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Evaluate the surface integral SSS 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 = 7 – x2 - y2 that lies above the square 0 sxs1,0 y s 1, and has upward orientation. Part 1 of 3 Since S is part of the paraboloid z = g(x, y) = 7 – x2 - y2, it is the graph of a function, and so we know that F. ds = QOQ + R) DA. ag da ax ay -S[(- - A. Thus, we have the following. 1 1 F ds = -xy(-2x) – yz(-2y) + zx| DA SIS = J 60*L* [) + ZX ] DA %*%*[2x2y + 2y247 – x2 - y2) + X07 – x2 - y2)) da 6*T*(2x2y * – xy2) dy dx + Х 1 0 = + G