Lesson 9-2: Prisms and Cylinders (Day 92)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

If the surface area of this sphere is 4cbrt(144pi) sq. units, what is its volume?

Let's find the radius first:

4pi r^2 = 4cbrt(144pi)
pi r^2 = cbrt(144pi)
r^2 = cbrt(144/pi^2)
r = cbrt(12/pi)

And now we can find the volume:

V = (4/3)pi r^3
   = (4/3)pi cbrt(12/pi)^3
   = (4/3)pi(12/pi)
   = 16

Thus the volume is 16 cubic units -- and of course, today's date is the sixteenth. This is another one of those measurement problems where the surface area must be some complicated irrational number involving the cube root of pi, just to make the volume come out to be a natural number, the date.

This is a Chapter 10 problem -- specifically Lessons 10-8 and 10-9 on spheres. So far we're still in Chapter 9, where we're just barely learning about two simpler space figures.

Lesson 9-2 of the U of Chicago text is called "Prisms and Cylinders." Our text refers to both prisms and cylinders as "cylindric surfaces."

This is what I wrote last year about today's lesson:

This lesson consists mainly of definitions. Terms defined in this lesson are surface, solid, box, rectangular solid, faces, opposite faces, edges, vertices, and skew lines -- and that's just the first page! Another term defined in this lesson is parallelepiped.

Let's get back to Euclid's Elements, since I've started discussing it yesterday. We can finally look at the first two definitions in Book XI:

https://mathcs.clarku.edu/~djoyce/java/elements/bookXI/bookXI.html

Definition 1.

A solid is that which has length, breadth, and depth.

Definition 2.

A face of a solid is a surface.

Both Euclid and the U of Chicago distinguish between the boundary and the interior.

Here are the definitions of the two surfaces mentioned in the lesson title:

Definition 13.

A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar, and parallel, while the rest are parallelograms.

Definition 21.

When a rectangular parallelogram with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder.

Definition 22.

The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned.

Definition 23.

And the bases are the circles described by the two sides opposite to one another which are carried round.

It's interesting to compare Euclid's definitions to the U of Chicago's. Euclid imagines a cylinder as a solid of revolution -- we take a rectangle and rotate it around one of its sides. (Don't forget that a rotation in 3D has an axis, not a center.) But in the U of Chicago text, we perform a very different isometry -- we begin with a circle and translate it out of the plane. David Joyce points out that Euclid's definition doesn't generalize -- it describes only right cylinders. Not only does the U of Chicago definition include oblique cylinders, but the idea of cylindric solids (or surfaces) extends to prisms as well.

By the way, Euclid defines a prism in terms of its faces. Since the lateral faces of Euclid's prisms are parallelograms (and not necessarily rectangles), oblique prisms are included.

Let's return to the theorems. Since we looked at Proposition 4 yesterday, let's try Proposition 5 today:

Proposition 5.

If a straight line is set up at right angles to three straight lines which meet one another at their common point of section, then the three straight lines lie in one plane.

Here's a modern version of the proof. Notice that Euclid writes "For suppose they do not...," which implies that this is an indirect proof.

Given: Line AB perpendicular to BC, BD, and BE.
Prove: Lines BC, BD, and BE are coplanar.

Indirect Proof:
Assume to the contrary that they aren't. By the Point-Line-Plane Postulate part f, any three points are coplanar, and so let P be the plane containing B, D, E, and let Q be the plane containing A, B, C.

By Point-Line-Plane part g, since planes P and Q intersect (in B), they intersect in a line. Let F be another point on this line. So lines AB, BC, and BF are coplanar as they all lie in plane Q.

We are given that AB is perpendicular to both BD and BE, and so by yesterday's Proposition 4, AB is perpendicular to the entire plane containing BD and BE, namely P. By the definition of a line perpendicular to a plane, AB is perpendicular to every line in P through B, which includes BF.

But we are given that AB is also perpendicular to BC. Therefore AB is perpendicular to two lines, BC and BF, both in plane Q. This is a contradiction, since through a point on a line, there is exactly one line in the plane perpendicular to the line. Thus the assumption that BC, BD, and BE aren't coplanar is false. Therefore BC, BD, and BE are coplanar. QED

This proof isn't as difficult as yesterday's, but it is an indirect proof -- and we don't cover indirect proofs in our text until Lesson 13-4. It also requires yesterday's Proposition 4 in order to prove -- and Proposition 4 has a difficult proof.

Here is the worksheet for today's Lesson 9-2: