# Volume of a Sphere

This article presents an intuitive approach to understanding how to derive the equation for the volume of a sphere.

## Introduction

### Finding the Volume of a Sphere: Your First Encounter

To start, I'd like to invite you to try to come up with a solution on your own. Take a few minutes and think about how you have found equations for the volumes of other round 3-dimensional shapes (cylinders, cones, etc.). Can you think of any way to reuse or extend one of those techniques? Can you create your own method? Please don't work too hard. If you find yourself getting frustrated, take a break or move on to the next section. There are some nice pointers to get you started. Please stop often and reassess if you have new ideas about how to solve the problem. The fewer hints you use, the stronger your creative and problem-solving skills will become. Remember there are lots of ways to solve these problems. If you invent one for yourself, which is not covered here, that is awesome!

### An Experimental Approach

The first digital manipulative can be (painstakingly) recreated with clay, plasticine, etc. and clear dividers. It is not recommended though. If you need to prove this for yourself, you can, but it will take a great deal of time and effort. If you must take the experimental approach, the following is easier.

The relationship can be guessed by filling a small, hollow cone (with liquid, sand, etc.) and seeing how many pours from the cone it takes to fill a larger sphere (hemisphere). Specifically, it can be instructive to find a cone and sphere (hemisphere) which have the same radius. The cone's height also must equal the radius. Then you can see how many full cones it takes to fill the sphere (hemisphere). Some people prefer the cone's height to be twice the radius, but that wouldn't fit nicely with the rest of this lesson. It's important to make sure that the shapes are manufactured to hold the correct volumes though. Based on the reviews, I believe that, if you want to take the "twice the radius" approach, this set meets the criteria. Amazon Link to Hollow 3-Dimensional Solids Measuring precisely and having the sphere (hemisphere) fill exactly makes for a better "ah-ha moment." If you have a higher tolerance for imprecision, other options like DIY/homemade solids can be used too (ie. made from cardboard and filled with sand). I'd recommend finding a suitable sphere (hemisphere) and making a cone to match, if you choose this route.

## Finding the Volume of a Sphere: A Visual Approach

Now that you have tried to find the relationship, take a look at this app. Leave the action button (beside the full-screen button) and hint button alone for now. You can use the full-screen button if you like. Please play with the dial. You can also tip the figure by dragging it.

What do you see? What does the dial do?

Please note, when the hints ask you to use the action button, you will see an animation. This animation depicts several layers changing shapes. It's important you know that the volume of each layer remains constant. You could (in theory) recreate this animation with clay, to prove that the volumes stay constant. That would take a lot of time and effort though, so I recommend that you just trust what the animation shows you.

## Finding the Volume of a Sphere: A More Rigorous Approach

You have the equation now, but perhaps you would like an explanation that justifies the equivalence of the volumes of the layers (rather than assuming it). If that is what you want, this section can help.

Imagine what it would look like to extend the previous app with a very large number of layers. Way way more than 50. Imagine you could have two copies beside each other, one set to the hemisphere and the other set to the cylinder with the cone taken out. The approximation of the hemisphere and the cylinder with the cone taken out would start to look very accurate. Increasing the number of layers can make the approximation as accurate as you want it to be. This means you can imagine your approximation getting infinitely accurate. The upshot of that is you can work with a model that is both the actual shape and retains the separate layers aspect of the approximation. The result might look a lot like the app in this section.

The goal here is to show that the volume of each layer of the hemisphere is equal to the volume of each layer of the cylinder with the cone taken out (instead of assuming it, like in the previous section). If you can show that, then it is clear that the volume of the two shapes are equal. Feel free to take a look at the app. Do not use the hint buttons yet, but the slider and fullscreen button can be used whenever you'd like. What do you see? What does the slider do?

### Caveat

You may be uncomfortable with the assertion that a shape and it's approximation are the same, if the approximation can be made as precise as you choose. The details for why this is true are part of Calculus and Calculus' parent, Analysis. Please trust that I have not led you astray. I have taken care of the details. These geometric arguments using intuition don't always work, but the one I showed you does work. If you want more, think about what it would mean to make an approximation (that can be made as precise as you choose) infinitely precise. Would it differ from the exact shape? Taming infinity is one of the great accomplishments of math.

## A Brief Discussion of Other Approaches (Extra Credit)

There are many other ways to derive the equation for the volume of a sphere. The most obvious is to learn Calculus and then use it directly (rather than waving hands). Can you think of any riffs on the approach discussed? When you are done thinking two will be discussed.

The first is to use a sphere directly by adding a second mirror image of the hemisphere below. Everything is the same as in the previous sections, but there is a mirror image to handle at the same time. You get a cylinder with the same radius as before, but twice the height. It has two cones (the same size as before) taken out, one from each end. You can then do the algebra and it works out. This is a lot of work to avoid doubling the equation!

The other option is to take a mirror image again and further complicate things. Instead of just subtracting the missing cones, you can take the ring layers and restack them. If you choose to stack them by increasing inner ring radius, you get a cylinder with a cone taken out, but the cylinder and cone now both have heights equal to twice the radius! Can you see why?

The cylinder was already that height, so no surprise there, but what happened with the cones? You stacked the layers of rings in pairs (with equal radii). This is just like stacking a single layer of each radius that is twice as tall. This results in the conical hole looking like the original but stretched to twice the height. The algebra can then be done and you get the right result.

I bring this up because it relates to the "Experimental Approach", using a cone with twice the height. A cone has one-third the volume of a cylinder with the same radius and height. If you make that height twice the radius (as above), then a sphere with the same radius has a volume of two-thirds the cylinder (a whole cylinder minus a third). Two-thirds of the volume of the cylinder is twice the volume of the cone. Therefore, it takes two volumes of the cone to match the sphere and it would clearly take one cone to match an appropriate hemisphere.