Cylinder and Sphere

An edge is where two surfaces meet. On a cylinder, the curved surface meets each circle along a round rim, so a cylinder has two curved edges, but no straight edge and no vertex.

A cylinder is a solid with two congruent, parallel circles for its ends, joined by one curved surface. Take a circle and slide it straight up without turning it, and the space it sweeps through is a cylinder: a pile of identical circles, just as a prism is a pile of identical polygons. The line joining the centers of the two circles is its axis; in the cylinders of this unit it stands at right angles to the circles (a right circular cylinder). Its size is set by two lengths: the radius r of its circles and its height h, the distance between them.

A tin of milk powder, a gas cylinder, a drum, a pipe, a round pencil and a log of wood are all cylinders, or very nearly.

Surface area of the cylinder

So the curved surface of a cylinder, opened out, is a rectangle $2\pi r$ long and h wide:

$$\text{Curved surface area (CSA)} = 2\pi r \times h = 2\pi rh$$

Add the two circles at the ends, which are congruent, each $\pi r^2$, and you have the whole surface:

$$\begin{aligned} \text{Total surface area (TSA)} &= 2\pi rh + 2\pi r^2 \\ &= 2\pi r(r + h) \end{aligned}$$

The two circles together, $2\pi r^2$, are called the plane surface area of the cylinder; one circle, $\pi r^2$, is the area of its base.

Not every cylinder you make has both ends. A pipe has neither (CSA only); a glass, a bucket or a pencil holder has a bottom but no top (CSA + $\pi r^2$); a closed tin or tank has both (TSA). Always ask which surfaces are really there.

Volume of the cylinder

$$\text{Volume of a cylinder } V = \pi r^2 h = A \times h$$

where $A = \pi r^2$ is the area of the base. It is the same rule as for a prism, base area × height, because a cylinder too is a pile of identical slices.

Question 1

The radius of a cylinder is x cm and its height y cm. Find:

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The radius of a cylinder is r units and its height h units. Find:

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Find the area of the plane surfaces, the curved surface area and the volume of these cylinders:

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Find the area of the base, the area of the plane surfaces, the curved surface area, the total surface area and the volume of each cylinder.

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The circumference of a cylinder’s base is 88 cm, and its radius and height add up to 24 cm. Find:

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A circle is a flat figure: every point on it is the same distance, the radius, from a fixed point, its center, and you can draw it in your copy. A sphere is the same idea in space: a solid whose surface is made of every point at the same distance r from a center. Balls, marbles and bubbles are spheres; so, very nearly, are the Earth and the Moon. Any line through the center from one side to the other is a diameter, d = 2r, and a cut through the center shows the great circle (the large circle), the biggest circle on the sphere, with radius r: its circumference is $2\pi r$ and its area $\pi r^2$.

So the surface of a sphere is the area of four of its great circles:

$$\text{Surface area of a sphere } A = \pi r^2 + \pi r^2 + \pi r^2 + \pi r^2 = 4\pi r^2$$

A sphere has only the one, curved, surface, so this is its whole surface (its total surface area).

Cut a sphere through its center and you get two hemispheres. Each has two surfaces: a curved surface, half the sphere’s, and a plane surface, the great circle where it was cut. Its radius is the sphere’s.

$$\begin{aligned} \text{Curved surface area (CSA)} &= \tfrac{1}{2} \times 4\pi r^2 = 2\pi r^2 \\ \text{Plane surface area} &= \pi r^2 \\ \text{Total surface area (TSA)} &= 2\pi r^2 + \pi r^2 = 3\pi r^2 \end{aligned}$$

As with a cylinder, ask which surfaces are really there: an open bowl is only the curved surface, $2\pi r^2$; a solid half-ball, or a dome with its floor, is $3\pi r^2$.

The cylinder round the ball has radius r and height 2r, so its volume is $\pi r^2 \times 2r = 2\pi r^3$. The ball fills two thirds of it:

$$\begin{aligned} \text{Volume of a sphere } V &= \tfrac{2}{3} \times \text{volume of the cylinder} \\ &= \tfrac{2}{3} \times \pi r^2 \times 2r \\ &= \tfrac{4}{3}\pi r^3 \end{aligned}$$

$$\text{Volume of a hemisphere } V = \tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi r^3$$

Unlike its surface, the volume really is just half. (The total surface of a hemisphere, $3\pi r^2$, is more than half the sphere’s $4\pi r^2$, because cutting makes a new flat face.)

A family wants to store 1,000 liters, 1 m³, and the tank has a lid. The cube, 1 m every way, needs 6 m² of plastic. Now a round tank. Try one just as tall as the cube, 1 m high: it must have $\pi r^2 \times 1 = 1$, so $r^2 = \frac{7}{22}$, $r \approx 0.564$ m, about 1.13 m across. Its plastic:

$$2\pi r^2 + 2\pi rh = 2 \times 1 + 2 \times \tfrac{22}{7} \times 0.564 \times 1 \approx 2 + 3.55 = 5.55$$

About 5.55 m², about 0.45 m² (7.6%) less than the cube, for exactly the same water. Can you do better still by making it wider or narrower?

The best closed round tank is exactly as tall as it is wide: about 1.08 m across and 1.08 m high, with about 5.54 m² of plastic. Wider or narrower, it needs more; and near the best, the plastic changes very little (1.00 m across and 1.27 m high needs about 5.57 m²), so a tank maker can make it a little taller, to fit through a door or onto a small stand, at almost no cost. (Why exactly “as tall as wide” is best, you will be able to prove with calculus in higher grades.)

And the answer to the last question: yes, one shape does better. A ball holding 1 m³ has $\frac{4}{3}\pi r^3 = 1$, so $r \approx 0.62$ m, and a surface of only $4\pi r^2 \approx$ 4.84 m². Of all shapes, the sphere holds the most for the least surface. But a ball rolls, needs a frame to stand on and is hard to make and fit on a roof, so for water the round tank, nearly as good, wins. Where every bit of material counts, as in the gas spheres above, engineers build the sphere.

Question 1

The radius of a sphere is x units. Find:

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The diameter of a sphere is d cm. Find:

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Find the surface area and volume of each sphere or hemisphere.

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Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Mensuration is 13 marks in three questions; the grid’s model cylinder question is 4 marks on an oil can:

Part	Level	Marks	What it asks
a.	knowledge	1	the formula for a cylinder’s curved surface area
b.	application	3	how many liters of oil the can holds, from facts about its surface

Here is one in that shape, with how it is marked.

Question 1

Find the area of each triangle.

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Find the total surface area and volume of each prism.

Question 10

Find the cost of painting each solid all over at Rs 180 per cm².