Problem #8: Find the flux of the vector field F = xi + e j + zk through the surface S given by that portion of the plane 3x+y+ 9z = 8 in the first octant, oriented upward.Given : F = xitej + zu S : 3 x + 4+9z = 8 in the first octant , oriented upward . To compute : flux = SS Finds S where in is the unit vector oriented upward on the surface S. Let = 3xty+9z-8, Vo 3 it itgu = = ( 3 it jitgi ) Vgti+81 V91 And ds = Next 2 tzuti dydx Here , Z = 9 (8 - 3 x -y ) 7 Z x= - and Zy = g as = g t 31 + 1 dydx = 191 g we have, F . he ( 3 x + es "+ gz) = (3xte + (8- 321 -4 ) )(': z= 1(8- 3x-4) on the surface s ) (3 x te +8- 3x-y) = (e' 7 8 - y) Substituting Fin and Is in the formula for fluxe and applying the limits of ofintegration Nzo , M = . , and yzo , y = 8 - 3x , we get 2=0 7 32+4=8 osys 8- 3x y= = =0 = 3x=8 Flux = 8/3 8-3x (e + 8 -4 ) . 191 X= 0 Jy =0 9 = g 8/3 8-3x ( e + 8 - y ) dyax 8/3 8- 3x [ my + 84 - 4 do 813 = . (e 3x (8-3x) +8(8- 3x ) - (8- 3x) 2 ax S (8- 3x ) edx = (8- 32) Jean - S du (8- 3 x) (Sex) dx (8- 324). 23 - [( - 3 ) . ez da- (8-32) -3x e 3x + 3 3 3 8/3 Now , flux = 9 X (8- 3x)e 3x 8 (8-3x) (8-3x) 3 + + 2x( 3 ) 3x2X(-3) Jo 8 = + 0 - 0 LO ( o + 8 256 256 W )- 3 + 3 g 485 9 = 81 [3e + 485] ANS