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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) = 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

  1. 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^2-y^2 for z
    and do the integrals from 0 to 1

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