The document verifies that the volume of a cone is equal to one-third the volume of a cylinder with the same base and height using calculus. It shows that the volume of a cylinder is πr^2h, while the volume of a cone is ∫π(sh)^2 dh = πs^2(1/3)h^3 = πr^2(1/3)h, where s is the slope of the cone's side.

1. VOLUME OF A CONE IS EQUAL TO ONE-THIRD OF THE
VOLUME OF A CYLINDER OF THE SAME BASE AND
HEIGHT.
MATHEMATICS HOLIDAY HOMEWORK
2014-2015

2. QUESTION TO BE ANSWERED
Verify that volume of a cone
is equal to one-third of the
volume of a cylinder of the
same base and height.

3. RELATION OF A CONE AND A CYLINDER

4. ANSWER
Using Calculus:
Firstly, the cylinder is easy, just have lots of little discs, each disc has a radius of "r" and
an area of πr², and we need to integrate over height:
V = ∫πr² dh = πr²h
That was fairly easy!
Now for the cone, the little disc's radius get smaller as you get higher. The rate they get
smaller is a constant, the slope of the sides, which we can call s.
For simplicity I will turn the cone upside-down, so the disc's radius get bigger with
height. Each disc will have a radius of "sh" and an area of π(sh)², and we need to
integrate over height:

5. Which is one third of the cylinder's volume!
Now, I recall seeing proofs without calculus that did something similar, they turn
the cone into "N" small discs, and add them up.
Disc number "i" will have a height of h/N, and a radius of r(i/N), with a volume of
πr²(i/N)²(h/N) = πr²i²h/N³
Summing over "i" : V = Σπr²i²h/N³ = πr²h/N³ Σi²
The only thing in the way now is "Σi²", which I know can be simplified to a few
terms, but am not sure how. If that could be done, hopefully we would see the
formula come out to be πr² (1/3)h
So, I haven't totally solved this for you, but I hope I have helped.

6. V = ∫π(sh)² dh = ∫πs²h² dh = πs² (1/3)h³
Now we know that the slope s = r/h, so: V = π(r/h)² (1/3)h³ = πr² (1/3)h
This or turn to the next slide for more

9. Thank You
HOPE YOU LOVED THE PROJECT