Attempt at Deriving the Volume of a Cone

In summary, the conversation revolved around the calculation of the volume of a cone compared to the volume of a cylinder using JavaScript. The code involved a summation of the area of a circle, where the height is the number of times the summation is iterated. The volume of the cone was found to be approximately one third of the volume of the cylinder. The conversation then delved into representing this relationship with calculus and the use of the disk method to solve for the volume of a cone. The conversation ended with the individual needing to do more research on integration.
  • #1
mburt
52
0
Hi, so this morning I made an attempt at this... With javascript (website programming language) I was able to successfully yield the ratio of the volume of a cone compared with the volume of a cylinder (1/3).

This is the source code:

<script>
var pi = Math.PI,
cone_v = 0,
h = 0,
r = 1;

for (var i = 0, h = 0;i < r; i = i+0.0001, h++) {
cone_v += pi*i*i;
}

var cyl_v = pi*r*r*h;

var cone_constant = cone_v/cyl_v;

//cone_constant = 1/3
</script>

And basically this is the idea. It's a summation of the area of a circle from radius 0 to 1 (adding on 0.0001 each time), where the height is the number of times the summation is iterated. So in this case it iterates 1000 times... Therefore this is the height, in units.

The volume of the cone = 10473.5 units^2
The volume of the cylinder = 31419.1 units^2

If you divide the volume of cone by cylinder you get approximately one third.

... Now I attempted represent this with calculus but I think I failed. This is as far as I got:

[tex]\sum_{0} i = i + \frac{1}{\infty}[/tex]
therefore
[tex]h = 1^\infty[/tex]

Vcone = [tex]\pi(5^\infty)^2 = 25\pi^\infty[/tex]

Vcylinder = [tex]\pi{(1)^2}1^\infty = \pi^\infty[/tex]

But clearly 25/1 doesn't equal 1/3. What am I doing wrong here? Is it possible to represent this relationship with a summation formula?

**EDIT: accidentally said "area" not volume. Whoops
 
Last edited:
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  • #2
You can solve for the volume of a cone by the disk method.
Let r = h so that r varies as h varies exactly.

Imagine the volume of a cone as stacked disks, the radius of every disk different.

The approx volume of a disk at any r is:

[tex] \Delta V = \pi r^2 \, \, \Delta r [/tex]
The exact volume of a disk at any r is:

[tex] dV = \pi r^2 \, \, dr [/tex]
Integrate to find volume of a cone

[tex] V = \int \pi r^2 \, \, dr [/tex]

[tex] =\frac {\pi r^3}{3} [/tex]

since h = r

[tex] =\frac {\pi r^2h}{3} [/tex]
 
Last edited:
  • #3
I'm just getting differentiation down now, I guess I need to do some research on integration. Thanks!

And the "disks" method is exactly how I had pictured it, I just didn't know how to describe it
 

1. What is the formula for calculating the volume of a cone?

The formula for calculating the volume of a cone is V = (1/3)πr^2h, where V represents volume, r represents the radius of the base, and h represents the height of the cone.

2. How is the formula for the volume of a cone derived?

The formula for the volume of a cone is derived by using the formula for the volume of a cylinder (V = πr^2h) and recognizing that a cone is essentially a third of a cylinder. By dividing the formula for the volume of a cylinder by 3, we get the formula for the volume of a cone.

3. What are the units for measuring volume of a cone?

The units for measuring volume of a cone can vary depending on the units used for the radius and height. Typically, the volume will be measured in cubic units (such as cubic meters or cubic inches).

4. Can the formula for the volume of a cone be used for any cone?

Yes, the formula for the volume of a cone can be used for any cone, as long as the measurements for the radius and height are consistent with the units used in the formula.

5. How can the formula for the volume of a cone be applied in real life?

The formula for the volume of a cone can be applied in various ways in real life, such as calculating the volume of a cone-shaped container or determining the amount of material needed to fill a cone-shaped structure. It is also commonly used in mathematics and engineering applications.

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