15_6_3 - 3 F = xi yj zk ^ The box has six faces F N = 0 on...

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3. F = x i + y j + z k . The box has six faces. F ˆ N = 0 on the three faces x = 0, y = 0, and z = 0. On the face x = a , we have ˆ N = i , so F ˆ N = a . Thus the flux of F out of that face is a × ( area of the face ) = abc . By symmetry, the flux of
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