Find the volume of the solid described by integrating its cross-sectional area.

Question

2. A regular tetrahedron with side length "a". I.e. a pyramid with a triangular base whose faces are all equilateral triangles with side length "a".

Answer 1

Find h. I found it by finding the length to the center of the base: a/sqrt(3). Then using the right triangle made with the slant height, the height and the length to the center we find that h = sqrt(a² - (a/sqrt(3))²) = sqrt(2/3)a

You could just use (1/3) Ah = (1/3)(sqrt(3)a²/4)sqrt(2/3)a = a³ sqrt(2)/12

but nooooo! You have to integrate in y from 0 to sqrt(2/3)a the triangular area as a function of y.

You can make a linear relationship between y and the s(y) of the triangle:

y/(sqrt(2/3)a) = (a-s)/s which you can see gives s(0) = a and s(sqrt(2/3)a) = 0

so s = a - sqrt(3/2)y

Now take the integral from y=0 to sqrt(2/3)a of (sqrt(3) (a-sqrt(3/2)y)²/4)dy

You do get the same result.

Please consider a tutor. Take care.