Engaging students: Finding the volume and surface area of spheres

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Cameron Story. His topic, from Geometry: finding the volume and surface area of spheres..

How could you as a teacher create an activity or project that involves this topic?

As a geometry teacher, manipulatives and visuals are important for conceptual understanding. Rather than handing out a formula sheet, it is far more rewarding to have your students derive volume and surface area formulas for themselves using some kind of physical representation. Not only is this more engaging for students, but the concepts behind the formula are emphasized. Yes, the volume of a sphere is $V = \frac{4}{3} \pi r^3$ , but why? Where does the fraction come from? These are important questions.

An example of an activity that could be useful when teaching the volume of a sphere is best shown by Megan Millan in the following YouTube Video:

Here, students fill up hollow solids with water and find ratios between the volumes of several different shapes.

Assuming students already know the formulas for cones and cylinders, it would make it much easier to visualize those volumes with water. Through pure experimentation, students conclude that the volume inside of a cone (whose height is twice the radius) plus the volume of a sphere is equal to the total volume of a cylinder equal height and radius.

From the student’s own experimentation (and some specifically sized manipulatives), the formula is found instead of given.

How has this topic appeared in the news?

An interesting news story by the Daily Galaxy reports that Saturn’s moon Titan has a methane cycle analogous to the water cycle on Earth; Titan has methane rain, methane clouds, and methane lakes. Ligeia Mare, Titan’s second largest methane lake, “occupies roughly the same surface area as Earth’s Lake Huron and Lake Michigan together,” (The Daily Galaxy, 2018). This news story is exciting as it hits on possible life outside earth, one that may even live in these liquid-methane lakes. As a math teacher, we can follow up this story with the following visual, illustrating the size of Earth compared the size of Titan. If these lakes are the same size, what fraction of the total surface area is the lake on Earth compared to the lake on Titan?

This can lead into how surface area changes as spheres grow or shrink. It also leads to some curiosity in the student. For example, what would Texas look like on Titan?

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

The Greek mathematician Archimedes discovered many things about solids and their properties long before calculus, and this is perfect for students in geometry; they can’t use calculus yet either. Archimedes is known for many mathematical discoveries, but in particular he is famous for finding that “…the volume of a sphere with radius r is two-thirds that of the cylinder in which it is inscribed,” (Toomer, 2018). This fact leads directly to the standard formula for the volume of a sphere: $V = \frac{4}{3}\pi r^3$ . Supposedly, Archimedes was proud enough of this discovery to “leave instructions for his tomb to be marked with a sphere inscribed in a cylinder,” (Toomer, 2018).

What I like about this bit of history is that your students can discover this formula on their own with some support from the teacher. The great mathematician Archimedes found the same formula and found it so important that he had it be inscribed in his final resting place, so your students will have a sense of pride knowing that they overcame the same challenge that only the best mathematicians from 2,000 years ago could tackle.

References:

YouTube video by Megan Millan – “Cylinder, Cone, and Sphere Volume” https://www.youtube.com/watch?v=RZkhnIzBC_k

Toomer, Gerald J. “Archimedes.” Encyclopedia Britannica, Encyclopedia Britannica, Inc., 28 Mar. 2018, www.britannica.com/biography/Archimedes#ref=ref383380&tocpanel=sectionId~toc214869,tocId~toc214869.

“Cassini’s Final Encounter with Saturn’s Giant Moon Titan –‘Like the Early Earth.’” The Daily Galaxy, The Daily Galaxy, 14 Sept. 2018, dailygalaxy.com/2018/09/cassinis-final-encounter-with-saturns-giant-moon-titan-like-the-early-earth/.

Engaging students: Finding the volume and surface area of spheres

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Austin DeLoach. His topic, from Geometry: finding the volume and surface area of spheres..

What interesting word problems using this topic can your students do now?

I found an interesting word problem that has to do with finding the size and density of Pluto using satellite images and data at https://spacemath.gsfc.nasa.gov/Geometry/6Page143.pdf that would be a good way for students to practice finding the volume of a sphere among other things. This problem could not be used at the very beginning of the section, but it is definitely interesting and could be very engaging for some students. There are multiple parts to the problem, but the third part has students calculate the volume of Pluto using the scale of measurement that they discovered in an earlier part. Students would then use their calculated volume to determine the density of the planet and compare it to other common things by using the given mass of the planet. Not only is this practice for the students to be able to calculate volume of spheres, but it helps them by showing further applications and how their calculated volume can be used to make more scientific discoveries. Problems like this are very good for students to see so that they can recognize real-world application for what they are learning in school, even if it is simplified for the sake of the class.

