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Stokes' Theorem

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Through Stokes' Theorem, I am given a formula and vector (see attached document), where V is a vector, and S is the right-circular cylinder (including the endcaps) which is bounded by (x^2) + (y^2) = 9, z=0, and z = 5.



    2. Relevant equations

    See attached document.



    3. The attempt at a solution

    I took the curl of V and set up the double integral. However, I do not know how to work out the normal unit vector. How do I decide what the limits are? In general, with Stokes' Theorem, how would one go about solving the surface double integral with a 3 dimensional solid?

    I am still not confident in Stoke's theorem. I would appreciate any educational information on this subject. Thanks.
     

    Attached Files:

  2. jcsd
  3. Sep 11, 2009 #2
    Correction: I am actually supposed to use Gauss' Theorem. But, in addition to that, I would also appreciate a Stokes' approach.

    Thanks.
     
  4. Sep 11, 2009 #3

    gabbagabbahey

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    Your attachment could take a while to get approved, why not just type out the problem?
     
  5. Sep 11, 2009 #4
    This was in the attachment:

    V n dS

    V = 2xy (i) - y² (j) + (z + xy) (z)
     
  6. Sep 11, 2009 #5

    Dick

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    It sounds like you are actually talking about the divergence theorem. That's the one that relates a volume integral to a surface integral of the sort you are talking about, and is also basically Gauss' theorem. If so, find the divergence of the vector field and integrate over the cylinder.
     
  7. Sep 11, 2009 #6
    I know I must integrate over the whole cylinder. But my question deals with the details of the integral. For instance, I assume I'm going to have to integrate over three surfaces: the top, bottom, and sleeve. The top and bottom integrals should be pretty straightforward. But when integrating the sleeve, what would my differentials be? I know I have the height of the cylinder (in the z-direction), but would I not also have dL in the horizontal plane (whose normal unit vector would be the gradient of the scalar [x^2 + y^2 = r^2] over the magnitude)?

    I am confused as to the latter details.

    Thanks.
     
  8. Sep 11, 2009 #7

    Dick

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    I wouldn't do the surface integral if you can help it. It's equal to the volume integral of the divergence isn't it? Just do that one. Unless you HAVE to do both.
     
  9. Sep 12, 2009 #8
    I would like to use Stokes' and Gauss' theorem. But okay, lets stick with Gauss' for now.

    So, I would have

    [tex]\int[/tex][tex]\int[/tex][tex]\int{\nabla .\vec{V} d\nu}[/tex]

    I have taken the divergence. But, would [tex]d\nu[/tex] just be the volume of a cylinder? And would I translate to polar coordinates?
     
  10. Sep 12, 2009 #9

    Dick

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    Use cylindrical coordinates, if you wish. dv is the volume element=r*dr*dphi*dz. But first calculate div(V). What did you get? You may find this problem is much easier than you think.
     
  11. Sep 12, 2009 #10
    The div V = 2x - 2y + 1.
     
  12. Sep 12, 2009 #11

    Dick

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    Wrong, div(V)=1. That's why it's easy.
     
  13. Sep 12, 2009 #12
    Oops! Typo, I meant to say 2y-2y+1. Yes, it is 1.

    And the integration limits are (0, r) (0, 2pi), (0, z)?
     
  14. Sep 12, 2009 #13

    Dick

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    Integrate 1 over the volume of cylinder. If you know a formula for the volume of a cylinder you don't have to do any integration at all. Just multiply the volume of the cylinder by 1. If you don't then the integration limits are r=[0,3], phi=[0,2*pi], z=[0,5], right?
     
  15. Sep 12, 2009 #14
    Ya, I got it: 45*pi (through integration) which is also what you would get if you directly took the volume of the cylinder and multiplied it by one as you've suggested.

    My instructor did a similar problem with Stokes' theorem with a vector F (which in fact was identical to the position vector r). And the answer he got was exactly the same as if one had just directly taken the volume of the cylinder with the basic dimensions.

    Hence, are Stokes' and Gauss' theorems just a way of finding volume? Or were these coincidences?
     
  16. Sep 12, 2009 #15

    Dick

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    If the divergence is 1, then sure, you are just finding volume. If the divergence is not 1 then you are finding flux through the surface of a variable charge distribution. I'm not sure what you are asking. Gauss (divergence) and Stokes CAN be used to find volume and area, but that's not their only use.
     
  17. Sep 12, 2009 #16
    Thanks, now I understand.

    If I wanted to use Stokes' Theorem on the 'sleeve' of the cylinder (i.e. the middle surface between top and bottom), how would I set up the differential elements on that sort of problem?

    P.S. Sorry for the load of questions, I'm just trying to understand the concept behind these formulas.
     
  18. Sep 12, 2009 #17

    HallsofIvy

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    Either do it directly, without Stokes theorem or use Stokes theorem to find the integral over the entire cylinder, then subtract of the integral over the "flat" ends.

    To integrate directly over the cylinder [itex]\{(x,y,z)| x^2+ y^2= 9, 0\le z\le 5}[/itex] use cylindrical coordinates with r set equal to 3, the radius of the cylinder. Then [itex]x= 3cos(\theta)[/itex], [itex]y= 3sin(\theta)[/itex], z= z. You can write the "position vector" of a point on that surface as [itex]\vec{r}= 3cos(\theta)\vec{i}+ 3sin(\theta)\vec{j}+ z\vec{k}[/itex]. The derivatives of that are [itex]\vec{r}_\theta= -3sin(\theta)\vec{i}+ 3cos(\theta)\vec{j}[/itex] and [itex]\vec{r}_z= \vec{k}[/itex] and are vectors in the tangent plane to the cylinder at each point. The cross product [itex]\vec{r}_\theta\times\vec{r}_z= 3cos(\theta)\vec{i}+ 3sin(\theta)\vec{j}[/itex] gives the "vector differential of surface area", [itex]3(cos(\theta)\vec{i}+ sin(\theta)\vec{j})d\theta dz[/itex]. Take the dot product of that with your vector function and integrate.

    Since here your vector function is [itex]2xy\vec{i}- y^2\vec{j}+ (z+ xy)\vec{k}[/itex][itex]= 2sin(\theta)cos(\theta)\vec{i}- sin^2(\theta)\vec{j}+ (z+ sin(\theta)cos(\theta))\vec{k}[/itex], that dot product is [itex]6cos^2(\theta)sin(\theta)- 3sin^3(\theta) and the integral would be
    [tex]3\int_{z= 0}^5\int_{\theta= 0}^{2\pi}(2cos^2(\theta)sin(\theta)- sin^2(\theta))d\theta dz[/tex]
    [tex]= 3\int_{z=0}^5\int_{\theta= 0}^{2\pi}(2cos^2(\theta)0- sin^2(\theta))sin(\theta)d\theta dz[/tex]
    [tex]= 3\int_{z=0}^5\int_{\theta= 0}^{2\pi}(3cos^2(\theta)- 1)sin(\theta)d\theta dz[/tex]
     
    Last edited: Sep 12, 2009
  19. Sep 12, 2009 #18
    This makes sense, thanks. Though, I am wondering if you are using Stokes' theorem or not. The reason for this query is that I am trying to learn how to interpret Stokes' theorem and its fundamental variables when applying it.

    I see how you have come up with the differential surface area. But, what about the curl of [tex]\stackrel{\rightarrow}{V}[/tex] and multiplying it with the unit vector [tex]\widehat{n}[/tex]? When I take the unit vector and dot it with the curl of the vector field, I get a scalar. Now, I want to multiply it by the surface area which you derived, but it is a vector. What would I do with this (in accordance with Stokes' theorem)?
     
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