Calculate the volume of a tetrahedron with vertices $(1,1,1)$, $(2,1,1)$, $(1,2,1)$, $(1,1,2)$

Question

Calculate the volume of a tetrahedron whose vertices are $(1,1,1),(2,1,1),(1,2,1),(1,1,2)$.

I don't know how to start, do I have to find the planes?

Answer 1

Shift the tetrahedron by $(-1, -1, -1)$ then the vertices becomes

$(0,0,0), (1, 0, 0), (0, 1, 0), (0, 0, 1) $

So the volume is one third the area of the base times the height, which

$$ V = \frac{1}{3} \left( \dfrac{1}{2} (1)^2 \right) (1) = \dfrac{1}{6} $$

And we're done.

However, if you want to calculate the volume using integration, then you can use

$V = \displaystyle \int_{z=0}^1 A(z) \ dz $

where $A(z) $ is the cross sectional area of the tetrahedron at elevation $z$. Using similar triangles, or vector methods, one can show that the cross-section area is given by

$A(z) = A(0) (1 - z)^2 = \dfrac{1}{2} (1 - z)^2$

Hence, the volume is

$ V = \displaystyle \int_0^1 \dfrac{1}{2} (1 - z)^2 = \dfrac{1}{6} $