Wednesday, April 1, 2015

Section 10-8: Volume of a Sphere (Day 139)

Today I decided to quit my job as a substitute teacher. Rather than continue to pursue my goal to become a math teacher, I will join the military instead. But I will continue to post to the Common Core Geometry blog anyway, whenever I have the chance, since I will bring the U of Chicago text with me. My hope, of course, is that I won't be in the middle of posting to the blog when I am captured...


Yes, today is April Fool's Day. Nothing in the first paragraph of this post is actually true. In fact, today I subbed mainly for an Earth Science class, but during the conference period (6th) I ended up in a Precalculus class. The class was beginning Chapter 7, on Linear Systems and Matrices. with Section 7.1 on Solving Systems of (Two Linear) Equations. Even though the text directed the students to use substitution to solve them, the teacher asked the students to solve them by graphing regardless of directions in the book. But unfortunately, the text included several whose y-intercepts are not integers, such as the system:

x - y = -4
x + 2y = 5

The first equation is easy to solve, y = x + 4, but the second gives y = -1/2 * x + 5/2. The y-intercept of this line is 5/2, which isn't easy to locate on a graph -- especially considering that it must be graphed accurately enough to locate (-1, 3) as the solution. Naturally, some of the students wanted to forget the graph and solve it algebraically, but I had to adhere to the teacher's lesson plan. The text directs the students to check the solution by graphing -- but considering that a picture of a graphing calculator window accompanies the problem, the text is expecting the students just to plug the solutions into a TI-83 or TI-84 to obtain the graph.

In the Earth Science classes, students were to watch a video on earthquakes. I decided to pull an April Fool's Day prank on the students -- instead of saying "Today there will be an earthquake video," I said, "Today there will be an earthquake drill." Many students were fooled and were expecting to evacuate the classroom. The only thing that would have made this prank better was if there had been a sudden announcement on the intercom right at that moment -- even if the announcement had nothing to do with earthquakes, a student might have jumped under the desk anyway! (I know, this may be confusing my readers outside of California who don't know about earthquake drills.)

The rest of this post will contain some serious math, but first, let me prove that mathematicians really do have a sense of humor.

I've mentioned the website Metamath before. This site gives many proofs in two-column format -- and most of these aren't geometry proofs at all. Here's a link to one page of this website:

As of today -- the list is always changing -- look at some of the proofs that appear on this page:

-- Fundamental Theorem of Algebra (often mentioned in Algebra II classes)
-- The Basel Problem (The series 1 + 1/4 + 1/9 + 1/16 + ... + 1/n^2 + ... adds up to pi^2 / 6.)
-- Bertrand's "Postulate" (There's a prime between n and 2n for every natural n. This is called a "postulate," but it's actually proved here in the link, so it's a theorem.)
-- Ostrowski's Theorem (a complicated theorem about absolute value)
-- various definitions from ring theory (But no proofs appear until the following page.)

The first three of these are easy to state -- even in a high school class -- but difficult to prove. The proofs that appear on the Metamath page are either covered in upper-division math or not at all.

But in between all of these theorems are some April Fool's Day proofs. Two of the proofs involve complex math symbols that end up spelling out phrases. The first spells out "April Fool" -- this was first posted to the Metamath website ten years ago:

And the other one spells out "Hello World." Computer scientists are more likely to get this joke, as the traditional first program that one learns is how to print "Hello World" on the screen:

Last year's April Fool's Day entry is a proof that division by zero is forbidden:

The actual statement that is proved is "(1 / 0) = 0" -- that is, 1 divided 0 is the empty set. Here I use a struck-through 0 to denote the empty set, but the line should be diagonal, not horizontal. When I was in that seventh grade classroom that used the McDougal Littell text -- although that text itself directs students to write "undefined" -- the teacher told the students to write the symbol for empty set, which she pronounced "o-slash." Before learning about o-slash, common errors made by the students include giving the answer to 1 / 0 as 1 or 0 -- that is, they added, subtracted, or multiplied the 1 and 0 when they were asked to divide it. After learning about o-slash, a common error is to state that the answer to 0 / 1 is also o-slash instead of 0.

How does Metamath "prove" that 1 / 0 is the empty set, anyway? We notice that the third line reads:

dom / = (C x (C \ {0}))

Here "dom" is short for "domain." So this line states that the domain of the division function is the set of complex numbers for the dividend and the set of complex numbers except 0 for the divisor. To put it simply, we prove that division by 0 is impossible because we defined division to exclude 0 from its domain (hence the reason that 1 / 0 is "undefined") -- which isn't very enlightening at all. If we ask the TI-83 to find the tangent of 90 degrees, it likewise returns a vague "domain" error, whereas 1 / 0 gives a more descriptive "divide by 0" error.

