Question

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

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