# 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

I have a vector field $\stackrel{\to }{u}=\left(x,{z}^{2},y\right)$ over a tetrahedron with vertices at $\left(0,0,0\right),\left(1,0,0\right),\left(0,1,0\right)$ and $\left(0,0,2\right)$ and I need to compute ${\int }_{S}\stackrel{\to }{u}\cdot d\stackrel{\to }{S}$ being the surface of the tetrahedron.
I have already calculated ${\int }_{V}\left(\mathrm{\nabla }\cdot \stackrel{\to }{u}\right)dV$, which I have found to be equal to $\frac{1}{3}$, 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 ${\stackrel{\to }{x}}_{s}=\left(s,t,2-2s-2t\right)$
Next I find $d\stackrel{\to }{S}$ as such:
$d\stackrel{\to }{S}=\left(\frac{\mathrm{\partial }\stackrel{\to }{{x}_{S}}}{\mathrm{\partial }s}×\frac{\mathrm{\partial }\stackrel{\to }{{x}_{S}}}{\mathrm{\partial }t}\right)dsdt$
Therefore: $\stackrel{\to }{u}\cdot d\stackrel{\to }{S}=10{s}^{2}+16st-16s+8{t}^{2}-15t+8$
But, upon calculating the integral ${\int }_{0}^{1}{\int }_{0}^{1}\stackrel{\to }{u}\cdot d\stackrel{\to }{S}dsdt$ I get $\frac{5}{2}$. 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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hottchevymanzm
Vector field $\stackrel{\to }{u}=\left(x,{z}^{2},y\right)$ and vertices of tetrahedron are $A\left(0,0,0\right),B\left(1,0,0\right)$ $C\left(0,1,0\right)$ and $D\left(0,0,2\right)$
a) For surface $ABC,z=0,\stackrel{\to }{n}=\left(0,0,-1\right)$
So the surface integral,
b) For surface $ABD$$y=0,\stackrel{\to }{n}=\left(0,-1,0\right)$
So the surface integral, c) For surface ACD $x=0,\stackrel{\to }{n}=\left(-1,0,0\right)$, hence the integral is simply zero.
d) Now for surface BCD $z=2-2x-2y,\stackrel{\to }{n}=\left(2,2,1\right)$

So when you add them up, it does come to $\frac{1}{3}$