Find the flux of F over Q where Q is bounded?

Question

Find the flux of F over Q where Q is bounded by 3x+2y+z=6 and the coordinate planes, F=<y^2*x, 4x^2*sin(z), 3>. (Answer: 27/5).

The divergence of the vector field is y^2, which is what I've found. And z=6-3x-2y, so how should I set up the integral?

Answer 1

div F = y^2

By the Divergence Theorem ∫∫F⋅N dS = ∫∫∫div F dV,

the flux of F over Q

= ∫∫∫(y^2) dx dy dz

= ∫∫∫{0, 6-3x-2y} (y^2)dz dx dy

= ∫∫{0, 2-(2/3)y}(6-3x-2y) (y^2) dx dy

= ∫{0, 3}y^2[(6-2y)(2-(2/3)y) - (3/2)(2-(2/3)y)^2] dy

= 27/5 <==Answer

Answer 2

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