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Alright, so I forgot how to do Surface Integrals

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data

    I have a vector function, and I need to take the surface integral of it over a hemisphere, top half only. I'm "confirming" the divergence theorem by doing a volume integral and surface integral. Already did the volume one so I have something to compare to already.

    3. The attempt at a solution

    Yeah yeah, "look at your book!"

    I've checked my calc book, my mathematical physics book, and my E&M book (which assigned the problem) and for some reason I just can't get it.

    EDIT: Also, I should mention that the vector isn't given in "i,j,k" form, but "r, theta, phi", which is even more confusing for me.

    The books either use a simple cube (ya thanks) or cylindrical coordinates, which still makes it easier for me to grasp.

    Let me see if I am thinking of this correctly:

    dS will be d(theta)d(phi) for the top half, and should be d(r)d(phi) for the disk (My phi goes from 0 to 2pi). However I think I am missing something (Jacobian?). Besides that, I need to dot the function vector with the normal vector, yes?

    The book says I need to take the gradient of the surface function (and normalize it) to make it a normal vector. So for the top half it would just be <r,0,0>? And I can't figure it out for the disk at the bottom. I guess I could use -k and then just transformed into spherical coordinates?
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 4, 2007 #2


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    Yes -- you're missing the area. dS is an area element, so if you parametrize it with, say, [itex]d\theta d\phi[/itex], then you need to include the area of the sphere element spanned by that differential.

    and [itex]d\vec{S}[/itex] is [itex]\bf{\hat{n}}dS[/itex], so yes.

  4. Oct 5, 2007 #3
    Limits of integration? Or are you talking about something else?

    Ok, so if I had <1,0,0>, that would work, right? Since it would be 1 unit in the "r" direction, so radially outward.
  5. Oct 5, 2007 #4
    I hate myself and I want to die.

    I started reading different sections of the book (going over curvilinear coordinates) and at the bottom of one of the pages, it gives me exactly what I needed... what da is when I'm integrating over a sphere, and when it's a disk.

    Should I switch my major?

    EDIT: Oh, and 5 minutes after my discovery, I am done with the problem. I spent over an hour on it before that.

    I really want to cry...
    Last edited: Oct 5, 2007
  6. Oct 5, 2007 #5


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    Just an hour? Pfft. I've spent days, maybe even weeks wondering about something before I discovered it was in one of the books I have lying around! :tongue:

    Let this be a lesson in research -- knowing where to find information is almost as good as knowing that information. :smile:

    "Look at your book" isn't idle advice: it's a vital habit for a mathematician!
    Last edited: Oct 5, 2007
  7. Oct 5, 2007 #6
    Yeah. Every time I think I know something, it turns out I don't. I was looking under the "divergence" and "divergence theorem" sections, when I should have been looking at the curvilinear section. I mean, I already know that stuff, and I didn't think the author would just idly throw in "Oh yeah, use *this* for da using these coordinates in these 2 circumstances."

    And actually I lost track of time. 'twas already two hours and I had also spent another 2 at school. :(
  8. Oct 5, 2007 #7
    Don't worry, I don't think I would be able to do a surface integral without symmetry. I would either have to think really hard about it for a day and derive it myself, or look it up. While looking something up is a time saver, there is always something to be said about deriving math.
  9. Oct 5, 2007 #8
    To sort of go against my thread on professors only doing proofs: yes, I find I learn something better when I do it the hard way.
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