Volume of a Sphere

This article presents an intuitive approach to understanding how to derive the equation for the volume of a sphere. It is inspired by the concept of limit from Calculus and leans on the intuition behind Cavalieri's principle.

For the SoME3 Judges and Peer Referees

What is Most Important to Investigate

Please explore the Digital Manipulatives (apps) thoroughly. The rest of the article is designed to support them and is expected to match SoME3 referees' expectations and instincts to some degree. Please do not feel compelled to read overly closely, if you already see where the lesson is going.

Intended Audience

This lesson is for kids. Please imagine yourself as a child learning this for the first time. Would this have been useful to you when you were learning years ago? Would you have benefited from encountering limits this early, even if in an imprecise manner? What about your peers, who perhaps struggled more?

What is this Hint System

The hint system is designed to allow students to engage with the problems with little to no support (Discovery/Inquiry-Based Math). The hints gradually give more and more information until we end up with a full explanation (Traditional Lecture). Since Math is not a "spectator sport," it is hoped that repeated use of this hint system, in multiple lessons, will encourage students to use fewer and fewer hints and learn to solve problems creatively for themselves.

Isn't this Basically Integral Calculus?

Yes! I am using this lesson as a Trojan Horse. Exposing students to these ideas early, so they have horizon knowledge about what is coming down the track later, is important. It gives their subconscious extra years to digest the ideas before they become everyday companions.

Why no Rigor?

Sorry, the treatment is not rigorous. To my Pure Math friends, sometimes rigor can get in the way of clarity when first encountering a concept. I promise a rigorous treatment of calculus, straight "outta" Real Analysis, will come in future years.

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, and if you want to take the "twice the radius" approach, this set seems to meet 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.

Digital Manipulative: Volume of a Sphere, Visually

Hint 1

The app loads with one large cylinder and the dial set to 1. As the dial changes, the figure updates. A larger dial value results in more cylinders stacked on one another. Do you see anything interesting about the heights of the cylinders?

Hint 2

There are three things to notice. First, the total height of the stack remains constant. Second, the heights of the individual cylinders shrink as the number of cylinders increases. Third, the height of all the cylinders are equal, for a given dial value. Can you describe the height of the cylinders more precisely?

Hint 3

The height appears to be the same as the original single cylinder, divided by the dial value. Said another way, the height of each cylinder is the total height of the stack divided by the dial number and the stack height is constant. For example, when the dial is set to 3, the height of each cylinder is one-third the height of the whole stack. What about the radius of each cylinder?

Hint 4

The radii of the cylinders shrink higher up the stack. Now try setting the dial to a large number, the maximum of 50, if your device can handle it. What do you see? If you don't see anything new, please be sure to drag the shape to get a better perspective.

Hint 5

The stack of cylinders looks a lot like a hemisphere (half a sphere). Try pressing the action button now. What do you see? If you don't see anything new, please be sure to drag the shape to get a better perspective.

Hint 6

The largest (bottom) red cylinder remains unchanged, but the rest change significantly. The red cylinders' outer radii grow until they match the largest red cylinder. At the same time, a cylindrical hole appears at the center of each red cylinder, except the largest one. The hole grows too. By the end of the animation, there is a stack of rings (cylinders with cylindrical holes). How do the thicknesses of these rings compare?

Hint 7

Starting from the top, they are thin and get thicker farther down. The bottom layer is so thick that there is no hole. Alternatively, you could describe the inner radii as large at the top and getting smaller towards the bottom (eventually reaching zero by the bottom layer). In fact, it appears that the thicknesses/inner radii change the same amount, as you go down (or up) the stack (layer by layer). You can see this if you look down into the shape from above. Can you think of a way to describe the overall shape of the stack?

Hint 8

The constant heights of the layers and the constant change of the radii between layers suggests a constant slope. Does this help you identify the shape of the stack?

Hint 9

Try pressing the hint button. Does this help you identify the shape of the stack?

Hint 10

The shape of the stack looks a bit like a big cylinder with a cone cut out of it. This cylinder and cone have the same radius and height. What do you think that height is?

Hint 11

Try pressing the action button a few times. Does this help?

Hint 12

The height of the cylinder and the cone appear to be the same as the radius of the hemisphere and the radius of the hemisphere seems to be the same as the radius of the cylinder and the cone. How do you think we can put this together to find the equation for the volume of the hemisphere?

Hint 13

As mentioned before the hints began, the volume of each layer remains constant. That means the volume of the hemisphere and the cylinder with a cone cut out of it are the same. How could you represent this with an equation?

Hint 14

The result could be represented with this equation $V_{h-sphere}=V_{cyl}-V_{cone}$. How might you represent this in terms of values you know?

Hint 15

Recall that the equation for the volume of a cylinder is $V_{cyl}=\pi r^{2}h$ and that the equation for the volume of a cone is $V_{cone}={\frac {1}{3}}\pi r^{2}h$. Does this help?

Hint 16

You could substitute these equations into the equation in hint 14 to get $V_{h-sphere}=\pi r^{2}h-{\frac {1}{3}}\pi r^{2}h$. How could you further simplify this equation?

Hint 17

You could collect like terms. What would that look like?

Hint 18

It would look like this $V_{h-sphere}=(1-{\frac {1}{3}})\pi r^{2}h$. Which simplifies to $V_{h-sphere}={\frac {2}{3}}\pi r^{2}h$. Can it be further simplified?

Hint 19

Yes, you could substitute $r$ for $h$ because the height is equal to the radius (see hint 12). This simplification would give the equation $V_{h-sphere}={\frac {2}{3}}\pi r^{3}$, after collecting the $r$ values. Now that you have the equation for the volume of a hemisphere, what do you think the equation for the volume of a sphere is?

