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Thread: Surface integral

  1. #1 Surface integral 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    How would I compute the surface integral
    , where , boundary of the cube , outward pointing normal.

    Thank you


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  3. #2  
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    Quote Originally Posted by Heinsbergrelatz View Post
    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?


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  4. #3  
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    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
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