Since S is part of the paraboloid z = g(x, y) = 7 - x2 - y , it is the graph of a function, and so we know that 1/5 . as = 1/ ( -P39 - Q.0 + R) dA. Thus, we have the following. 1/5 F . as = 16 10" [-xy(-2x) - 42(- 24) + 2x/ dA = [2x24 + 242(7 -x2 - 12) + x (7- x2 - y2) dA - 16 16 ( 2x2 + 14)2 - 2x2,2-24+ + 7x- 3 14y2 - 243 x2 - 2y' + 7x - 23 - xy) dy dx Step 2 Now, we have the following. (2x 2y + 14y2 - 2x2/2 - 2y4 + 7x- x3 - xy?) dy= 12 + 64 - + X 15 Submit Skip (you cannot come back)