I hate to be the one to break it to you but the flux over the bottom is NOT -3. The unit, outward, normal to the bottom, z= 0, is <0, 0, -1> so you are integrating <2x,-3y,2z>\cdot<0, 0, ->= <2x, -3y, 0>\cdot<0, 0, -1>= 0, since z= 0 there, over the xy-plane, 0< x< 1, 0< y, 1. That flux is 0. It is correct that the flux over the top, z= 1 is
\int_{x=0}^1\int_{y= 0}^1 <x, y, 3>\cdot<0, 0, 1> dxdy= 3[/itex].<br />
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The "left" face is x= 0 and an outward unit normal is <-1, 0, 0>. On that face, x= 0 so the integrand is &lt;0, -3y, 3z&gt;\cdot&lt;-1, 0, 0&gt;= 0 so the flux there is also 0.<br />
The "right" face is x= 1 and an outward unit normal is <1, 0, 0>. On that face, x= 1 so the integrand is &lt;2, -3y, 3z&gt;\cdot&lt;1, 0, 0&gt; integrate that for 0<y<1, 0< z< 1.<br />
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The "front" face is y= 0 and an outward unit normal is <0, -1, 0>. On that face, y= 0 so the integrand is &lt;2x, 0, 3z&gt;\cdot&lt;0, -1, 0&gt;.<br />
The "back" face is y= 1 and an outward unit normal is <0, 1, 0>. On that face, y= 1 so the integrand is &lt;2x, -3, 3z&gt;\cdot&lt;0, 1, 0&gt;.<br />
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Of course, it would be easy to very easy to use the divergence theorem to get the answer.
