# Surface Area of a Sphere

This article presents an intuitive approach to understanding how to derive the equation for the surface area of a sphere. It is inspired by the concept of limit from Calculus (particularly multivariate integration).

## For Teachers and Parents

### 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. This is, of course, only one possible arrangement of these hints. An instructor will be better able to adapt the presentation of the hints to the progress a specific student has made. This is beneficial to the student until they learn to use the system and self-regulate. If the student needs hints, they can choose to set aside their particular approach temporarily while filling in the gaps in the order the text presents them. Later they can analyze how the approach they took compares to that of the text.

### 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?

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 Surface Area 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 and surface areas of other 3-dimensional shapes (cylinders, cones, pyramids, etc.). Also, consider the equations and ways you found those equations for relevant 2-dimensional shapes (like the circle). 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!

### Before Using the Digital Manipulative

Before using the digital manipulative (in the next section), take a moment to review the following lessons (if you haven't already).

It is of particular importance that you are aware of the approach taken in the manipulatives and the final equations found. You will use these ideas to find the equation for the surface area of a sphere (in the next section).

## Finding the Surface Area 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 and zooming in and out (using the pinching gesture on mobile devices and the mouse scroll wheel otherwise). As a note, setting the dial to 128 or above, will slow the frame rate while panning and zooming (as well as the animation when you are asked to use the action button), quite a bit on mobile devices. Likewise for a dial setting of 512 and above on any device without a discrete graphics card.

What do you see? What does the dial do?

Please note, that when the hints ask you to use the action button, you will see an animation. If you are particularly eagle-eyed, you might notice an imperfection in the animation. If you don't notice it, then don't worry, it doesn't change anything. The animation is perfect at the beginning and at the end, but in between there is some minor overlapping of shapes. The shapes are meant to be solid, and thus they should not be capable of passing through each other. The potential changes to the code I considered (to remedy this imperfection) either would slow the animation significantly or make it a great deal less pretty, so I decided to leave it as is. Hopefully, this warning suffices to prevent any confusion among those who spot the issue.