How could you as a teacher create an activity or project that involves your topic?

One idea that I think is interesting and engaging for students is taking an orange, measuring the diameter, then seeing how many circles of the same diameter the removed orange peel can fit in. There is a short demonstration video at https://www.youtube.com/watch?v=FB-acn7d0zU to see what I mean. This is a good activity because it is very hands-on for students to be actively engaged, and it also helps students recognize that mathematical formulas are not just thrown together, but there is reasoning behind all of them. This will also help the students remember the formula for the surface area of a sphere, as they will be able to think back to this activity and remember the time that they discovered the formula on their own. There is potential to be messy with this activity, but because it is such a memorable activity and will genuinely engage the students and let their curiosity about mathematics come to life, it is worth it if you can set aside the time for clean-up afterwards.

How has this topic appeared in pop culture?

A big place for volume and surface area of spheres to come up in pop culture is in sports. One recent situation can be seen at http://www.espn.com/espn/wire/_/section/ncw/id/18605942 where the Charleston women’s basketball team had to forfeit two victories because their basketballs for those games were not regulation size. The team accidentally used NCAA men’s basketballs (which have a circumference of 29.5-30 inches) instead of the standard women’s basketballs (circumference of 28.5-29 inches). Because the balls were not regulation size, the victories did not count. Students could use the given circumferences to find the surface area and volumes of each ball and see how significant the difference is, then discuss with their peers what the significance of different sized basketballs is. Although this is not an advanced practice idea, it is still a way for students to compute volume and surface area, as well as discover the significance of each of those properties in a way that could interest them, as many students are interested in sports and do not often think of math as playing a significant role in them. Computing the volume and surface area of the basketballs would also help them recognize the relationship between those and circumference.

Lessons from teaching gifted elementary students (Part 7b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprising depth of mathematical knowledge.

Here’s a question I once received after a really big hailstorm:

How big would a 1000-pound hailstone be?

The guesses from the students ranged from the size of a small car to the size of a large pick-up truck. Here’s how I gave a reasonably accurate answer using only mental arithmetic. This kind of describes the way that I try to size up things when only an approximation (and not an exact answer) is necessary.

First, I had some real-world experience that quickly told me the answer was going to be deceptively small. When I was young — maybe 10 or 12 years old — I was getting ready for a picnic, and I was assigned cut a block of ice — maybe a cubic foot of ice, if memory serves — into smaller chunks. (The party organizer bought a block of ice instead of a bag of ice to economize.) I remembered how incredibly heavy that block of ice was even though it wasn’t much larger than a basketball… several of us kids had a lot of trouble lifting the block of ice as we prepared to chop it into pieces. So, for the sake of argument, if that cubic foot of ice weighed about 100 pounds, then 8 cubic feet would weigh 800 pounds. So, based on that chance encounter with a block of ice when I was a kid, my guess would have been that the hailstone would measure 2 feet across.

Back to the problem at hand.

First, I converted to metric. I knew that there are about 2.2 pounds in a kilogram, and so I knew that the block would weigh something like 400 or 450 kilograms. I knew that I would be making plenty of crude approximations, so I just went with 400 kilograms and didn’t worry too much about immediately calculating $1000/2.2$ .

Next, I knew that metric units were originally defined so that a cubic centimeter of water weighs a gram, so that a 10 cm-by-10 cm-by 10 cm cube of water weighs one kilogram. Ice (hail) is slightly less dense than water (after all, ice floats in water), but for crude approximation purposes, I ignored this.

So, if a cube of ice with a side length of 1 decimeter (10 cm) weighs 1 kilogram, then a cube of ice with a side length of $\sqrt[3]{400}$ decimeters would weigh about 400 kilograms.

How big is $\sqrt[3]{400}$ ? Well, I have memorized that $7^3 = 343$ and $8^3 = 512$ , so it’s between 7 and 8 someplace… say 7.5. So the answer would be a cube of side length 7.5 decimeters. Also, I have memorized that 1 decimeter (10 cm) is approximately 4 inches, so the cube would have side length $7.5 \times 4 = 30$ inches.

Finally, hailstones are more spherical in shape than cubic, and a sphere of diameter $d$ has less volume than a cube of side length $d$ . So the answer should be a bit larger than 30 inches, so I just rounded up to a nice even number: 36 inches (one yard).

