Compute the outward flux of the vector field F(g;, 1/72) = 51d + lyj + 32k across the boundary of the right cylinder with radius 4 with bottom edge at height 2 = 2 and upper edge at z = 4. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the cylinder to be positive. Part 1 - Using a Surface Integral First we parameterize the three faces of the cylinder. Note: in the following type 0 as theta. The bottom face: 01(r, 9) = (w1(7",t9),y1(r,9), 21(13 0) where 0 g 6' g 21:, 0 g 'r g 4 and 391(7", 9) = 111 (r. 9) = 21(719) = 7 The top face: 02039) = (1,-2(1-,0),y2(1r,9),12(1'1 0) where 0 g 9 5 2m 0 g 1- g 4 and $20, 0) = 112 (7', 9) = 22 ('r, 0) = 4 The lateral face: 03(9, 1) = (m3(0,z),y3(0,z),23(9,z) where 0 g 9 5 2m 72 g 2 g 4 and {133(9) = 113(9) = 23 (z) = Then (mind the orientation) 2 2 2 7r 001x801 0 [1/4 (802x 302)T 9 fwf:(al13x3t73) 9 fandS: f/W(1)( Wr)dd+ F(c72) 51' 67dd+ F073) 82 66 dzd + II 4. Part 2 - Using the Divergence Theorem lelndS=fdeideV / / f