Surface integral

  • Thread starter squenshl
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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #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.
 
  • #3
479
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That's easy. Cheers.
 

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