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\fThe surface S consists of the top and the four sides (but not the bottom) of the cube with one corner at ( - 1, - 1, - 1) and a diagonal corner at (5, 5, 5), with outward orientation. If F (x, y, x) = (xyz, xy, x2 yz) then: J curl ( F) . as = 144 (Suggestion: Use the Curl Theorem, twice.) Show Detailed Solution The boundary of S is a square in the plane z = - 1 with vertices at ( - 1, - 1, - 1), (5, - 1, - 1), (5, 5, - 1) and ( - 1, 5, - 1), oriented counterclockwise when viewed from above. Applying the Curl Theorem: curl ( F) . as = F . ds This curve OS is also the boundary of the square region Q in the plane z = - 1 with vertices at ( - 1, - 1, - 1), (5, - 1, - 1), (5, 5, - 1) and ( - 1, 5, - 1). Applying the Curl Theorem again: where Q is oriented upward, so that n - k for Q. Computing curl ( F ) : curl ( F) = (x2z, xy - 2xyz, y - xz) yields: ( cul ( F) . as = ( 2, ay - 2042, - zz ty) . (0, 0, 1) dA which simplifies to: LS (1x + y) dy do = 144