I have a vector field u → = ( x , z 2 , y ) over a tetrahedron with vertices at ( 0 , 0 , 0 ) , ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) and ( 0 , 0 , 2 ) and I need to compute ∫ S u → ⋅ d S → being the surface of the tetrahedron.I have already calculated ∫ V ( ∇ ⋅ u → ) d V, which I have found to be equal to 1 3 , now I need to calculate the previous integral and assumedly show that the answers agree (Gauss' Theorem).So far I have parameterised the surface to say x → s = ( s , t , 2 − 2 s − 2 t )Next I find d S → as such: d S → = ( ∂ x S → ∂ s × ∂ x S → ∂ t ) d s d tTherefore: u → ⋅ d S → = 10 s 2 + 16 s t − 16 s + 8 t 2 − 15 t + 8But, upon calculating the integral ∫ 0 1 ∫ 0 1 u → ⋅ d S → d s d t I get 5 2 . I'm confident in my first integral, but I don't feel as confident in this one as I feel there may be some theoretical misunderstanding with my parameterisation. Any help is appreciated.

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I have a vector field             u      →        =  (  x  ,      z    2    ,  y  ) over a tetrahedron with vertices at   (  0  ,  0  ,  0  )  ,  (  1  ,  0  ,  0  )  ,  (  0  ,  1  ,  0  ) and   (  0  ,  0  ,  2  ) and I need to compute       ∫    S              u      →        ⋅  d            S      →       being the surface of the tetrahedron.I have already calculated       ∫    V    (  ∇  ⋅            u      →        )  d  V, which I have found to be equal to       1    3  , now I need to calculate the previous integral and assumedly show that the answers agree (Gauss&#039; Theorem).So far I have parameterised the surface to say                     x        →              s    =  (  s  ,  t  ,  2  −  2  s  −  2  t  )Next I find   d            S      →       as such:  d            S      →        =  (            ∂                                                  x              S                        →                                      ∂              s              ×            ∂                                                  x              S                        →                                      ∂              t              )  d  s  d  tTherefore:             u      →        ⋅  d            S      →        =  10      s    2    +  16  s  t  −  16  s  +  8      t    2    −  15  t  +  8But, upon calculating the integral       ∫    0    1        ∫    0    1              u      →        ⋅  d            S      →        d  s  d  t I get       5    2  . I&#039;m confident in my first integral, but I don&#039;t feel as confident in this one as I feel there may be some theoretical misunderstanding with my parameterisation. Any help is appreciated.

hottchevymanzm · Accepted Answer

Vector field             u      →        =  (  x  ,      z    2    ,  y  ) and vertices of tetrahedron are   A  (  0  ,  0  ,  0  )  ,  B  (  1  ,  0  ,  0  )   C  (  0  ,  1  ,  0  ) and    D  (  0  ,  0  ,  2  )a) For surface   A  B  C  ,  z  =  0  ,            n      →        =  (  0  ,  0  ,  −  1  )So the surface integral,       I          A      B      C        =            ∫      0      1              ∫      0              1        −        x              −    y         d    y         d    x    =    −          1      6      b) For surface   A  B  D  y  =  0  ,            n      →        =  (  0  ,  −  1  ,  0  )So the surface integral,       I          A      B      D        =            ∫      0      1              ∫      0              2        −        2        x              −          z      2             d    z         d    x    =    −          2      3      c) For surface ACD   x  =  0  ,            n      →        =  (  −  1  ,  0  ,  0  ), hence the integral is simply zero.d) Now for surface BCD   z  =  2  −  2  x  −  2  y  ,            n      →        =  (  2  ,  2  ,  1  )      I          B      C      D        =            ∫      0      1              ∫      0              1        −        x              (    2    x    +    2    (    2    −    2    x    −    2    y          )      2        +    y    )         d    y         d    x    =          7      6      So when you add them up, it does come to             1      3

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