Hint 20

It's just double the hemisphere volume. What would that look like?

Hint 21

It would look like this $V_{sphere}={\frac {4}{3}}\pi r^{3}$. This is the equation for the volume of the sphere.

Finding the Volume of a Sphere: Now With Fewer Assumptions

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?

Digital Manipulative: Volume of a Sphere, More Rigorously

Hint 1

There's a cylinder with a cone taken out of it and a hemisphere. The setup appears analogous to the previous app. When the slider is moved, the layer that is lit-up changes. It appears that the layers at the same height in the two shapes are both lit up at the same time. The layers are so thin that they appear as if they are two-dimensional. From the previous section, what do you expect about the tiny heights of each of these pairs of highlighted layers?

Hint 2

The heights of each of the pairs of highlighted layers were chosen to be equal. For the volumes to be equal another quantity (simpler than the volume) must also be equal between the two highlighted layers. What is it?

Hint 3

The thin cylinder and the thin ring are both prism-like shapes. This means their volumes must follow the equation $V=hA_{face}$. Does this help?

Hint 4

If the two volumes are to be equal, then it must be true that $hA_{ann}=hA_{circ}$. $A_{circ}$ is for the area of the circle and $A_{ann}$ is for the area of the annulus (ring). Can you simplify this?

Hint 5

Yes! The heights are the same, therefore the equation can be simplified to $A_{ann}=A_{circ}$. This means that the area of the faces must be the same if the volumes are to be the same. Can you think of another way to describe this area? Especially given how thin the layers are.

Hint 6

Because of how thin the layers are, the area of the faces are essentially the cross-sectional areas of the overall hemisphere or cylinder with the cone taken out of it respectively. If you can show that all the cross-sectional areas are the same between the two shapes (at equivalent heights), you will have shown that the volumes are equivalent. This is called Cavalieri's Principle. How would you go about characterizing these cross-sections?

Hint 7

First it might be useful to include the dimensions of the overall shapes. Please sketch a diagram of the app and label it too. Feel free to use the first hint button if you want some help. You should put your cross-sections at an arbitrary height (say about halfway up).

Hint 8

Compare your diagram to the app with the first hint button on. Do they look similar? If you haven't included the radius of the circular cross-section, add it now and label it $x$. If you haven't included the inner radius of the annulus-shaped cross-section, add it now. The outer radius of the annulus is always just $R$ which is clear from the $R$ labels already on the diagram. Does anything jump out at you?

Hint 9

Look at the radius of the sphere and the radius of the circular cross-section. Is there anything interesting to add there?

Hint 10

You could draw a line segment vertically from the center of the bottom face of the hemisphere. It could be used to form the third side of a right triangle, by connecting the two existing line segments. draw this line and label it $y$. What does $y$ represent?

Hint 11

$y$ is the height at which the circular cross-section is taken. Does it appear anywhere else?

Hint 12

Yes. It is the height at which the annulus-shaped cross-section is taken too. Put that line segment on your diagram as well and label it $y$. Finally, what should you label the inner radius of the annulus?

Hint 13

What is the height and what is the radius of the cone?

Hint 14

They are both $R$. What does that mean?

Hint 15

The slope of the cone must be 1. This means you must move over the same amount you move up or down. What does this mean for the inner radius of the annulus?

Hint 16

The inner radius of the annulus must be the same as the height at which the annulus-shaped cross-section is taken. Label it $y$ too. Now press the final hint button and compare. Try shifting the slider around. Now it is time to show the areas are equal. What is the equation for the area of the annulus?

Hint 17

It is $A_{ann}=\pi R^{2}-\pi y^{2}$, because you take the area of the big circle (outer radius $R$) and subtract the area of the circular hole (inner radius $y$). Can you simplify this expression?

Hint 18

It can be simplified by collecting the $\pi$ terms. It then becomes $A_{ann}=\pi (R^{2}-y^{2})$. What about the area of the other cross-section?

Hint 19

It's just $A_{circ}=\pi x^{2}$. How do you show that the areas are equal?

Hint 20

For them to be equal it must be the case that $x^{2}=R^{2}-y^{2}$, because if they were equal the $\pi$ on each side would cancel out. Why would this new relation be true though?

Hint 21

Try rewriting it as $x^{2}+y^{2}=R^{2}$. Is it easier to show that this relationship is true?

Hint 22

What does the Pythagorean Theorem say about the triangle you drew in the hemisphere?

Hint 23

It says $x^{2}+y^{2}=R^{2}$ which is exactly the result you need! You have now shown that the two shapes have the same volume. That is $V_{h-sphere}=V_{cyl}-V_{cone}=\pi r^{2}h-{\frac {1}{3}}\pi r^{2}h={\frac {2}{3}}\pi r^{2}h={\frac {2}{3}}\pi r^{3}$. Double the hemisphere to a sphere and you have $V_{sphere}={\frac {4}{3}}\pi r^{3}$.

Caveat

You may be uncomfortable with the assertion that a shape and its 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 single 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 a conical hole like in the first app but each layer (and the entire shape) are 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", in the case where one chooses to use a cone with a height equal to twice the radius. A cone has one-third the volume of a cylinder with the same radius and height. If you make both heights 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), as we saw in the previous paragraph. Two-thirds of the volume of the cylinder is twice the volume of the cone. Therefore, it takes two volumes of this cone to match the sphere and it would clearly take one cone to match an appropriate hemisphere.

SoME1 & SoME2 Submissions

If you enjoyed this lesson, you may also like my previous SoME submissions

• Area Contained in a Circle
• Volume of a Pyramid

or perhaps this lesson on

• Pythagorean Theorem