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Evaluate the surface integral

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I've been reading up on surface integrals for several hours now but I can't get my head around this at all.

    There are just too many things flying around.

    Do I project dS onto a single plane and work it out as dA = dS/n.k? (where n is unit normal vector of surface and k is unit vector in z direction). This was my original plan but when I go through the steps I end up with a.dS = 0.

    Or is it a Stoke's Theorem thing?

    Or Divergence theorem?

    I thought I had these things nailed last night, but now I'm quite overwhelmed.

    Thanks for any pointers.
  2. jcsd
  3. May 5, 2014 #2
    Picturing it in 2D with a disk may be the best way to learn this, then apply the concepts to 3D. Imagine a flat disk in the XY plane. dS would be a vector along the z-axis whose magnitude is equal to a small area of the disk. If integrating this dS, to just S (the surface), one would have a vector whose magnitude is equal to the area of the disk. dS can point either along the +z axis, or -z axis, so you must be wary of direction. The vector "a" would simply represent some entity, and when dotted with dS, would be "how much of 'a' acts through this surface.

    It's the Divergence Theorem. The rate at which "a" changes throughout the volume is equal to the amount of "a" through the surface.
  4. May 5, 2014 #3
    Ok sure. I think I see now how I must be the divergence theorem. The integral is a vector over a closed surface, which essentially describes a divergence.

    Ok, so going by the divergence theorem:


    In the case of my problem I have:

    ∇.a = 1

    So I have

    ∫∫∫1dV , over the whole volume??? Which is just a plain volume integral for the given surface?

    Surely I've missed something here?
  5. May 5, 2014 #4
    No, that's right. With the disk, the volume would be that of a cylinder. So you're divergence = 1 implies "a" is growing linearly in a single direction. So the RHS of the equation would be something like (using the disk analogy), z*2pi*R, which is the volume of a cylinder. Both sides hold.
  6. May 5, 2014 #5
    Excellent! So much easier than I thought.

    To conclude (and to check), my final answer is:

    ∫∫∫1dV = ∫∫∫R2Sinø dRdθdø



    = 2/3 πa3
  7. May 5, 2014 #6
    You're off by a factor of 2. You're integral for Phi should be 0< Phi < pi.
  8. May 5, 2014 #7
    My fault, I didn't read the problem. You're right, as that is the volume of a single hemisphere.
  9. May 5, 2014 #8


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    Unfortunately, you are not given a closed surface, so the divergence theorem doesn't hold. And if you close up the surface by adding the circular disk in the xy plane, you have to account for the extra surface in the calculations. And in that case you only integrate over half the total sphere.
  10. May 6, 2014 #9
    So the previous answer from the other guy was wrong?

    In which case I'm still totally unsure of where to go with this, haha.
  11. May 6, 2014 #10
    It seems to me he did account for the surface? I picked a disk and told him to use the divergence theorem because I assumed it an easier route to take.
  12. May 6, 2014 #11
    My cylinder example was derived from a line charge along an axis. I hadn't read the problem until he attempted the integration. A disk is still applicable however, as to close the surface on the plane intersecting the normal would be the same.
  13. May 6, 2014 #12


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    Yes, it was wrong. I would just work the problem directly. Your problem uses the letter ##a## for two different things, the radius of the sphere and the vector in the integral. To avoid confusion, lets call ##\vec F = \langle x,0,0\rangle## and you are calculating ##\iint_S \vec F \cdot d\vec S## over the upper half of ##x^2+y^2+z^2 = a^2## oriented upward. You can parameterize the sphere as ##\vec R(\phi,\theta) = \langle a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi\rangle##. Then you can calculate ##d\vec S = \pm \vec R_\theta \times \vec R_\phi##, choosing the sign for outward orientation. Then work the integral in terms of ##\phi## and ##\theta## with appropriate limits for the hemisphere. Note that with the given ##\vec F## and symmetry of the hemisphere, you would expect an answer of zero.
  14. May 6, 2014 #13
    Now I would like some clarification...because I don't see how the projection on the surface could be zero when there is no symmetry argument for the point at the top of the hemispherical bowl.
  15. May 6, 2014 #14
    Great! I now see how it's not a divergence theorem problem because the flux is not through a closed surface. Haha.

    Yeah that method is far better in this case. I already learned how to parameterise for line integrals, and now I see how it can easily be done for surface integrals too, since a surface function is essentially a 3D version of a line function! (R(x,y) , S(x,y,z) , clear similarity..)

    I tried projecting it as a flat circle to and converting dS to dA but it got very messy. The parameterisation is far better

    Thanks very much :thumbs:

    Oh and it came to zero, heh
  16. May 6, 2014 #15
    It kind of makes sense when you think about it.

    It's a hemisphere lying flat on the x-y plane. And the vector through it is a constant xi.

    And keep in mind that the hemisphere is centred around x=0.

    So for points in the negative direction, the flux into it is exactly matched by the flux out on the positive side.

    It might be easier if you picture it from above. Then you just have a circle on an x-y plane with a bunch of straight arrows going through it from one side to the other.
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