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 $\int_S\vec{u}\cdot d\vec{S}$, $S$ being the surface of the tetrahedron.
I have already calculated $\int_V(\nabla\cdot\vec{u})dV$, 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 $\vec{x}_s=(s,t,2-2s-2t)$
Next I find $d\vec{S}$ as such:
$$d\vec{S}=(\frac{\partial{\vec{x_S}}}{\partial{s}}\times\frac{\partial{\vec{x_S}}}{\partial{t}})dsdt$$
Therefore:
$\vec{u}\cdot d\vec{S}=10s^2+16st-16s+8t^2-15t+8$
But, upon calculating the integral $\int_0^1\int_0^1\vec{u}\cdot d\vec{S}dsdt$ 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.
Best Answer
Vector field $\vec u = (x, z^2,y)$ 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 $ABC, z = 0, \vec n = (0, 0, -1)$
So the surface integral, $I_{ABC} = \displaystyle \int_0^1 \int_0^{1-x} - y \ dy \ dx = - \frac{1}{6}$
b) For surface $ABD$, $y = 0, \vec n = (0, -1, 0)$
So the surface integral, $I_{ABD} = \displaystyle \int_0^1 \int_0^{2-2x} - z^2 \ dz \ dx = - \frac{2}{3}$
c) For surface $ACD$, $x = 0, \vec n = (-1, 0, 0)$, hence the integral is simply zero.
d) Now for surface $BCD$, $z = 2 - 2x - 2y, \vec n = (2, 2, 1)$
$I_{BCD} = \displaystyle \int_0^1 \int_0^{1-x} (2x + 2(2-2x-2y)^2 + y) \ dy \ dx = \frac{7}{6}$
So when you add them up, it does come to $\displaystyle \frac{1}{3}$.