$Volume of a sphere.$

Question

$Volume of a sphere.$

I was trying to find the volume of a sphere using my own method. The answer I got ($2\pi r^3 - 2(2\pi )(r^2-\frac{1}{4}\pi r^2)$ or, $\frac{1}{2} \pi ^2r^3$) is incorrect except I'm not sure where my logic fails.

I began with the volume of a cylinder, given by the formula $2\pi r^3$ , where the cylinder has a height equivalent to its diameter, believing that the excess could be trimmed to approximate a sphere. My next step was to create a geometric shape to remove from this cylinder to approach the spherical shape.

This shape is defined by the expression $2(\pi r)(r^2-\frac{1}{4}\pi r^2 )$ . Here’s how it breaks down:

$r^2$ represents the area of a square whose side is the radius of the sphere.
Subtracting $\frac{1}{4}\pi r^2 $ removes a quarter of a circle’s area from this square, with the same radius as the sphere, aiming to reflect the curvature of the sphere.
Multiplying this adjusted area by $\pi r$ , you proposed to extrude this two-dimensional shape into the third dimension, extending it by a length equal to half the sphere's circumference.
The 2 outside the brackets is there since without that it would only create half a sphere with the other half would still be the cylinder.

My thought was to create numerous thin, triangular sections from this extruded shape, similar to the method used to find the area of a circle with infinitely many triangles. These triangles would then hypothetically be wrapped around the exterior of the cylinder, meant to subtract the same volume that distinguishes a sphere from a cylinder, thereby converting the cylinder's volume into that of a sphere.

To illustrate my thought process in attempting to derive a formula for the volume of a sphere, I have created a series of images. The initial shape, represented in the first image, is the result of taking the area of a square (with sides equal to the radius of the intended sphere) and subtracting a quarter of the area of a circle with the same radius. This difference in area is visualized by the shaded region.

Following this, I extruded this shape along a path equal to half the circumference of the sphere, as seen in the next three images. This creates a three-dimensional object that I then imagined could be dissected into an infinite number of triangular prisms, as depicted in the subsequent two images from different perspectives.

Geometry

Mathgnome

45

Matchmaticians Volume of a sphere. File #1

File #1 (pdf)

Report

Mathe

0

I think you have the wrong formula for the volume of the cylinder. It should be pi*r^2*r. That's area of the base times it's height. What you have is the superficial area of the cylinder times height, but that's incorrect.
Mathgnome

0

I was trying to make a cylinder with the height equivalent to the diameter of the circular part for instance pi*r^2 = to the circular part multiplied by the height which is 2r. Am I wrong?

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Mathe

0

I think you have the wrong formula for the volume of the cylinder. It should be pi*r^2*r. That's area of the base times it's height. What you have is the superficial area of the cylinder times height, but that's incorrect.
Mathgnome

0

I was trying to make a cylinder with the height equivalent to the diameter of the circular part for instance pi*r^2 = to the circular part multiplied by the height which is 2r. Am I wrong?

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Your approach is quite creative! However, the error lies in the step where you subtract the shape created by $2(\pi r)(r^2-\frac{1}{4}\pi r^2)$. This shape doesn't accurately represent the excess volume of the cylinder over the sphere. What you have, represents the excess volume of a cube over the sphere!

Your shapes do not represent the curvature of a sphere. A sphere has a uniform curvature at every point on its surface, meaning that every point is equidistant from the center. The shapes in the image have flat surfaces and sharp edges, which do not convey the continuous, smooth surface of a sphere.

Instead, you should subtract the volume of a cone with height and radius equal to the radius of the sphere. This cone precisely accounts for the extra volume of the cylinder that isn't part of the sphere. This is the famous ratio determined by Archimedes. The ratio of the volume of a sphere to the volume of its circumscribed cylinder is 2:3.

Kav10

1.9K

Mathgnome

0

Thanks for the reply. What do you mean by "What you have, represents the excess volume of a cube over the sphere!" since if you create a cube with the same lengths as the diameter of the sphere 8r^3 you still don't get the volume for a sphere. Additionally when you refer to the equidistant of a sphere are you saying that even if you divide the shape into many small pieces no matter how small you make them it still will not reach the exact shape of the sphere? Thanks again.
- Kav10
  
  0
  
  Sure. What I meant is that the shape you created and posted the images for, doesn't represent the excess volume of the cylinder over the sphere. Imagine a cube over the sphere, and the cube can be divided into 8 identical cubes, now calculating the excess volume of the smaller cube (1/8 of the main cube) over 1/8 of the sphere, will give the volume of the sphere. In other words, the sharp edges that you have in your shape does not represent the curvature of cylinder’s and sphere’s.
- Kav10
  
  0
  
  So, no matter how small you make those shapes, you won’t get to sphere’s volume because the shape is not representing the excess volume correctly.

$Volume of a sphere.$

Answer

Related Questions

Search