But then again, this is just an April Fool's Day proof. Here's a more common proof to illuminate the reasons that division by 0 is undefined:

This problem is a classic "proof" that 1 = 2. In this proof, we are given that a = b in Step 1, and in Step 8, we divide by a - b (actually given as a^2 - ab, but factoring gives a - b). As a = b, a - b must be zero. So we actually divided by 0 -- and this is why division by zero  must be undefined, since otherwise we could prove that 1 = 2. I once tutored a geometry student (not my current student) whose teacher assigned the classic "proof" of 1 = 2 and the student was asked to find the error. It took a while, but I think the student did eventually figure out that a - b = 0, so that they were dividing by 0.

Let's conclude the April Fool's Day part of this post by giving a link to some math jokes:

Here are just a few of the jokes from this site:

    Top ten excuses for not doing homework:
  • I accidentally divided by zero and my paper burst into flames.
  • Isaac Newton's birthday.
  • I could only get arbitrarily close to my textbook. I couldn't actually reach it.
  • I have the proof, but there isn't room to write it in this margin.
  • I was watching the World Series and got tied up trying to prove that it converged.
  • I have a solar powered calculator and it was cloudy.
  • I locked the paper in my trunk but a four-dimensional dog got in and ate it.
  • I couldn't figure out whether i am the square of negative one or i is the square root of negative one.
    • Warning! It is against the rule to use these excuses in my classes! A. Ch.

Note: I left out the last three excuses because they refer to college-level math. Notice that Isaac Newton was born on Christmas Day -- a day when schools are already closed. The joke about the proof that can't fit in the margin refers to Fermat's Last Theorem. I told the story about Fermat's proof and the margin on this blog back during the first week in September. We're currently learning about three-dimensional figures in geometry -- not four-dimensional objects. A four-dimensional object is supposed to be powerful enough to get inside of a three-dimensional locked object.

    Two male mathematicians are in a bar. The first one says to the second that the average person knows very little about basic mathematics. The second one disagrees, and claims that most people can cope with a reasonable amount of math.
     The first mathematician goes off to the washroom, and in his absence the second calls over the waitress. He tells her that in a few minutes, after his friend has returned, he will call her over and ask her a question. All she has to do is answer one third x cubed.
     She repeats "one thir -- dex cue"?
    He repeats "one third x cubed".
    Her: `one thir dex cuebd'? Yes, that's right, he says. So she agrees, and goes off mumbling to herself, "one thir dex cuebd...".
     The first guy returns and the second proposes a bet to prove his point, that most people do know something about basic math. He says he will ask the blonde waitress an integral, and the first laughingly agrees. The second man calls over the waitress and asks "what is the integral of x squared?".
    The waitress says "one third x cubed" and while walking away, turns back and says over her shoulder "plus a constant!" 

This is a calculus joke. I mention it only because the "1/3" that shows up in this calculus problem is also the source of the "1/3" that appears in the volume of a pyramid or cone -- if we were to use calculus to derive the formula.

Just above these two, we have:

    The Evolution of Math Teaching
  • 1960s: A peasant sells a bag of potatoes for $10. His costs amount to 4/5 of his selling price. What is his profit?
  • 1970s: A farmer sells a bag of potatoes for $10. His costs amount to 4/5 of his selling price, that is, $8. What is his profit?
  • 1970s (new math): A farmer exchanges a set P of potatoes with set M of money. The cardinality of the set M is equal to 10, and each element of M is worth $1. Draw ten big dots representing the elements of M. The set C of production costs is composed of two big dots less than the set M. Represent C as a subset of M and give the answer to the question: What is the cardinality of the set of profits?
  • 1980s: A farmer sells a bag of potatoes for $10. His production costs are $8, and his profit is $2. Underline the word "potatoes" and discuss with your classmates.
  • 1990s: A farmer sells a bag of potatoes for $10. His or her production costs are 0.80 of his or her revenue. On your calculator, graph revenue vs. costs. Run the POTATO program to determine the profit. Discuss the result with students in your group. Write a brief essay that analyzes this example in the real world of economics.

This joke was written back in 2000. (What was I saying earlier about the Precalculus text that expected students to graph lines on the calculator again?) But some of the comments listed under "1990's" often appear in criticisms of Common Core. The fact that this joke is dated 1990's emphasizes the point that the traditionalists are actually attacking the progressive philosophy of teaching. So it predates Common Core -- but of course, the Common Core is an extension of the progressive philosophy.

