How would I compute the surface integral
, where , boundary of the cube , outward pointing normal.
Thank you

How would I compute the surface integral
, where , boundary of the cube , outward pointing normal.
Thank you
Because the cube has "edges" and is not smooth, do each of the six faces separately.
1) On the face x= 0, the unit normal is <1, 0, 0>, on the face x= 3 the unit normal is <1, 0, 0>, and F has 0 xcomponent so those integrals are 0.
2) On the face y= 0, the unit normal is <0, 1, 0>, on the face y= 3 the unit normal is <0, 1, 0>, and F has 0 ycomponent so those integrals are 0.
3a) On the face z= 0, the unit normal is <0, 0, 1> and F has z component so this integral is
3b) On the face z= 3, the unit normal is <0, 0, 1> and F has z component so this integral is
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