[SOLVED] Theoretical Proof for volume of a cone in relation to the volume of a cylind

May 2010
1
0
Hi,
This is just out of curiosity. I know that the formula to calculate the volume of a sphere is \(\displaystyle V=\pi r^2 h\)
And the volume of a cone is \(\displaystyle V=\frac{1}{3} \pi r^2 h\)
This means that given the same heights and radii, a cone's volume will be a third of a cylinder.

My question is this. Why is it not that the volume such a cone be half of the volume of such a cylinder?



The highlighted edges are congruent, so as the edges at the bottom of the two figures. It could be said that the area of the triangle is half of that of the rectangle.

I think that if we rotate the rectangle at the left around its highlighted egde, we will have a cylinder. And if we do the same to the rectangle at the right, we will get a cone.

In my understanding, the volume of a figure in space would be the sum of all the areas of the "slices" that comprise it. Then if the area of each of the triangles are half of that of the rectangles, then why is not the volume of the cone half the volume of a cylinder of same radius and height? Is there a viable way to explain this?

I searched the internet, yet the results are always using experiments that relies on fluidic substances to measure the volume. I really want to know that can it be proven theoretically.

Thanks.
 
Last edited:
Dec 2009
3,120
1,342
Hi,
This is just out of curiosity. I know that the formula to calculate the volume of a sphere is \(\displaystyle V=\pi r^2 h\)
And the volume of a cone is \(\displaystyle V=\frac{1}{3} \pi r^2 h\)
This means that given the same heights and radii, a cone's volume will be a third of a cylinder.

My question is this. Why is it not that the volume such a cone be half of the volume of such a cylinder?



The highlighted edges are congruent, so as the edges at the bottom of the two figures. It could be said that the area of the triangle is half of that of the rectangle.

I think that if we rotate the rectangle at the left around its highlighted egde, we will have a cylinder. And if we do the same to the rectangle at the right, we will get a cone.

In my understanding, the volume of a figure in space would be the sum of all the areas of the "slices" that comprise it. Then if the area of each of the triangles are half of that of the rectangles, then why is not the volume of the cone half the volume of a cylinder of same radius and height? Is there a viable way to explain this?

I searched the internet, yet the results are always using experiments that relies on fluidic substances to measure the volume. I really want to know that can it be proven theoretically.

Thanks.
Hi kietjohn,

When you rotate the triangle around it's left edge,
you are tracing out circles from the base to the tip.

The circles at the base are "much larger" than the circles at the top,
in your sketch.
The cone volume is the integral of the areas of the circles.

\(\displaystyle \int_{0}^H\left({\pi}r^2\right)dh\)

For convenience, turn the triangle upside-down,
select a smaller radius about halfway up the triangle,
letting "h"=the height of the smaller triangle formed.
Then, in my sketch, the circles are much larger at the top.

\(\displaystyle \frac{r}{h}=\frac{R}{H}\ \Rightarrow\ r=\frac{hR}{H}\)

\(\displaystyle \int_{0}^H\left({\pi}\frac{h^2R^2}{H^2}\right)dh=\frac{{\pi}H^3R^2}{3H^2}=\frac{{\pi}R^2H}{3}\)
 

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