Since the given vector field depends only on z, slice the triangular region into strips along lines with z=constant, as shown. The vector displacement along each such line is j−i at the base, and has a magnitude which varies linearly with height; this vector displacement is therefore (j−i)(1−z). The slant height between nearby slices is dr=dxi+dyj+dzk which reduces to dr =(−i/2−j/2+k)dz since x=y=(1−z)/2 up the middle. The (upwards-oriented) area vector for a single strip is therefore the (cross) product of these (perpendicular) vectors, namely dA=(i+j+k)(1−z)dz. Dotting this with the given vector field E=zk yields z(1−z)dz, and integrating this as z goes from 0 to 1 yields 1/6.
(The slant height [really dr] does not need to be taken up the middle so long as the cross product is used. Alternatively, the argument up the middle can be recast without vectors at all.)
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