A sphere and cone fit tightly into two identical cylinders. Show that when the volume of the cone is added to the volume of the sphere, the result is equal to the volume of the cylinder.
2r cm r cm
r cm
2r cm TRY THIS
FOCUS ON WORKING MATHEMATICALLY F
OCUS ON WORKING MATHEMATICALLY
T
HE SURFACE AREA OF A SOCCER BALLIntroduction
Measuring the area of a curved surface of a sphere is an intriguing mathematical problem.
Is it related to the area of a circle? Is there an exact formula? Archimedes (287–212 BC), one of the greatest mathematicians of antiquity, answered yes to both questions with brilliant mathematical reasoning. It was just a small part of his geometrical discoveries about the volumes and surface areas of spheres, cones and cylinders. Interestingly they were discovered from the study of mechanics, and he used original ideas which were developed into the concepts of calculus centuries later by Newton and Leibnitz.
In this activity we will measure the area of a soccer ball, assumed to be a sphere, to test a hypothesis about the formula. The Challenge activity will enable you to explore how Archimedes solved it, and his most famous theorem.
L
E A R N I N G A C T I V I T Y 1Practical measurement
Plan for a double period. The first activity may take a single 50 min period and is best carried out in groups. Each group will need a soccer ball, suitable grid paper, scissors, calculator, tape and callipers. If soccer balls are not available and time is limited go straight to question 4.
1 Examine the shapes which cover the surface. Does it surprise you that the surface can be covered in this way? Is this a tessellation?
F
O C U S O NW
O R K I N GM
A T H E M A T I CA L L Y0 F
O C U S O N W 0 R K I N G M A T H E M A T I C A L L Y2
FOCUS ON WORKING MATHEMATICALLY
F
OCUS ON WORKING MATHEMATICALLY 2 Decide how you will count and measure the area of each basic shape to get an estimate of
the area of the ball. Decide on a unit of measurement, say square centimetres.
3 Draw up a table to show your results. Change the table if you have different shapes. Record the number of each shape on the surface of the ball and multiply to get the total area.
Surface area of ball = cm2
4 A quicker alternative is to wrap a rectangular piece of fairly stiff paper around a suitable ball so that it just fits. The width of the paper is cut so that it is equal to the diameter of the ball. Unwrap the paper and calculate the rectangular area. This will approximate the surface area of the ball.
L
E A R N I N G A C T I V I T Y 2Reasoning and hypothesising
The second activity is to make a hypothesis about the formula and to test it against the calculated value of the surface area of the ball.
1 Firstly, consider the cross sectional area of the ball through the centre. What shape do you get? Does it seem reasonable that the surface area is some how connected to the area of this circle? Why?
2 Calculate the area of the great circle whose centre is the centre of the sphere. Note that it is easier to measure the diameter of the ball than the radius.
3 Compare the area of the great circle with the calculated surface area of the ball. How many times greater is it, roughly?
4 If your measurements were accurate you should have found that the surface area is close to 4 times the area of the circular cross section whose centre is the centre of the sphere. Make a hypothesis about the formula for the surface area of a sphere in terms of its radius r.
Record this in your book and draw a diagram to illustrate it.
Shape Area Number Total area (cm2)
Pentagon Hexagon
paper width
= diameter of ball
perimeter
diameter
2
FOCUS ON WORKING MATHEMATICALLY F
OCUS ON WORKING MATHEMATICALLY
C
H A L L E N G E A C T I V I T I E S 1 The big question now is how did Archimedes prove thatthe surface area of a sphere of radius r is exactly 4πr2? It came as a consequence of his discovery of the volume of a sphere. Copy the figure shown into your book and write down the formula for the volume V of the whole sphere.
2 The idea is to imagine the volume of the sphere as made up of small polyhedra with base area ∆A. A typical one is shown in the diagram. The volume of this ‘element of volume’
is approximately (area of the base x perpendicular height).
As the subdivision is made smaller and smaller, the perpendicular height h gets closer to the radius r, and the sum of all the elements approaches the volume of the sphere. Furthermore the sum of the base areas is the surface area A of the sphere.
Using the symbol ∑ to mean the sum of the elements of volume, in the limit, as the subdivision is made finer and finer, h approaches r, and ∑( × ∆A × h) gets closer and closer to the volume of the sphere πr3.
That is × A × r = πr3, and so A= 4πr2.
3 Now copy this figure into your book.
The cylinder is solid and the base and top are tangential to the sphere.
4 Using known formulae, prove that the volume of a sphere is the volume the cylinder which encloses it. Prove that surface area of a sphere is also of the cylinder which encloses it. That the ratio of volumes and the ratio of surface areas were identical was considered by Archimedes as his most profound theorem.
8
P
Q
O r
r
S R
∆A
O
P Q
S R r h r
r 1
3
---1 3 ---4
3 ---1
3--- 4
3
---2 3
---2 3