Evaluate the surface integrals 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) = xyi + yzj + zxk, 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

2. Let S denote the piece of parabolic surface x2 + y? = z, 0 < z < 4. For the vector field F(x, y, z) = (y + xy, xz, z). calculate the integral (V x F) · n dA where the unit vector n is chosen so that its z component is positive.

Find the flux of the vector field F(x, y, z) = [x4, −4yx3, 7z + sin(xey))] through the outwards oriented surface which is the boundaryof the solid 1 ≤ x2 + y2 < 4, 1 ≤ z ≤ 3.