Find the volume V of a regular tetrahedron whose face is an equilateral triangle of side s. The tetrahedron has height h = squareroot(2/3)*s

Answer 1

The apex point of tetrahedron is perpendicularly above median of base equilateral triangle, so as problem makes given the height (h) must be s*√(2/3).

don't confuse h of tetrahedron with altitude of dA triangle

For the integral, from z=0 to h of ∫dAdz (I consider the base on the xy plane)

correction: dV=A(z)dz, should be ∫dV=∫A(z)dz

A(z)=(1/2)*((1-(z/h))*s*((√3)/2)*(1-z/h)*s

area = (1/2) times triangle base times triangle altitude

notice the triangle shrinks as z increases to h

the integral calculates to Volume=((√3)/4)*s²*(z-(z²/h)+z³/3h²) evaluated from z=0 to z=h

Volume=((√3)/4)*s²*h/3

h=s*√(2/3).

Volume=((√2)/12)*s³

Answer 2

Cool problem: The volume is equal to the integral of the (area at y)dy from 0 to the height.

But the area of the cross-sectional equilateral triangles is easily calculated from s²sqrt(3)/4 . Also, we know that the expression h = sqrt(2/3) S holds for all heights in the triangle because the sides of the tetrahedron are lines - s and h are directly proportional or y = sqrt(2/3)s where s is a variable on the slant height that goes from 0 to S.

You can do the integral in terms of s or y: (dy = sqrt(2/3)ds)

s is probably easier: integral for 0 to S of (s²sqrt(3)/4)(sqrt(2/3)ds

That leads to the correct volume of a tetrahedron which you can Google.