lecture notes 13

.Example. Find the flux of F(x, y, z) = (x, y, 3) out of the region D. bounded by the paraboloid z = x 2 + y 2 and the plane z = 4.Solution. Let S 1 be the circular top { (x, y, 4) | x 2 + y 2 ≤ 2 2 }, and S 2be the parabolic part parameterized by z(x, y) = x 2 + y 2 withx 2 + y 2 ≤ 4 as shown in the diagram. In this case, the outward unitnormal vector field is given by on S 1 : n 1 (x, y, z) = k, and on S 2 byn 2 (x, y, z) = ∇(−z + x2 + y 2 ) (2x, 2y, −1)∥∇(−z + x 2 + y 2 = √ . In fact, one can)∥ 4x 2 + 4y 2 + 1check n 2 (0, 0, 0) = −1, so F 1 · n 1 = (x, y, 3) · (0, 0, 1) = 3, and(2x,2y,−1)F 2 · n 2 = (x, y, 3) · √ = 2x2 +2y 2 −3√ . On S4x 2 +4y 2 +1 4x2, the surface area2 +4y 2 +1√element dS = 1 + z 2 x + z 2 y dxdy = √ 4x 2 + 4y 2 + 1 dxdy for∫∫D = { (x, y) | x 2 + y 2 ≤ 4}. Then F · n dS =∫∫∫∫∫∫ S ∫∫F · n 1 dS + F · n 2 dS = 3 dxdy + (2x 2 + 2y 2 − 3) dxdyS 1 S 2 DD∫∫∫ 2π ∫ 2[ 2r= (2x 2 + 2y 2 ) dxdy = 2r 2 4· r dr dθ = 2π = 16π.4D00] 20. . . . . .