A cone, a sphere and a cylinder: 1, 2, 3
Tonight's idea
Three volume formulas, for a radius \(r\) and height \(h\):
\[V_{\text{cylinder}} = \pi r^2 h, \qquad V_{\text{cone}} = \tfrac13 \pi r^2 h, \qquad V_{\text{sphere}} = \tfrac43 \pi r^3.\]
Tonight we use them rather than prove them (grade 10's solids sheet shows where the \(\tfrac13\) and \(\tfrac43\) come from). Take a sphere of radius \(r\) and the tightest tin around it: a cylinder of radius \(r\) and height \(2r\). Stand a cone in the same tin, radius \(r\) and height \(2r\), point at the lid. Three solids, one radius, one height.
Put \(h = 2r\) into the formulas:
\[V_{\text{cone}} = \tfrac13 \pi r^2 (2r) = \tfrac23 \pi r^3, \qquad V_{\text{sphere}} = \tfrac43 \pi r^3, \qquad V_{\text{cylinder}} = \pi r^2 (2r) = \tfrac63 \pi r^3.\]
Cone : sphere : cylinder \(= \tfrac23 : \tfrac43 : \tfrac63 = 1 : 2 : 3\). The sphere fills two thirds of its tin, and cone plus sphere fill it exactly.
Archimedes proved the sphere result around 250 BC, without algebra, and prized it above his other work: he asked for a sphere in a cylinder, with their ratio 2 : 3, to be carved on his tomb, and Cicero later claimed to have found the tomb by that carving.
Watch one: \(r = 3\) cm
Then \(h = 6\) cm.
\[V_{\text{cone}} = \tfrac13 \pi \cdot 9 \cdot 6 = 18\pi, \qquad V_{\text{sphere}} = \tfrac43 \pi \cdot 27 = 36\pi, \qquad V_{\text{cylinder}} = \pi \cdot 9 \cdot 6 = 54\pi.\]
In cubic centimetres, about 56.5, 113.1 and 169.6. The ratio \(18 : 36 : 54\) is \(1 : 2 : 3\). Leave \(\pi\) in: the ratio is exact, and decimals hide it.
A tin of three tennis balls is one third air, whatever the radius: the balls have volume \(3 \times \tfrac43 \pi r^3 = 4\pi r^3\) and the tin, of height \(6r\), has \(\pi r^2 \cdot 6r = 6\pi r^3\).
Your turn
- Take \(r = 5\) cm and \(h = 10\) cm. Compute the three volumes, leaving \(\pi\) in, and check that they are in the ratio \(1 : 2 : 3\).
- A cone has radius 4 and height 8. (a) Without computing, how does its volume compare with a sphere of radius 4 and a cylinder of radius 4, height 8? (b) Compute all three to one decimal place.
- A cylindrical glass of radius 6 cm and height 12 cm is full of water. A steel ball of radius 6 cm is lowered to the bottom; it just fits. What fraction of the water is left, and how many cm³?
- (a) Halve everything: a cone of radius \(r\) and height \(r\), a hemisphere of radius \(r\), a cylinder of radius \(r\) and height \(r\). Show the volumes are again \(1 : 2 : 3\). (b) Keep the sphere but make the cone and cylinder height \(3r\). Find the new ratio cone : sphere : cylinder.
Stretch: the ice cream
A wafer cone has radius 3 cm and height 6 cm. It is filled with ice cream and topped with a hemisphere scoop of radius 3 cm. Show that the total ice cream equals the volume of a single sphere of radius 3 cm. Then explain why, using only the \(1 : 2 : 3\) and no arithmetic.
Answers
- \(V_{\text{cone}} = \tfrac13 \pi \cdot 25 \cdot 10 = \tfrac{250}{3}\pi\), \(V_{\text{sphere}} = \tfrac43 \pi \cdot 125 = \tfrac{500}{3}\pi\), \(V_{\text{cylinder}} = \pi \cdot 25 \cdot 10 = 250\pi = \tfrac{750}{3}\pi\). Ratio \(250 : 500 : 750 = 1 : 2 : 3\).
- (a) The height is \(2r\), so the sphere is twice the cone and the cylinder three times it. (b) \(V_{\text{cone}} = \tfrac13 \pi \cdot 16 \cdot 8 = \tfrac{128}{3}\pi \approx 134.0\); sphere \(\tfrac{256}{3}\pi \approx 268.1\); cylinder \(128\pi \approx 402.1\).
- The ball displaces its own volume, \(\tfrac43 \pi \cdot 216 = 288\pi\) cm³, which spills over. The glass holds \(\pi \cdot 36 \cdot 12 = 432\pi\) cm³, so \(144\pi \approx 452\) cm³ remains, exactly one third. The ball is two thirds of the tin; the water left is the other third, whatever the radius.
- (a) Cone \(\tfrac13 \pi r^2 \cdot r = \tfrac13 \pi r^3\); hemisphere \(\tfrac12 \cdot \tfrac43 \pi r^3 = \tfrac23 \pi r^3\); cylinder \(\pi r^3 = \tfrac33 \pi r^3\). Again \(1 : 2 : 3\), each solid being the top (or bottom) half of before. (b) Cone \(\tfrac13 \pi r^2 \cdot 3r = \pi r^3\), sphere \(\tfrac43 \pi r^3\), cylinder \(3\pi r^3\). Multiply by 3: \(3 : 4 : 9\).
Stretch. Cone: \(\tfrac13 \pi \cdot 9 \cdot 6 = 18\pi\). Hemisphere: \(\tfrac12 \cdot \tfrac43 \pi \cdot 27 = 18\pi\). Total \(36\pi\) cm³, and a sphere of radius 3 is \(\tfrac43 \pi \cdot 27 = 36\pi\). Without arithmetic: the cone has height \(2r\), so by \(1 : 2 : 3\) it is half the sphere of radius \(r\); the hemisphere is the other half. Half plus half is a whole sphere. If it melts, it will not fit back in the cone: the cone holds only half.