta, rsin\theta \ d\phi)=rsin\theta \ d\phi$$

So $$\oint_C \vec{F} \cdot d\vec{\ell} = Rsin\theta \int_0^{2 \pi} d\phi= 2 \pi R sin \theta$$

Is this correct?

b. We have $F_r=F_{\theta}=0$ and $F_{\phi}=1$, so for the curl I find $$\nabla \times \vec{F}=(\frac{1}{r}tan\theta, 0, -\frac{1}{r})$$

I think this is right so far. But in calculating $\int_H \nabla \times \vec{F} \cdot d\vec{a}$, I'm not sure what to use as $d\vec{a}$. Any help? Thanks in advance.

2. Sep 12, 2013

### pasmith

"Verify Stoke's theorem" means "observe that the line integral is indeed equal to the surface integrals"

No. The circle in question is $r = R$, $\theta = \pi/2$ and $\phi \in [0, 2\pi)$. Thus the element of line length is $\mathrm{d}\vec{l} = R\hat\phi \mathrm{d}\phi$, and the integral is
$$\oint_C \vec F \cdot \mathrm{d}\vec{l} = \int_0^{2\pi} R\mathrm{d}\phi$$

I get $\nabla \times \vec F = r^{-1}\tan\theta \hat r - r^{-1}\hat\theta$.

$d\vec a$ is the normal to the surface in question times the element of area. Given the choice of parametrization of the line integral, it's the normal with the non-negative $\hat z$ component.

3. Sep 12, 2013

### wifi

Ah, I see. So it's simply the perimeter of a circle. Makes sense. $$\int_0^{2\pi} R\mathrm{d}\phi = 2\pi R$$

Whoops, yeah that's what I meant.

Well in class my prof used $r^2sin \theta d\theta d \hat{\phi}$. Is that correct? If so, I would like to understand why that's the case.

Thanks, pasmith.

Last edited: Sep 12, 2013
4. Sep 12, 2013

### vela

Staff Emeritus
No, that's not correct. $d\vec{a}$ points in the direction normal to the surface. For the hemisphere, this would be radially outward, so $d\vec{a}$ points in the $\hat{r}$ direction. The magnitude of the area element is $R^2\sin\theta\,d\theta\,d\phi$.

5. Sep 12, 2013

### wifi

That makes perfect sense. So it's $d\vec{a}=R^2sin \ \theta \ d\theta \ d\phi \ \hat{r}$, right?

6. Sep 12, 2013

### vela

Staff Emeritus
Yup.

7. Sep 12, 2013

### wifi

So when I calculate $\int_H \nabla \times \vec{F} \cdot d\vec{a}$ does $r \rightarrow R$ in $\nabla \times \vec{F}=(\frac{1}{r}tan \ \theta, -\frac{1}{r}, 0)$? (ie. is it $\nabla \times \vec{F}=(\frac{1}{R}tan \ \theta, -\frac{1}{R}, 0)$?)

8. Sep 12, 2013

### vela

Staff Emeritus
Yes. You're evaluating the curl of $\vec{F}$ on the surface of the sphere, so you can set r=R.

9. Sep 12, 2013

### wifi

Okay. I have to integrate $sin \ \theta \ tan \ \theta$. This is going to be a mess...

10. Sep 12, 2013

### wifi

Uh oh, on wolfram alpha it's saying this integral does not converge. Something has to be wrong.

11. Sep 12, 2013

### wifi

$\int_H \nabla \times \vec{F} \cdot d\vec{a} = \int (\frac{1}{R}tan \ \theta, -\frac{1}{R},0) \cdot (R^2 sin \ \theta \ d\theta \ d\phi, 0, 0) = \int_0^{2 \pi} \int_0^{\pi} R sin \ \theta \ tan \theta \ d\theta \ d\phi$. Yes?

12. Sep 12, 2013

### vela

Staff Emeritus
Mathematica gives
$$\vec{\nabla} \times \vec{F} = \frac{\cot \theta}{r}\,\hat{r} - \frac{1}{r}\,\hat{\theta}.$$ That should fix things for you.

13. Sep 12, 2013

### wifi

Yeah I get $\int_0^{2 \pi} \int_0^{\pi} R cos \ \theta \ d\theta \ d\phi = 2 \pi R$.

Hopefully I can get part c. with no difficulty. I'll keep you posted.

14. Sep 12, 2013

### wifi

So, for the disk we have $d\vec{A}= - R^2 sin \ \theta \ d\theta \ d \phi \ \hat{\theta}$?

15. Sep 12, 2013

### vela

Staff Emeritus
That integral as written should evaluate to $4\pi R.$

Nope. For one thing, r isn't constant over the disk. What should the magnitude of the area element be? Which way should it point?

16. Sep 12, 2013

### wifi

Ah, $\theta$ should range from 0 to pi/2. Then the integral is correct.

I'm not sure about the magnitude, but it should point in the +z direction.

17. Sep 13, 2013

### wifi

$d\vec{A}=-rdrd\phi \hat{\theta}$. Using this in the integral (with r:0 to R and Φ:0 to 2π), I get $2 \pi R$, as expected. Is this correct?

Last edited: Sep 13, 2013
18. Sep 13, 2013

### vela

Staff Emeritus
Yup.