# Calculus I - Volume by Slicing

by GreenPrint
Tags: calculus, slicing, volume
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Imagine the tetrahedron sitting on one of its triangular faces. The fourth vertex is connected to the other three. If you slice parallel to the base, you get a triangle that is "similar" to the base. Since the three sides of the base all have length 4, it is an equilateral triangle and so are all slices. You may know, or can calculate, that the area of an equilateral triangle, with side length s, is $\sqrt{3}s^2/4$. Finally, The side length of the slices depends linearly upon the height. It is 4 when the height, z, is 0 and is 0 at the vertex. You will, of course, need to calculate the height of that vertex. You can, for example, use the Pythagorean theorem to determine the distance from a vertex of an equilateral triangle to its centroid (where the three altitudes cross): Let the length of a side be "s". Then the length of all altitudes is $\sqrt{3}s/2$. Call the distance from the centroid to the base of the altitude "a". Then the length of the rest of the altitude is $\sqrt{3}s/2- a$. And that is the length of the hypotenuse of a right triangle with legs of length a and s/2. The Pythagorean formula gives $(\sqrt{3}s/2- a)^2= a^2+ (s/2)^2$. Solve that for a in terms of s. Now use the Pythagorean formula again to find the height of the pyramid. The side length of each slice will be kz+ 4, where "k" is chosen to give "0" when z= the height. Integrate the area of each slice, dz, from 0 to that height.