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Stokes's theorem in spherical coordinates

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  • #1
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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 [tex]\vec{F} \cdot d\vec{\ell}=(0,0,1)(dr, d\theta, rsin\theta \ d\phi)=rsin\theta \ d\phi[/tex]

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

Is this correct?

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

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.
 

Answers and Replies

  • #2
pasmith
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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.)

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

Moving on...

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

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

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

b. We have ##F_r=F_{\theta}=0## and ##F_{\phi}=1##, so for the curl I find [tex]\nabla \times \vec{F}=(\frac{1}{r}tan\theta, 0, -\frac{1}{r})[/tex]
I get [itex]\nabla \times \vec F = r^{-1}\tan\theta \hat r - r^{-1}\hat\theta[/itex].


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.
[itex]d\vec a[/itex] 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 [itex]\hat z[/itex] component.
 
  • #3
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No. The circle in question is [itex]r = R[/itex], [itex]\theta = \pi/2[/itex] and [itex]\phi \in [0, 2\pi)[/itex]. Thus the element of line length is [itex]\mathrm{d}\vec{l} = R\hat\phi \mathrm{d}\phi[/itex], and the integral is
[tex]
\oint_C \vec F \cdot \mathrm{d}\vec{l} = \int_0^{2\pi} R\mathrm{d}\phi
[/tex]
Ah, I see. So it's simply the perimeter of a circle. Makes sense. [tex]\int_0^{2\pi} R\mathrm{d}\phi = 2\pi R[/tex]

I get [itex]\nabla \times \vec F = r^{-1}\tan\theta \hat r - r^{-1}\hat\theta[/itex].
Whoops, yeah that's what I meant.

[itex]d\vec a[/itex] 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 [itex]\hat z[/itex] component.
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:
  • #4
vela
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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.
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
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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##.
That makes perfect sense. So it's ##d\vec{a}=R^2sin \ \theta \ d\theta \ d\phi \ \hat{r}##, right?
 
  • #6
vela
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Yup.
 
  • #7
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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
vela
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Yes. You're evaluating the curl of ##\vec{F}## on the surface of the sphere, so you can set r=R.
 
  • #9
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Okay. I have to integrate ##sin \ \theta \ tan \ \theta##. This is going to be a mess...
 
  • #10
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Uh oh, on wolfram alpha it's saying this integral does not converge. Something has to be wrong.
 
  • #11
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## \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
vela
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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
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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
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So, for the disk we have ## d\vec{A}= - R^2 sin \ \theta \ d\theta \ d \phi \ \hat{\theta}##?
 
  • #15
vela
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Yeah I get ##\int_0^{2 \pi} \int_0^{\pi} R cos \ \theta \ d\theta \ d\phi = 2 \pi R##.
That integral as written should evaluate to ##4\pi R.##

So, for the disk we have ## d\vec{A}= - R^2 sin \ \theta \ d\theta \ d \phi \ \hat{\theta}##?
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
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That integral as written should evaluate to ##4\pi R.##
Ah, ##\theta## should range from 0 to pi/2. Then the integral is correct.

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?
I'm not sure about the magnitude, but it should point in the +z direction.
 
  • #17
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##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?
 
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  • #18
vela
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Yup.
 

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