Instructions

As the lecture progresses, complete the questions and fill in the blanks. At the end of the lecture, you will have 10 minutes for a discussion of the lecture materials.

Content

Worksheet Background on Archimedes

During today's lecture you will be learning about Archimedes, one of the greatest mathematical minds of all time. During the lecture we will cover Archimedes' method for finding the surface area and volume of a cylinder and sphere.

It took Archimedes his entire life to discover these formulas, and he was so proud of them that he had a picture of a sphere inscribed with a cylinder, and the formula for the volume of a sphere, carved into his headstone. In the 13th century, Archimedes' works were translated into Latin, opening them up to study by Europeans. The invention of the calculus, arguably the next significant advance in mathematical thinking, would not occur until nearly 2,000 years later.

Today we are going to cover 2,000 years of mathematics in 30 minutes.

Lecture

In the lecture, we start with a definition of Pi (C/D) and nothing else. How would we use what we know now about integration to derive the volume and surface area of a sphere?

PowerPoint for part 0 on Archimedes and who he was

Lecture for Parts 1/2/3

Part 0: Archimedes

Part 0: Who was Archimedes

Born in Syracuse, lived around 250 BC
Give me a lever and I can move the world
Density, water screw, fire ships
Spirals, curves, method of exhaustion (limit without the concept of infinity), logarithms
Integral of parabola
Volume of a sphere
Surface area of a sphere

Part 1: Area of a Circle

Part 1: How to obtain the area of a circle from the circumference of a circle (via integration)

Start with a circle, and no other information than the definition of pi: "pi is defined as the ratio of the circumference of the circle to the diameter of the circle:

$\pi ={\frac {C}{D}}={\frac {C}{2R}}$

To obtain an area, imagine a very thin circular strip, width dr, one side is length C(r) (circumference of r), other side is C(r+dr), circumference at r + dr. Sum up all these strips to get area approximation:

$A\approx \sum _{i=1}^{n}C(r)\Delta r$

Now use infinite strips:

$A=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}C(r^{\star })\Delta r=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}2\pi r^{\star }\Delta r$

This is the definition of the integral. The limits of integration are for circle radii from 0 to R:

$A=\int _{0}^{R}2\pi rdr$

which, on integrating, yields the area of a circle.

$A=2\pi \left({\frac {r^{2}}{2}}\right){\bigg |}_{0}^{R}$

or,

$A=\pi R^{2}$

Part 2: Volume of a Sphere

Part 2: How to obtain the volume of a sphere from the area of a circle (via disk method)

Disk Method Formula: $V=\int _{a}^{b}\pi f(x)^{2}dx$

The circle equation is given by $x^{2}+y^{2}=r^{2}$ or $y=\pm {\sqrt {r^{2}-x^{2}}}$ .

Limits of integration for x are -r to r, which make the disk method:

$V=\int _{-r}^{r}\pi {\sqrt {r^{2}-x^{2}}}^{2}dx$

$V=\pi \left(r^{2}x-{\frac {x^{3}}{3}}\right){\bigg |}_{-r}^{r}$

$V=2\pi \left(r^{3}-{\frac {r^{3}}{3}}\right)={\dfrac {4}{3}}\pi r^{3}$

This yields the formula for the volume of a sphere:

$V={\dfrac {4}{3}}\pi r^{3}$

Part 3: Surface Area of a Sphere

Part 3: How to obtain the surface area of a sphere from the volume of a sphere (via differentiation)

To find surface area, we can note that surface area is equal to the rate of change of volume as we shrink the radius:

$SA={\dfrac {dv}{dr}}$

Now, let's start with the equation for the volume of a sphere:

$V={\frac {4}{3}}\pi r^{3}$

Differentiate both sides. We'll find the change in volume that corresponds to a shell of thickness dr:

$dV={\frac {4}{3}}\pi 3r^{2}dr=4\pi r^{2}dr$

or,

${\frac {dV}{dr}}=4\pi r^{2}$

Questions

1) Where and when did Archimedes live?

2) What is the integral that relates the circumference of a circle to the area of a circle?

3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of $\Delta x$ and a cross-sectional area of $\pi R^{2}$ . Mr. Blue then computes the area of the sphere using the integral $\int _{0}^{R}{\pi R^{2}dx}$ , yielding the volume of a sphere of $V=\pi R^{3}$ . What conceptual error did Mr. Blue make?