# Thread: Surface integral

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

Thank you  2.

3. Originally Posted by Heinsbergrelatz How would I compute the surface integral , where , boundary of the cube , outward pointing normal.

Thank you
What have you tried so far? and where did you get stuck?  4. 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 x-component 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 y-component 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   Bookmarks
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