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Surface area problem in 3-d calculus

  1. Apr 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the area of the surface.
    The surface z = (2/3)(x^(3/2) + y^(3/2)), 0 </= x </= 1, 0 </= y </= 1

    2. Relevant equations

    Double integral over S of the magnitude of dr/du cross dr/dv dS, which equals the double integral over D of the magnitude of dr/du cross dr/dv dA.
    (SSs |dz/dx X dz/dy|dS = SS D |dz/dx X dz/dy|dA)

    3. The attempt at a solution
    _ 1 1
    I found the integral to be SS D (1+x+y)^(1/2) dA = SS (1+x+y)^(1/2) dxdy
    _ 0 0
    but my answer keeps coming out wrong. I might be making a mistake with algebra because I get a lot of different answers when i do it different ways. This is from an NC State Calculus 3 homework assignment, if anyone may have seen this problem before and remember how to do it. Help???
    Last edited: Apr 15, 2008
  2. jcsd
  3. Apr 15, 2008 #2


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    Staff Emeritus
    Science Advisor

    What you have shown is completely correct. Now, how did you do that integration and what answer did you get?
  4. Apr 15, 2008 #3
    I got the right answer finally, 1.4066. It was an algebraic error when I was evaluating the integral. Thanks for telling me I was doing it right though, that helped me know it wasn't a mistake in my set-up.
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