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

 Let r(x,y) = (x, y, 9 - x^2 - y^2) 

So, dr/dx x dr/dy = (2x, 2y, 1) 

So, integral(S) F * dS 
= integral(x in [0,1], y in [0,1]) (xy, y(9 - x^2 - y^2), x(9 - x^2 - y^2)) * (2x, 2y, 1) dy dx 
= integral(x in [0,1], y in [0,1]) (2x^2y + 18y^2 - 2x^2y^2 - 2y^4 + 9x - x^3 - xy^2) dy dx 
= integral(x in [0,1]) (x^2 + 6 - 2x^2/3 - 2/5 + 9x - x^3 - x/3) dx 
= integral(x in [0,1]) (28/5 + x^2/3 + 26x/3 - x^3) dx 
= 28/5 + 40/9 - 1/4 
= 1763/180