This calculation took me about a minute to do in my head and another half-minute to re-do to make sure I didn’t botch the arithmetic. So I held my hands about a yard apart (perhaps the crudest part of this calculation), pretending to hold a ball of diameter a yard across, and announced, “The hailstone would be about this big.”

Of course, a more thoughtful analysis produces the actual answer. The density of ice at the freezing point is 0.9167 grams per cubic centimeter, and 1 pound converts to 0.453592 kilograms. So:

$\displaystyle \frac{4\pi}{3} r^3 \times 0.9167 \frac{\hbox{grams}}{\hbox{cm}^3} = 1000 \hbox{pounds} \times \displaystyle 0.453592 \frac{\hbox{kilograms}}{\hbox{pounds}} \times 1000 \frac{\hbox{grams}}{\hbox{kilograms}}$

$r^3 \approx 118127 \hbox{cm}^3$

$r \approx 49.1 \hbox{cm}$

$r \approx 49.1 \hbox{cm} \times \frac{1 \hbox{inch}}{2.54 \hbox{cm}}$

$r \approx 19.3 \hbox{inches}$

Therefore, the sphere would have a diameter of twice that, or 38.6 inches.

Lessons from teaching gifted elementary school students (Part 7a)

Here’s a question I once received after a really big hailstorm:

How big would a 1000-pound hailstone be?

My head hurts thinking about hail that large. After about a minute of thinking, without using a calculator or even a pencil, I gave my answer: about a yard across.

I’ll reveal how I got this answer — which turns out to be a lot close than I had any right to expect — in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own without using a calculator.

Sphere Packing Solved in Higher Dimensions

I enjoyed reading this bit of mathematical news: https://www.quantamagazine.org/20160330-sphere-packing-solved-in-higher-dimensions/

The opening paragraphs:

In a pair of papers posted online this month, a Ukrainian mathematician has solved two high-dimensional versions of the centuries-old “sphere packing” problem. In dimensions eight and 24 (the latter dimension in collaboration with other researchers), she has proved that two highly symmetrical arrangements pack spheres together in the densest possible way.

Mathematicians have been studying sphere packings since at least 1611, when Johannes Kepler conjectured that the densest way to pack together equal-sized spheres in space is the familiar pyramidal piling of oranges seen in grocery stores. Despite the problem’s seeming simplicity, it was not settled until 1998, when Thomas Hales, now of the University of Pittsburgh, finally proved Kepler’s conjecture in 250 pages of mathematical arguments combined with mammoth computer calculations.

Engaging students: Finding the volume and surface area of spheres

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Avery Fortenberry. His topic, from Algebra: finding the volume and surface area of spheres.

How does this topic extend what your students should have learned in their previous courses?

The topic of volume and surface area of spheres is building upon the students’ knowledge of area and circumference of a circle. A sphere is similar to a circle in that a circle is a closed shape with all points equidistant from the centerpoint (the distance is the radius) and a sphere is a closed object with all points at an equal distance from its centerpoint (the distance is also r). Students will be familiar with the area of a circle formula, which is A=πr² and will be able to easily use and understand the formula for volume of a sphere V=(4/3)πr³. The same is true for circumference in relation with surface area.

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

Archimedes was the first mathematician to discover the most important ratio in all of mathematics, π. He did this by finding the area of a circle using shapes that were incrementally closer and closer to the same size as that circle. In other words, he would start with a circle and enclose it within a square, then a pentagon, then a hexagon, and so on until he came extremely close to the same shape. He used this same method to find the volume of a sphere by enclosing it within a cylinder of a known volume and cutting out piece by piece and measuring until he found the parabolic segment is 4/3 that of an inscribed triangle.

Source: http://www.storyofmathematics.com/hellenistic_archimedes.html

What are the contributions of various cultures to this topic?

This topic had many cultures contribute to the understanding of it. These contributions came from Greek, Chinese, and Arabic mathematicians. The Greek contribution came mainly from Archimedes, which I discussed in D1. The Arithmetic Art in Nine Chapters is a Chinese book written in the 1^st century that gave a formula that was close but not exact to finding the volume of a sphere. The author of the book calculated pi as being equal to 25/8 or even as just 3 at times. Ancient Arabic mathematicians submitted very similar ideas to the Chinese in terms of the volume of a sphere. While it is known the Chinese derived some ideas from the Greeks, it is still unclear today how the ideas were spread to the Arabic mathematicians.

