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Surface area proof (vector analysis)

  1. Mar 15, 2013 #1
    edit: vector* analysis; sorry for the typo
    1. The problem statement, all variables and given/known data
    Given that A = ||ru||2, B = ru[itex]\bullet[/itex]rv, C = ||rv||2
    surface area of S is Area(S) = [itex]\int^{d}_{c}[/itex][itex]\int^{b}_{a}[/itex] = sqrt (AC - B2) dudv

    The Dirichlet energy can be thought of as a function as follows

    E(S) = 1/2[itex]\int^{d}_{c}[/itex][itex]\int^{b}_{a}[/itex] = (||ru||2+||rv||2) dudv

    Show that Area(S) ≤ 1/2 E(S) and that equality holds if
    ||ru||2 = ||rv||2 and ru [itex]\bullet[/itex] rv = 0
    2. Relevant equations



    3. The attempt at a solution
    Given that ||ru||2 = ||rv||2 and A = ||ru||2 I know that
    E(s) = 1/2[itex]\int^{d}_{c}[/itex][itex]\int^{b}_{a}[/itex] = (2A) dudv
    = [itex]\int^{d}_{c}[/itex][itex]\int^{b}_{a}[/itex] = (A) dudv


    Since A = C, and B = 0, then Area(S) = Area(S) = [itex]\int^{d}_{c}[/itex][itex]\int^{b}_{a}[/itex] = sqrt (A2) dudv

    which is just
    [itex]\int^{d}_{c}[/itex][itex]\int^{b}_{a}[/itex] = A dudv

    So I showed that they're equal, but how do I justify that Area(S) is less than 1/2 E(S)?
     
    Last edited by a moderator: Mar 16, 2013
  2. jcsd
  3. Mar 15, 2013 #2

    haruspex

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    Looks to me that it should not depend on the integration, i.e. If it is going to be true then it should be provable by comparing the integrands.
    So investigate (||ru||2+||rv||2) 2
     
  4. Mar 15, 2013 #3
    Hmm...I'm not sure where that leads me in the proof, could you please elaborate?
     
  5. Mar 16, 2013 #4

    haruspex

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    (Just noticed that in every one of your integrals the integral sign is separated from the integrand by an equals sign. i'm assumng that's a typo. If not, there's somethng in your notation I'm not familiar with. Also, you seem to have too many 1/2s floating around. Either there's a 1/2 in the definition of E(S) or there's a 1/2 in the inequality to be proved, not both.)

    Compare the squares of the integrands, i.e. compare (||ru||2+||rv||2) 2/4 with AC-B2. Can you see that one is guaranteed at least as large as the other?
     
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