Hint 1
The app loads with a figure and the dial set to 8. What does this 8 mean?
Hint 2
How many sides does the figure have when the dial is set to 8?
Hint 3
It has 8 sides. What might the dial number represent?
Hint 4
It represents the number of sides. If you increased the dial one increment (to 32) you would find 32 sides, and 128, 512 then 2048 if you continued increasing. You can check, but it gets to be a lot of counting, so feel free to trust me. What does increasing the number of sides accomplish?
Hint 5
Does the shape start to look like something important, when you increase the number of sides?
Hint 6
What is the goal of this lesson?
Hint 7
As the dial value increases, the shape starts to look like a sphere. The goal of the lesson is to find the equation for the surface area of a sphere. How might you use this approximation of a sphere to approximate the surface area?
Hint 8
How have you found surface areas previously?
Hint 9
Previously, surface areas have been found by adding the area of each of the faces together. It would be messy to do that directly here (and would take a long time for 2048 faces). What might you do to leverage this relationship while avoiding doing thousands of computations?
Hint 10
Let's agree on a convention. All the faces will be uniquely labeled with the numbers ${\displaystyle 1}$ up to the total number of faces (call it ${\displaystyle N}$). For the 8-sided figure, that would be ${\displaystyle 1,2,3,4,5,6,7,8}$. This can be written more concisely as ${\displaystyle 1,2,...,8}$. Then it can be generalized to work for any number of faces (${\displaystyle N}$) by writing ${\displaystyle 1,2,...,N}$. You can then write the area for each face as ${\displaystyle A_{1},A_{2},...,A_{N}}$. The sum of these areas is the surface area. This can be written ${\displaystyle A_{Shape-Surface}=A_{1}+A_{2}+...+A_{N}}$. No computation is needed. It's just a lot of notation. You now have an equation for the surface area of your approximation. To avoid the computation completely though, you must find a way to finesse this sum into something simpler. What could that be?
Hint 11
Check out the Circle Area (Segment Method) and Circle Area (Inscribed Regular Polygon Method) articles for inspiration. Can you think of a way to do something like this in 3-dimensions?
Hint 12
The trick in the circle apps is to relate the perimeter (circumference) to the area. Does this give you any ideas?
Hint 13
A sphere is sort of like a 3-dimensional circle, and a pyramid is sort of like a 3-dimensional triangle. Does this give you any ideas?
Hint 14
The general notion is to relate the boundary to the interior. Does this give you any ideas?
Hint 15
A 3-dimensional version of this might be found by relating the surface area to the volume. Does this give you any ideas?
Hint 16
Bring the dial down to 8 and try pressing the action button. Try running it again using other dial values. What do you see? Please be careful not to overload your device. The graphics power required to run the app (when the dial is a large number) can outstrip the capabilities of most devices.
Hint 17
The approximation unfolds into a shape that looks like flower petals. How is this useful?
Hint 18
The shape has an area equal to the surface area of the approximation. How can you produce a useful volume (the volume of the approximation)?
Hint 19
Try using the hint button and watching the animation for a few dial values. What do you see?
Hint 20
The approximation unfolds into a bunch of pyramids arranged like flower petals. How is this useful?
Hint 21
The sum of the volumes of the pyramids is the volume of your approximation. Let's use a little more notation. Label each pyramid ${\displaystyle 1,2,...,N}$. Please choose the most appropriate option. That is, use the number (assigned in Hint 10) from the face that forms the base of each pyramid to label the pyramid. Then label the volume of each pyramid ${\displaystyle V_{1},V_{2},...,V_{N}}$ accordingly. The sum of these volumes is the volume of the approximation. This can be written ${\displaystyle V_{Shape}=V_{1}+V_{2}+...+V_{N}}$. You are getting close. This sum is similar, but not the same. The volumes are associated with the areas you want to replace. How could you transform these volume terms to include areas?
Hint 22
What is the equation for the volume of a pyramid?
Hint 23
The equation for the volume of a pyramid is ${\displaystyle V_{pyramid}={\frac {1}{3}}A_{base}h}$ and can be found in the Volume of a Pyramid article. Apply the equation to each pyramid in the volume sum (Hint 21). What does the result look like?
Hint 24
Because you chose to match the numbering system of the areas and volumes the result looks like this ${\displaystyle V_{Shape}={\frac {1}{3}}h_{1}A_{1}+{\frac {1}{3}}h_{2}A_{2}+...+{\frac {1}{3}}h_{N}A_{N}}$. This uses the notation where the height of each pyramid has a matching index. You have found another place where those areas are summed. Now you must clean up those heights and factors of one-third so the sum matches the one used for surface area exactly. What happens to all those heights as the dial value increases (as the approximation gets better)?
Hint 25
When the pyramids are clumped (not unfolded into the flower shape), to approximate the sphere, where are the peaks?
Hint 26
They are all at one special point. Think relative to the sphere that is being approximated.
Hint 27
All the peaks are at the center of the sphere. When the approximation is improved, the base of the pyramid shrinks. What value from the sphere do the heights approximate?
Hint 28
The heights all tend towards the radius of the sphere as the approximation improves. If you improve it infinitely, then they become the same. What does this do to the volume equation?
Hint 29
The equation becomes ${\displaystyle V_{Shape}={\frac {1}{3}}rA_{1}+{\frac {1}{3}}rA_{2}+...+{\frac {1}{3}}rA_{N}}$. Can you simplify this expression?
Hint 30
All the terms have a factor ${\displaystyle {\frac {1}{3}}r}$. Does this help?
Hint 31
Common factor it out. What do you get?
Hint 32
You get ${\displaystyle V_{Shape}={\frac {1}{3}}r(A_{1}+A_{2}+...+A_{N})}$. What could you do next?
Hint 33
The area sum from Hint 10 appears in the brackets. Can you do something with it?
Hint 34
You could replace it with the value it is equal to, which is ${\displaystyle A_{Shape-Surface}}$. That would look like this ${\displaystyle V_{Shape}={\frac {1}{3}}rA_{Shape-Surface}}$. Thank goodness that sum is gone! The approximation has been made infinitely precise though. How does that affect ${\displaystyle V_{Shape}}$ and ${\displaystyle A_{Shape-Surface}}$?
Hint 35
They become ${\displaystyle V_{Sphere}}$ and ${\displaystyle A_{Sphere-Surface}}$ respectively. These are the volume of the sphere and the surface area of the sphere respectively. Plugging them in gives the equation ${\displaystyle V_{Sphere}={\frac {1}{3}}rA_{Sphere-Surface}}$. Finally! The surface area of a sphere has appeared in the equation, but you can simplify it further. Can you see how?
Hint 36
What is the equation for the volume of a sphere?
Hint 37
The equation for the volume of a sphere is ${\displaystyle V_{sphere}={\frac {4}{3}}\pi r^{3}}$ and can be found in the Volume of a Sphere article. How does substituting this change the equation?
Hint 38
The equation becomes ${\displaystyle {\frac {4}{3}}\pi r^{3}={\frac {1}{3}}rA_{Sphere-Surface}}$. Almost there! Let's switch the order so surface area appears on the left side of the equals sign ${\displaystyle {\frac {1}{3}}rA_{Sphere-Surface}={\frac {4}{3}}\pi r^{3}}$. What else can be done to simplify?
Hint 39
You can isolate the surface area by multiplying both sides by ${\displaystyle 3}$ (to eliminate the ${\displaystyle {\frac {1}{3}}}$) and dividing both sides by ${\displaystyle r}$. What does that produce?
Hint 40
You get ${\displaystyle {\frac {3{\frac {1}{3}}rA_{Sphere-Surface}}{r}}={\frac {3{\frac {4}{3}}\pi r^{3}}{r}}}$. The cancellations leave you with ${\displaystyle A_{Sphere-Surface}=4\pi r^{2}}$. This is the equation for the surface area of a sphere.

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

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

One option would be to use Cavalieri's Principle to dodge that nasty sum. You could then smash the pyramids together into one big pyramid with a height equal to the radius and a base with an area equal to the surface area of the sphere and do the algebra. There are many options, but the base might look like a sinusoidal map projection, if certain choices were made. It would be a lot of geometric manipulation to keep track of and the animation would grind any machine without a discrete GPU to a halt, even with only 8 pyramids.

The other option is the same idea, but you get there by peeling spherical shells off of the sphere and mushing each layer into a shape similar to the sinusoidal map projection (or any other choice that respects the rules of Cavalieri's Principle), then stacking them to make a pyramid. The same reasons for not doing it apply (complexity and compute power).