Use a double integral to compute the volume of a regular tet

Use a double integral to compute the volume of a regular tetrahedron with side length s.

Solution

According to given we will have some foloowing steps

step 1) Compute the distance x from a vertex to the centroid of the base.
drop a perpendicular from the centroid to a base edge. Call this length y .
Half an edge, ., s/2 , the length y, and x form a 30-60-90 triangle, so
s/2 = sqrt(3)y ; x = 2y ; so x = s/sqrt(3)

step 2) Now we will find the height of the regular tetrahedron

Compute the altitude h of the tetrahedron.
An edge, x, and the altitude h form a right triangle, so
.h^2 = s^2 - (s/sqrt(3))^2 = (2/3)s^2 , so h = sqrt(2/3)s

3rd step) now we will set up integrals

Set up the integral
. Let z be the (variable) height. Let s be the side of the tetrahedron of height z
.The triangle described in the previous step gives the ratio
of the variable distances s and z : s = sqrt(3/2)z
.The volume element is: s^2sqrt(3)/4 dz = (3sqrt(3)/8)z^2 dz

step 4) Integrate from 0 to sqrt(2/3)s we get

=s^3 sqrt(2) / 12 answer