I'm struggling with a math problem in my Math for Physics class. We just introduced Gauss's law and now I'm supposed to calculate the flux through a tetrahedron:
Let $$\vec{A}: \mathbb{R}^3 \space \rightarrow \space \mathbb{R}^3 \space , \space \vec{A}(x,y,z):=(x-2z, 3z-4x, 5x+y)$$ be a vector field. Calculate the flux of $A$ through the surface of a tetrahedron with vertices $\{(0,0,0),(1,0,0),(0,1,0),(0,0,1)\}$.
If $T$ denotes the space enclosed by the tetrahedron and $(T)$ the surface of the Tetrahedron, then my Gauss's law we can say that
$$ \int_{T} \vec{\nabla} \cdot \vec{A} \space dV = \int_{(T)} \vec{A} \cdot d\vec{F} $$
I attempted to split the integral up and calculate the flux through each of the triangular surfaces of the tetrahedron. In order to do that, I tried to find a parametrization of these triangles in $\mathbb{R}^3$. As that was not shown in my classes and I couldn't find anything in my lecture notes, I had to research quite a lot. I managed to calculate the flux, but I struggle with my understanding of why it works. Here is what I came up with:
Parametrization of a triangular plane
Let $A,B,C$ be the vertices of the triangle and $\vec{v_1} = C - B$, $\vec{v_2} = D - B$ vectors in the triangular plane. The corresponding parametrization is
$$ \Phi(x,y,z) = B + \alpha \vec{v_1} + \beta \vec{v_2} \quad \text{with $\alpha \in [0,1]$ and $\beta \in [0, 1-\alpha]$} $$
The normal vector of the surface is given by $\vec{v_1} \times \vec{v_2}$. Hence, for the differential surface element we have that $dF = || \vec{v_1} \times \vec{v_2} || \space d\alpha \space d\beta$. It follows that the surface integral over this triangle with surface $S$ is given by
$$ \int_{S} A_i (x,y,z) \space dF = \int_{0}^{1} \int_{0}^{1-\alpha} A_i(\Phi (x,y,z)) \cdot || \vec{v_1} \times \vec{v_2} || \space d\alpha \space d\beta .$$
My problem: I don't really understand why I can just set $dF = || \vec{v_1} \times \vec{v_2} || \space d\alpha \space d\beta$. Where does the $|| \vec{v_1} \times \vec{v_2} ||$ come from and how do I know that it is the right coefficient for my differentials $d\alpha \space d\beta$? When doing a parametrization, I usually find the coefficient using the Jacobian determinant of $\Phi$. However, I don't see how I can get from my parametrization $\Phi$ to the corresponding jacobian matrix. Let's take for example the triangular plane with vertices $(1,0,0), (0,1,0) (0,0,1)$. I get $$ \Phi(x,y,z) = (\alpha, \beta, 1 - \alpha - \beta) $$ As the Jacobian Matrix is defined as $J_{\Phi} = \left( \frac{d}{dx_j} \Phi_i \right) _{i,j}$, I don't see how I can do that with my $\Phi(x,y,z)$.
Are the cross product and the jacobian determinant somehow related? Is there maybe another approach to parameterize the tetrahedron and solve this practise problem? I'm sure that my path is probably not the easiest and/or fastest. I'm open for any suggestions.