An engineer, a physicist and a mathematician are staying in a hotel. 
The engineer wakes up and smells smoke. He goes out into the hallway and sees a fire, so he fills a trash can from his room with water and douses the fire. He goes back to bed. 
Later, the physicist wakes up and smells smoke. He opens his door and sees a fire in the hallway. He walks down the hall to a fire hose and after calculating the flame velocity, distance, water pressure, trajectory, etc. extinguishes the fire with the minimum amount of water and energy needed. 
Later, the mathematician wakes up and smells smoke. He goes to the hall, sees the fire and then the fire hose. He thinks for a moment and then exclaims, "Ah, a solution exists!" and then goes back to bed. 

This joke refers to the fact that many proofs in mathematics are existence proofs -- they don't actually tell how to find any solutions. We don't deal with this that much in Geometry -- but in Algebra II, the aforementioned Fundamental Theorem of Algebra tells us that an n-degree equation has n solutions, but it doesn't tell us how to find any of these n solutions -- as opposed to something like the Quadratic Formula, which does tell us the solutions of the equation.

    A physicist and a mathematician are sitting in a faculty lounge. Suddenly, the coffee machine catches on fire. The physicist grabs a bucket and leap towards the sink, filled the bucket with water and puts out the fire. Second day, the same two sit in the same lounge. Again, the coffee machine catches on fire. This time, the mathematician stands up, got a bucket, hands the bucket to the physicist, thus reducing the problem to a previously solved one.
      Another version:
    A mathematician and an engineer are on desert island. They find two palm trees with one coconut each. The engineer climbs up one tree, gets the coconut, eats. The mathematician climbs up the other tree, gets the coconut, climbs the other tree and puts it there. "Now we've reduced it to a problem we know how to solve." 

Mathematicians are also fond of reducing new problems to previously solved old problems. Chapter 11 on Coordinate Proofs reduces many Geometry problems to Algebra I problems. And right here in Chapter 10, we reduced the problem of a cone's volume to that of a pyramid, which in turn is reduced to that of a prism, which in turn is reduced to that of a box.

A mathematician, a physicist, and an engineer are all given identical rubber balls and told to find the volume. They are given anything they want to measure it, and have all the time they need. The mathematician pulls out a measuring tape and records the circumference. He then divides by two times pi to get the radius, cubes that, multiplies by pi again, and then multiplies by four-thirds and thereby calculates the volume. The physicist gets a bucket of water, places 1.00000 gallons of water in the bucket, drops in the ball, and measures the displacement to six significant figures. And the engineer? He writes down the serial number of the ball, and looks it up.

I mention this joke only because it gives the volume of a sphere, which is today's topic. Section 10-8 of the U of Chicago text derives the volume of a sphere. It is the only figure whose volume is given before its surface area.

The McDougal Littell text, in Section 10.7, demonstrates the sphere volume formula the same way that it does the cone volume formula. We take a cone whose height and radius are both equal to the radius of the sphere, and we find out how many conefuls of sand fill the sphere. The text states that two conefuls make up a hemisphere, and so four conefuls make up the entire sphere.

But of course, we want to derive the formula more rigorously. Recall that Dr. David Joyce states that a limiting argument is the best that can be done at this level -- but I disagree. Dr. Franklin Mason, meanwhile, enthusiastically gives another derivation of the sphere volume formula, and Dr. M's proof also appears in the U of Chicago text. Recall that Dr. M considers this day on which the sphere volume formula -- Section 12.6 of his text -- is derived to be one of the three best days of the year. So it's fitting that we give the formula on a day already associated with fun -- April Fool's Day.

The U of Chicago text mentions that this proof uses Cavalieri's Principle. But it was hardly the mathematician Cavalieri who first proved the sphere volume formula. Indeed, according to Dr. M, this proof goes all the way back to Archimedes -- the ancient Greek mathematician who lived a few years after Euclid. (It's possible that their lives overlapped slightly.)

Here is a Square One TV video about Archimedes:

We mentioned earlier that Archimedes used polygons to determine the value of pi (also known as Archimedes' constant) -- hence the line in the song, "He was busy calculating pi." He was also famous for using the principle of buoyancy (also known as Archimedes' principle) to determine whether the king's gold crown was a fake -- and this is also mentioned in the song. Legend has it that the Greek mathematician was so excited when he discovered his principle -- he had been in a public bath at the time -- that he ran down the streets naked and shouted out "Eureka!" to announce his discovery. The Greek word eureka, meaning "I have found," is the motto of my home state of California. The last joke about about how a physicist measures volume refers to Archimedes' principle.

