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Double Integration - Surface Area

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the surface area of the surface defined by:

    z = (2/3)(x^(3/2) + y^(3/2)), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

    2. Relevant equations

    SA = ∫∫ √(fx)^2 + (fy)^2 + 1 dA

    3. The attempt at a solution

    Solved for the partial derivatives and plugged them into the formula, but got a really messy answer in the end. I think I was supposed to do a change of variables somewhere but I'm not sure how...
     
  2. jcsd
  3. Apr 29, 2013 #2

    SammyS

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    Use adequate parentheses.

    Do you mean SA = ∫∫ √((fx)^2 + (fy)^2 + 1) dA , or perhaps something else.

    Added in Edit:
    What did you get for ∂f/∂x ?

    The integrand turns out to be fairly simple.
     
  4. Apr 29, 2013 #3
    Yes, sorry - I meant ∫∫ √((fx)^2 + (fy)^2 + 1) dA .

    I had √(x) for ∂f/∂x and √(y) for ∂f/∂y.
     
  5. Apr 29, 2013 #4

    Zondrina

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    It doesn't matter what order of integration you go in this case. If you plug in your ##f_x## and ##f_y## into your integral it becomes very easy with a regular substitution.

    i.e Lets say you integrate x first. So u = x + y + 1 implies du = dx.
     
  6. Apr 29, 2013 #5
    Ah, now it makes perfect sense! Thanks!
     
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