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) = xy i + yz j + zx k
S is the part of the paraboloid
z = 2 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
and has upward orientation
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1 answer

Flux = int over S of
F dot [1i  (dz/dx)^2j (dz/dy)^2k]
F dot [i 4x^2 j  4y^2 k]
[xy  4x^2yz 4y^2xz]dx dy
now put in 2 x^2y^2 for z
and do the integrals from 0 to 1 👍
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answered by Damon
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