I have a vector field vec(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 int_S vec(u) * d vec(S) , S being the surface of the tetrahedron. I have already calculated int_V(grad vec(u))dV, 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 vec(x)_s=(s,t,2−2s−2t) Next I find d vec(S) as such: ((del vec(x_s))/(del s) xx (del vec(x_s))/(del t))dsdt Therefore: vec(u) * d vec(S) =10s^2+16st−16s+8t^2−15t+8

Ashlyn Krause 2022-07-20 Answered
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 t
Therefore: u d S = 10 s 2 + 16 s t 16 s + 8 t 2 15 t + 8
But, 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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Answers (1)

hottchevymanzm
Answered 2022-07-21 Author has 15 answers
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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