lecture notes 13

.∫∫Example. Evaluate the surface integral (∇ × F) · n dS, whereF(x, y, z) = 3zi + 5xj − 2yk, and S is the parabolic surface z = x 2 + y 2that lie below the plane z = 4 and whose orientation is given by the. upper unit normal vector.Solution. The boundary of S is the circle C parameterized byr(t) = (2 cos t, ∫∫ 2 sin t, 4), where 0 ≤∮t ≤ 2π. It follows ∮ from the Stokes’theorem that (∇ × F) · n dS = F · T ds = 3z dx + 5x dy − 2y dz==∫ 2π∫02π0SC3 · 4(−2 sin t dt) + 5 · (2 cos t) · (2 cos t dt) + 2 · (2 sin t) · (0 dt)(−24 sin t + 20 cos 2 t) dt =∫ 2π0SC10(1 + cos 2t) dt = 20π.. . . . . .