Please help me with the problem below.
Find the volume of a tetrahedron with vertices:
$O(0,0,0)$, $A(1,2,3)$, $B(-2,1,5)$, $C(3,7,1)$ by using triple integral.
Hint: First find the the equations of the planes.
Note: My professor told me that I can use change of variables (which I need
to calculate the Jacobian and then the volume of a "simpler" tetrahedron).
Best Answer
Just to convert my comment to an answer.
If you want to use triple integral to find the volume. There are two ways to do this.
First method: Direct triple integral. You have to find the plane equations for all faces of this tetrahedron. Due to the piecewise nature, the limits of the integral are somewhat messy.
Second method: We can transform the points $ A(1,2,3)$, $B(-2,1,5)$, $C(3,7,1)$ to $A'(1,0,0)$, $B'(0,1,0)$, $C'(0,0,1)$. So that the volume of the new tetrahedron is easy to compute. The transformation matrix from $A',B',C'$ to $A,B,C$ is: $$ T = \begin{pmatrix} 1&-2 &3 \\ 2 &1 &7 \\ 3 &5 &1\end{pmatrix} . $$ Because the linearity this is also the Jacobian matrix, so $$ \mathrm{Volume} = \iiint_{OABC} 1 \,dxdydz = \iiint_{OA'B'C'} 1 \,{|\det (T)|}\,dx'dy'dz' = \frac{17}{2}. $$
Another tip with a formula to compute $n$-simplex, I took it from my computational geometry notes: $$ |V| = \frac{1}{3!}\left|\det \begin{pmatrix} x_1 & x_2 & x_3 & x_4\\ y_1 & y_2 & y_3 & y_4\\ z_1 & z_2 & z_3 & z_4\\ 1 & 1 & 1 & 1\\ \end{pmatrix} \right|, $$ where $(x_i,y_i,z_i)$ are the coordinates for the $i$-th vertex. It bears the same form for the area formula of a triangle with three vertices give $$ |T| = \frac{1}{2!}\left|\det \begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \\ \end{pmatrix} \right|. $$