Surface Integral: Finding K = $\int\int_S z/2 dA$

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SUMMARY

The integral K = $\int\int_S \frac{z}{2} dA$ is computed over the parametrized surface S defined by R(u, v) = (u², v², u + v) for 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1. The integral is expressed as $\int_0^1\int_0^1 \frac{u+v}{2}\sqrt{E(u,v)G(u,v)-(F(u,v))^2}dudv$, where E, F, and G are derived from the partial derivatives of R. This calculation involves evaluating the dot products R_u·R_u, R_u·R_v, and R_v·R_v, which are essential for determining the surface area element dA.

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  • Understanding of parametrized surfaces in multivariable calculus
  • Familiarity with surface integrals and their applications
  • Knowledge of vector calculus, particularly dot products
  • Proficiency in evaluating double integrals
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  • Study the derivation of surface area elements in parametrized surfaces
  • Learn about the application of the Jacobian in surface integrals
  • Explore advanced techniques for evaluating double integrals
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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.
 
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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.
 
That's easy. Cheers.
 

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