But Archimedes himself actually considered the discovery of the sphere volume formula to be his crowning achievement -- to the extent that he requested it to be engraved on his tombstone. So let's finally derive that formula the way that Archimedes did over 2000 years ago. And no, he didn't simply drop a ball into water to determine the formula. Archimedes' sphere formula has nothing to do with Archimedes' principle of buoyancy.

We begin by considering three figures -- a cone, a cylinder, and a sphere. We will use the known volumes of the cone and cylinder to determine the unknown volume of the sphere -- thereby reducing the problem to a previously solved one.

Our cylinder will have the same radius as the sphere, while the height of the cylinder will equal the diameter (i.e., twice the radius) of the sphere. This way, the sphere will fit exactly in the cylinder.

Our cone, just like the cone mentioned in McDougal Littell, will have the its height and radius both equal to the radius of the sphere. Such a cone could fit exactly in a hemisphere. But we want there to be two cones, so that their combined height is the same as that of the cylinder. We set up the cones so that they have a common vertex (i.e., they are barely touching each other) and each base of a cone is also a base of the cylinder. The two touching cones are often referred to as a "double cone" -- Dr. M uses the term "bicone." (A bicone is also used to justify to Algebra II students why a hyperbola is a conic section with two branches. A hyperbola is the intersection of a bicone and a plane, such that the plane touches both cones.)

The focus is on the volume between the cylinder and the bicone. The surprising fact is that this volume is exactly equal to the volume of the sphere! Here is the proof as given by the U of Chicago:

"...the purple sections are the plane sections resulting from a plane slicing these figures in their middles. These purple sections are congruent circles with area pi * r^2. At h units above each purple section is a section shaded in pink. In the sphere, by the Pythagorean Theorem, the pink section is a small circle with radius sqrt(r^2 - h^2). The area of this section is found using the familiar formula for the area of a circle.

"Area(small circle) = pi * sqrt(r^2 - h^2) = pi(r^2 - h^2)

"For the region between the cylinder and the cones, the section is the pink ring between circles of radius r and h. (The radius of that circle is h because the acute angle measures 45 degrees, so an isosceles triangle is formed.)

"Area(ring) = pi * r^2 - pi * h^2 = pi(r^2 - h^2)

"Thus the pink circles have equal area. Since this works for any height h, Cavalieri's Principle can be applied. This means that the volume of the sphere is the difference in the volume of the cylinder (B * 2r) and the volume of the two cones (each with volume 1/3 * B * r).

"Volume of sphere = (B * 2r) - 2 * (1/3 * B * r)
                            = 2Br - 2/3 * Br
                            = 4/3 * Br

"But here the bases of the cones and cylinder are circles with radius r. So B = pi * r^2. Substituting,

"Volume of sphere = 4/3 * pi * r^2 * r
                            = 4/3 * pi * r^3." QED

The sphere volume is indeed the crowning achievement of Chapter 10. We began the chapter with the volume of a box and end up with the volume of the least box-like figure of all, the sphere. The start of Section 10-8 summarizes how we did this:

"It began with a postulate in Lesson 10-3 (volume of a box). Cavalieri's Principle was then applied and the following formula was deduced in Lesson 10-5 (volume of a prism or cylinder). A prism can be split into 3 pyramids with congruent heights and bases. Using Cavalieri's Principle again, a formula was derived in Lesson 10-7 (volume of a pyramid or cone). In this lesson, still another application of Cavalieri's Principle results in a formula for the volume of a sphere."

So take that, Dr. Katharine Beals! After all, she was the one who derided Cavalieri's Principle as progressive fluff that the Common Core tests on instead of actual math. But without Cavalieri's Principle, we'd be stuck finding the volumes of only boxes and their unions. Well, I suppose if we simply declared the volume formulas by fiat (i.e., as postulates) rather than actually deriving them, then Cavalieri's Principle is not needed. But if we want to prove them, then the Principle gives us an elegant proof of the sphere volume formula that was discovered over 2000 years before there ever was a Common Core -- a proof that, if mastered, should permit one to date a mathematician's daughter (as Beals mentioned on her website regarding the Quadratic Formula proof).

Sadly, we don't know whether Archimedes ever dated anyone's daughter, or whether he ever had daughters of his own. His life ended tragically -- indeed, my April Fool's joke at the start of this post, describing my capture by an enemy army actually refers to Archimedes. Legend has it that he was busy working on a geometry problem when the Roman army captured him. His last words before he was killed are said to be, "Noli turbare circulos meos" -- Latin for "Do not disturb my circles!"

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