Surface Integral of Vector Fields

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SUMMARY

The discussion focuses on evaluating the surface integral ∫∫S F · dS for the vector field F(x, y, z) = y i + x j + z^2 k across the helicoid surface defined by the vector equation r(u, v) = ucos(v)i + usin(v)j + v k, where 0 ≤ u ≤ 3 and 0 ≤ v ≤ 3π. The integral setup involves the expression ∫∫D F*(r_u x r_v)dA, with the user attempting to integrate the resulting expression. The recommended approach is to first integrate with respect to v and then with respect to u, with a suggestion to simplify the integrand to u*(sin(v)^2 - cos(v)^2).

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  • Understanding of vector calculus, specifically surface integrals.
  • Familiarity with the concept of flux across surfaces.
  • Knowledge of parameterized surfaces and their vector equations.
  • Proficiency in performing double integrals.
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  • Study the method of evaluating surface integrals in vector calculus.
  • Learn about the divergence theorem and its applications to flux integrals.
  • Explore simplification techniques for integrands in double integrals.
  • Practice integrating parameterized surfaces using examples similar to helicoids.
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drecklia
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Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x, y, z) = y i + x j + z^2 k
S is the helicoid (with upward orientation) with vector equation r(u, v) = ucos(v)i + usin(v)j + v k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 3π.

∫∫S f*dS=∫∫D F*(r_u x r_v)dA

i got int[0,3] and int[0,3pi] usin(v)^2-ucos(v)^2+uv^2dvdu
i do not know how to integrate this(yeah, shame on me) and i don't know if this integral is correct
 
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It looks ok to me. What's stopping you from doing it? First integrate dv and then the result du. No special tricks needed. Though you could simplify u*(sin(v)^2-cos(v)^2).
 

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