# volume of a frustum in different way

maybe i will sound very stupid but i don't care. This question has hammered my head for days. i had to calculate the volume of the frustum (square based) having top base side(s), bottom base side(S) and height(H).

i didn't remember the frustum volume standard formula so instead of googling for it i tried to find it my way. In my head i tried to visualize it as the sum of

1.the volume obtained by multiplication of the top base surface by the height

2.the volume obtained by HALF of the multiplication of the difference of the two areas (so SB - Sb) by the height

the final formulas would be (( S^2 - s^2 ) * h )/ 2 + s^2 * h.

now the result of this would bè different from the application of the standard formula and my question is why? what is wrong in my way of think? please help me..

• What are you visualizing in (2) that leads to this formula? – Ethan Bolker Mar 26 '16 at 18:51
• Hint: A cone has volume $1/3\times \textrm{area of the base}\times \textrm{height}$. Your (1) is the volume of a cylinder, for your (2) I suppose you expect to find the difference between the frustum and the cylinder. But you are computing the difference between two cylinders (one whose base is the lower surface, one on the top one). – Jean-Claude Arbaut Mar 26 '16 at 18:54
• the frustum is derived from a pyramid in my case (squared based). So in (1) i'm visualizing the volume of a Parallelepiped and in (2) i visualize half of the difference between the two parallelepiped (one SBh and the other Sbh). in 2d this work perfectly (with a isosceles trapezoid). if i add the surface of b*h to ((B-b)*h)/2 it works... why for volume it didn't works? – Aer Mar 26 '16 at 19:56
• Ultimately, because when you integrate $x$, you get $x^2/2$, while when you integrate $x^2$, you get $x^3/3$. Notice the difference, $1/2$ then $1/3$. It's fairly easy to write resp. the area and the volume, with an integral (with only one integration variable, the height). By the way, that the base be square, circular or other doesn't change anything to the result. – Jean-Claude Arbaut Mar 26 '16 at 20:38
• yes you are right. Thank you! few moment after writing the comment i realized my mistakes.. and that is also stupid consider different the square/circular base.. – Aer Mar 26 '16 at 22:15

In the 3 dimensional case your ftustrum may be thought of as having four pieces.

1) a parallelepiped with two faces along the lateral faces of the frustrum, a third face along the small base of the frustrum, and the other faces parallel to these. The volume of this piece is $s^2h$.

2, 3) Two triangular prisms that are each "on their sides", so to speak. The volume of each one is the area of the triangular face ($(1/2)(S-s)h)$ times the length $s$ between the triangular faces.

4) The final piece is a square pyramid of base $(S-s)^2$ and altitude $h$. The volume of this piece is $1/3$ the product of these quantities.

Stop right there. The multiplying factor for that last piece is $1/3$, not $1/2$. That is why you can't just take the average of the two prism volumes like you do in two dimensions. Nor can you use a 2:1 weighted average because of the prisms giving a coefficient of $1/2$. Three dimensions is subtly more complicated than two.

Now go ahead and add up the four component volumes, remembering that there are two triangular prisms. You'll get the right formula.

• thank you. you are right.. to be quite frank i realized my mistake by the time i write the comment to my question.. you all are really kind. thank you – Aer Mar 26 '16 at 22:19