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Evaluate the following surface integral

  1. Jul 15, 2013 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_S 2z+1 dS [/tex]
    where S is the surface
    [tex] z = 16-x^2-y^2 \quad z>0[/tex]

    2. Relevant equations

    [tex]\int_S f dS = \int_S f(S(u,v)) | \frac{\partial S}{\partial u }\times \frac{\partial S}{\partial v } | dudv[/tex]

    3. The attempt at a solution
    let
    [tex] x= u \quad y=v[/tex]
    [tex] z = 16 - u^2-v^2[/tex]
    [tex]S=(u,v,16-u^2-v^2)[/tex]
    [tex] -4 \le u \le 4 \quad -4\le v \le 4[/tex]
    [tex]\frac{\partial S}{\partial u } = (1, 0, -2u) \quad \frac{\partial S}{\partial v } = (0,1,-2v) [/tex]
    [tex]| \frac{\partial S}{\partial u }\times \frac{\partial S}{\partial v }| = \sqrt(4u^2+4v^2+1)[/tex]
    [tex]\int_{-4}^4 \int_{-4}^4 (2(16-u^2-v^2)+1) \sqrt (4u^2+4v^2+1) dudv [/tex]

    using wolfram the answer I get is 2771.99. However this answer is wrong. This is my first time doing surface integrals and parameterizing surfaces so I can't figure out where I am wrong.
     
  2. jcsd
  3. Jul 15, 2013 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    No point in doing that. Just use ##x## and ##y## in the first place.

    Those limits would describe a square. But the boundary of the surface in the xy plane is ##x^2+y^2 = 16##, which isn't a square, whether you call the variables uv or xy. You might consider polar coordinates to evaluate the integral.
     
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