tetrahedron volume using points

Question

I can imagine that I should use the xy-plane and integrate over f(x,y,z)=1 but I can't find the integrational limits I should use. I have given (2,0,0) (0,3,0) (2,3,0) and (2,3,4). I have been trying to solve it for hours but I am only getting more desparate.

Answer 1

There is a formula for the the tetrahedron volume

Let $\mathbf{A} = (2,0,0), \mathbf{B} = (0,3,0) , \mathbf{C} = (2,3,0), \mathbf{D} = (2,3,4)$, then the volume is given by

$V = \dfrac{1}{6} | \begin{vmatrix} \mathbf{B-A, C-A, D-A} \end{vmatrix} |$

This is equal to

$V = \dfrac{1}{6} | \begin{vmatrix} -2 && 0 && 0 \\ 3 && 3 && 3 \\ 0 && 0 && 4 \end{vmatrix} | $

This evaluates to (expanding the determiant about the third row)

$V = \dfrac{1}{6} | 4 (-6) | = 4 $

Answer 2

For brevity assume $A=(2,0,0), B=(0,3,0), C=(2,3,0), D=(2,3,4)$. Let's look closer at the tetrahedron. Note, that $ABC$ face is in $xy$ plane, and both $ACD$ and $BCD$ are "vertical", i.e. included in planes $x=2$ and $y=3$ respectively. So the volume is equal to integral of the graph of plane including $ABD$ over region $ABC$. let $k=\{(x,y,z)\in \mathbb{R}^3|\lambda_1x+\lambda_2y+\lambda_3z+\lambda_4=0\}$ be plane including $ABD$. Further solution is as follow:

1. Find $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ using points $A,B,D$ (solution on wikipedia).
2. Find function $f:\mathbb{R}^2\to\mathbb{R}$ of which graph is $k$, i.e. $(x,y,z)\in k\iff z=f(x,y)$
3. Find function $g:\mathbb{R}\to\mathbb{R}$ of which graph is line $AB$ on $xy$ plane.
4. Integrate: $$|ABCD|=\int_{\Delta ABC}f(x,y)dydx=\int_{x=0}^2\int_{y=g(x)}^3f(x,y)dydx$$