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Surface integral

  1. Mar 17, 2010 #1
    Let S be a parametrised surface given by (x, y, z) = R(u, v) := (u2, v2, u + v), for 0 [tex]\leq[/tex] u [tex]\leq[/tex] 1 and
    0 [tex]\leq[/tex] v [tex]\leq[/tex] 1. How do I find the integral K := [tex]\int\int_S[/tex] z/2 dA.
     
  2. jcsd
  3. Mar 17, 2010 #2

    quasar987

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    By definition, this integral is

    [tex]\int_0^1\int_0^1 \frac{u+v}{2}\sqrt{E(u,v)G(u,v)-(F(u,v))^2}dudv[/tex]

    where
    [tex]E=R_u\cdot R_u[/tex]
    [tex]F=R_u\cdot R_v[/tex]
    [tex]G=R_v\cdot R_v[/tex]

    And this you know how to do.
     
  4. Mar 17, 2010 #3
    That's easy. Cheers.
     
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