Question

Find the flux of the field F(x,y,z)=z² i +xj - 3z k outward through the surface cut from the parabolic cylinder z=1-yby the planes x = 0, x=2, and z=0. The flux is (Simplify your answer.)

Answer 1

hiven : f(x, y, z) z² itu aj - 3 z The region bounded Z= by 1-y² , n=0, x=2 and z=0 Az (0,-1,0) z=1-g² (2,0,0) 10,1,0) ry By divergence theorem, Flor = 17.06 - ) duf.de I (22) d(x) + du + (-32)) av dy dz مالی (2) ۰۰۰ هور = + 0 + (-3)) du il -3 3) av

For Jav, we see that z น 2=0 to 2- changes from z=1-42 y changes in O=1-y2=) y = -1 to y=1 a changes from On n=2 3 o flux=-3 1-y? -3 av 3 2 3 1-91 [z] dy dn -1 I dady ale // 1) Congo) dy ale :] [0-41 2 · 3 11 3 dx (-1) + 6-1) du 2 =-3 Jadu -4 [n]. 2 -4 43) du -4 (2-0) -8 -

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