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Surface integral use stokes/divergence/whatever is convenient

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the closed surface S consisting of the graph [itex] z=1-x^2-y^2 [/itex] with [itex] z \ge 0 [/itex] and also the unit disc in the xy plane. Give this surface an outer normal. Compute: [itex] \int \int_s \mathbf{F} \cdot d \mathbf{S} [/itex]


    2. Relevant equations

    Stokes theorem, divergence theorem

    3. The attempt at a solution

    Well the divergence of F is 5.
    So I should calculate [tex] \int \int \int_S 5 dV [/tex]

    I'm not really sure where to go with this.
     
  2. jcsd
  3. Feb 18, 2012 #2

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    That expression already looks a lot simpler then the double integral! ;)

    So how to calculate a 3 dimensional integral?
    Perhaps you should switch to cylindrical coordinates to make it easier?
     
  4. Feb 18, 2012 #3
    So I really just calculate [tex] \int \int \int_S 5 dV [/tex] ?

    so.. would it be [tex] 5 \int \int \int r \; dz \; dr \; d \theta [/tex] ?

    How do I bound z?
     
  5. Feb 18, 2012 #4

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    Yes and yes. :)

    Your problem statement already gives a lower bound for z.
    What is the highest value of z where it still has an actual value (and does not become imaginary)?
     
  6. Feb 18, 2012 #5
    oh, 1.
    so I get 5 pi ?
     
  7. Feb 18, 2012 #6

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    Almost.
    I get a slightly different result.
    Perhaps one of us made a calculation error?
     
  8. Feb 18, 2012 #7
    I calculated [tex] 5 \int_0^{2\pi } \int_0^1 \int_0^1 r \; dz \; dr \; d \theta [/tex]
     
  9. Feb 18, 2012 #8

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    That's not quite right.
    The bounds of r depend on z.
     
  10. Feb 18, 2012 #9
    Are we integrating over a cylinder with height 1 and radius 1? If we are, we are allowed to use volume formulas. We aren't supposed to actually calculate stuff.
     
  11. Feb 18, 2012 #10

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    No, it's not a cylinder.

    The graph of ##z=1−x^2−y^2## is a paraboloid that extends downward.
    You can think of it as a kind of rounded cap of which you need the volume.

    Can you convert the equation of that graph to cylindrical coordinates?
     
  12. Feb 18, 2012 #11
    z=1-r^2?
     
  13. Feb 18, 2012 #12

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    Yep.
    The r in this equation is the upper bound for the r in your integral.

    Can you write this upperbound of r as a function of z?
     
  14. Feb 18, 2012 #13
    I shouldn't have to. I'm doing something wrong. If I actually calculate an integral I've done it wrong and missed the point here. Perhaps some other theorem would be a better approach?
     
  15. Feb 18, 2012 #14

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    Seems unlikely.

    You would not supposed to be calculating the double integral with the dot product.
    That is indeed a lot of work.

    But you would need to calculate the volume integral.
    Do you have a formula handy for the volume of a paraboloid?

    Otherwise, you will have to calculate it.
    There is an easier way however if you consider the volume to consist of a number of stacked circle disks.
    Each circle disk has volume ##\pi r^2 dz## with ##r^2## being equal to ##1-z##.
    This should be easy to integrate.
     
  16. Feb 18, 2012 #15
    Okay, I'm going to let this problem go for a while. I'm not getting any of the other problems either, so I'm going to go back to the drawing board and study the theorems some more. thanks for the help so far :)
     
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