Question

Question 11 (2 marks) Attempt 7 S is the half cylinder 2+y2=9, y>0, 0<z<2 Given the vector field: E=(12+7yz)i+(622+72)j+(1322-823)k, Calculate the outward flux (away from the z-axis) of E through surface S. That is find J- 1-48 S Your answer should consist of a single number accurate to five decimal digits or as an exact rational expression. For example: 10.13906368 OR rounded to 10.13906 OR 3*Pi+5/7 J =

Answer 1

We will use divergence theorem to calculate the flux through the surface {(r, y, )y 2}, S3 S. But to do that we need the surface to be closed. Let S 0, E3,3, {(, y, z)22 <9,y 2 0, z 0}. Then observe that SU [U_ S] is surface. Let the closed space be V. Hence by applying the divergence theorem we get 2€ [0, 2]}, Sz 3 {(г, у, 2)|x? +у? < 9, у > 0, 2 closed a - EdS V E dV V -y2 (12 0+13r2 -y2 24z2) d dy dz 9-7 9-у2 3 2 г? dx dy — 24 = 12Vol(V) 13 dr dy V9-y2 dz J0 9- 9-1y 3 9- /3]V dy 24[z3/3]x Area(S) = 12Vol(V) 26 -y2 0 3 Area(S (9 y2/ dy 64 x = 12V ol(V) + 26 x 2/3 0 (9y) dy (3cos2 0)3/23cos0 d0 now -п/2 cos' COS 0 de = 81 T/2 =81/4 (1cos(20))2 d0 S T/2 (1 2 cos(20 cos2(20)) d0 =81/4 0 п /2 =81/4[0sin(20) 81/16 (1cos(40) d =81n/881/160+ sin(40)/4]= 4057/32 Е - dS — 72т + 26 х 2/3 х 405л/32— 2881 sU[US

Now as al of the surfaces are disjoint, Hence E dS E-dS E dS sulUS S i-1 0 SP [As the z component is zero for all points in S3 y (2622 - 64) dr dy -V9-y now S3 9 E dS S2 (9 23/2 dy - 288T = 26 x 2/3 x 405T/32 2887 = 26 x 2/3 --S 3 E dS 7yz dy dz JO = -7|yj2/21322/2] -63 E dS 727T+63 S