Source: http://www.muslimheritage.com/article/volume-sphere-arabic-mathematics-historical-and-analytical-survey#sec2.2

Surface area

Source: http://www.xkcd.com/1389/

Area of a Triangle and Volume of Common Shapes: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on different ways of finding the area of a triangle as well as finding the volumes of common shapes.

Part 1: Deriving the formula $A = \displaystyle \frac{1}{2} bh$ .

Part 2: Cavalieri’s principle and finding areas using calculus.

Part 3: Cavalieri’s principle and finding the volume of a pyramid and then the volume of a sphere.

Part 4: Finding the area of a triangle using the Law of Sines.

Part 5: Finding the area of a triangle using the Law of Cosines.

Part 6: Finding the area of a triangle using the triangle’s incenter.

Part 7: Finding the area of a triangle using a determinant and the coordinates of the vertices.

Part 8: Finding the area of a triangle using Pick’s theorem.

Lessons from teaching gifted elementary school students (Part 3b)

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B \times B = C$

$C \times C \times C \times C= D$

If the pattern goes on, and if $A = 2$ , what is $Z$ ?

In yesterday’s post, we found that the answer was

$Z =2^{26!} = 10^{26! \log_{10} 2} \approx 10^{1.214 \times 10^{26}}$ ,

a number with approximately $1.214 \times 10^{26}$ digits.

How can we express this number in scientific notation? We need to actually compute the integer and decimal parts of $26! \log_{10} 2$ , and most calculators are not capable of making this computation.

Fortunately, Mathematica is able to do this. We find that

$Z \approx 10^{121,402,826,794,262,735,225,162,069.4418253767}$

$\approx 10^{0.4418253767} \times 10^{121,402,826,794,262,735,225,162,069}$

$\approx 2.765829324 \times 10^{121,402,826,794,262,735,225,162,069}$

Here’s the Mathematica syntax to justify this calculation. In Mathematica, $\hbox{Log}$ means natural logarithm:

Again, just how big is this number? As discussed yesterday, it would take about 12.14 quadrillion sheets of paper to print out all of the digits of this number, assuming that $Z$ was printed in a microscopic font that uses 100,000 characters per line and 100,000 lines per page. Since 250 sheets of paper is about an inch thick, the volume of the 12.14 quadrillion sheets of paper would be

$1.214 \times 10^{16} \times 8.5 \times 11 \times \displaystyle \frac{1}{250} \hbox{in}^3 \approx 1.129 \times 10^{17} \hbox{in}^3$

By comparison, assuming that the Earth is a sphere with radius 4000 miles, the surface area of the world is

$4 \pi (4000 \times 5280 \times 12) \hbox{in}^2 \approx 8.072 \times 10^{17} \hbox{in}^2$ .

Dividing, all of this paper would cover the entire world with a layer of paper about $0.14$ inches thick, or about 35 sheets deep. In other words, the whole planet would look something like the top of my desk.

What if we didn’t want to print out the answer but just store the answer in a computer’s memory? When written in binary, the number $2^{26!}$ requires…

$26!$ bits of memory, or…

about $4.03 \times 10^{26}$ bits of memory, or…

about $latex 5.04 \times 10^{25} bytes of memory, or …

about $5.04 \times 10^{13}$ terabytes of memory, or…

about 50.4 trillion terabytes of memory.

Suppose that this information is stored on 3-terabyte external hard drives, so that about $50.4/3 = 16.8$ trillion of them are required. The factory specs say that each hard drive measures $129 \hbox{mm} \times 42 \hbox{mm} \times 167 \hbox{mm}$ . So the total volume of the hard drives would be $1.52 \times 10^{19} \hbox{mm}^3$ , or $15.2 \hbox{km}^3$ .

By way of comparison, the most voluminous building in the world, the Boeing Everett Factory (used for making airplanes), has a volume of only $0.0133 \hbox{km}^3$ . So it would take about 1136 of these buildings to hold all of the necessary hard drives.

The cost of all of these hard drives, at $100 each, would be about $1.680 quadrillion. So it’d be considerably cheaper to print this out on paper, which would be about one-seventh the price at $242 trillion.

Of course, a lot of this storage space would be quite repetitive since $2^{26!}$ , in binary, would be a one followed by $26!$ zeroes.

Sphere

Source: http://www.xkcd.com/1248/