Spheres, composite solids and capacity · Solid mensuration

Spheres and hemispheres

Sphere

V = \frac43\pi r^3,\ S = 4\pi r^2

Hemisphere

V = \frac23\pi r^3

A hemisphere’s curved surface is half a sphere’s, $2\pi r^2$ ; a solid hemisphere also has its flat circle, making $3\pi r^2$ in all.

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WAEC 2023 · Paper 1 · Q18The radius of a sphere is $3\text{ cm}$ . Find, in terms of $\pi$ , its volume.
WAEC 2020 · Paper 1 · Q38The diameter of a sphere is $12\text{ cm}$ . Calculate, correct to the nearest $\text{cm}^3$ , the volume of the sphere. $\left[\text{Take }\pi = \frac{22}{7}\right]$
JAMB 1993 · UME · Q36Find the radius of a sphere whose surface area is $154\text{ cm}^2$ . $\left[\pi = \frac{22}{7}\right]$
JAMB 2016 · UTME · Q15Find the radius of a sphere whose surface area is $154\text{ cm}^2$ . $\left[\text{Take }\pi = \frac{22}{7}\right]$
NECO 2024 · Paper 1 · Q46The volume of a spherical ball is $114\text{ cm}^3$ . Find its radius to the nearest whole number.
NECO 2022 · Paper 1 · Q48A plastic spherical ball of radius 6 cm was melted and reshaped into a cube. Find the length of the cube, correct to the …
NECO 2022 · Paper 2 · Q8If $x$ and $y$ are the surface area and volume of a sphere respectively and $\frac{x}{y} = \frac27$ , find the values of $x$ …
JAMB 2002 · UME · Q16A solid hemisphere has radius 7 cm. Find its total surface area. $\left[\pi = \frac{22}{7}\right]$
WAEC 2009 · Paper 2 · Q12Three positive numbers are in arithmetic progression (A.P.). The sum of the squares of the three numbers is 155, while the …
WAEC 2012 · Paper 2 · Q13The internal diameter of a spherical bowl, half full of water, is $20\text{ cm}$ . The content is poured into an empty cylindrical …
JAMB 1985 · UME · Q39A solid sphere of radius 4 cm has a mass of 64 kg. What will be the mass of a shell of the same metal whose internal and …

Calculus questions use these formulas too: the rate at which a sphere’s volume grows with its radius, or the volume made by spinning a region round an axis (see calculus).

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Composite solids

For a solid made of two shapes, such as a cone on a hemisphere:

volume: add the volumes;
surface area: add only the surfaces you can see. The flat circle where the two shapes join is inside, so leave it out.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q7 (a)

The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is $48\text{ m}$ and the base radius is $14\text{ m}$ . Calculate, correct to three significant figures, the surface area of the structure. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Which surfaces show?
No. The outside is the cone’s curved surface and the hemisphere’s curved surface.
Think first. Is the circle where the cone meets the hemisphere on the outside?
The slant height
$l = \sqrt{48^2 + 14^2} = \sqrt{2500} = 50$ m.
Add the two surfaces
Cone: $\pi r l = \frac{22}{7} \times 14 \times 50 = 2200\text{ m}^2$ . Hemisphere: $2\pi r^2 = 2 \times \frac{22}{7} \times 196 = 1232\text{ m}^2$ .
Total $= 3432\text{ m}^2$ , which is $3430\text{ m}^2$ to three significant figures.

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A bucket: the frustum

A bucket is a frustum, a cone with its tip cut off. Call the two radii $R$ and $r$ and the height $h$ .

Slant height. Drop a line straight down from the edge of the small circle. It makes a right-angled triangle with sides $h$ and $R - r$ , so $l^2 = h^2 + (R - r)^2$ .
Volume. The whole cone minus the small cone that was removed. Find the full height $H$ first, by similar triangles: $\frac{H}{R} = \frac{H - h}{r}$ .

A frustumPut the tip back to find H; the slant height comes from the small triangle

For $R = 5$ cm, $r = 2$ cm and $h = 4$ cm: the slant height is $\sqrt{16 + 9} = 5$ cm. For the volume, $\frac{H}{5} = \frac{H - 4}{2}$ gives $2H = 5H - 20$ , so $H = \frac{20}{3}$ and the removed tip is $\frac{8}{3}$ cm tall:

\begin{aligned} V &= \tfrac13\pi \times 5^2 \times \tfrac{20}{3} - \tfrac13\pi \times 2^2 \times \tfrac83 \\ &= \tfrac13\pi \times \tfrac{500 - 32}{3} = 52\pi\text{ cm}^3 \end{aligned}

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Similar solids

If every length of one solid is $k$ times the matching length of another, then its areas are $k^2$ times as big and its volumes $k^3$ times as big. For two spheres with surface areas in the ratio $4 : 25$ , the radii are in the ratio $2 : 5$ and the volumes $8 : 125$ .

Similar solidsLengths × k, areas × k², volumes × k³

Solids that are not similar still compare through their formulas. Two cylinders with radii in the ratio $2 : 1$ and heights in the ratio $1 : 3$ have volumes in the ratio $2^2 \times 1 : 1^2 \times 3 = 4 : 3$ .

More: comparing solids

Capacity: litres and cubic metres

1000\text{ cm}^3 = 1\text{ litre}, \qquad 1\text{ m}^3 = 1000\text{ litres}

When liquid is poured from one container to another, or drawn off, the volume of liquid stays the same. The level changes by (volume) ÷ (area of the base).

How far does the level drop?Change the tank and the amount drawn off

36,000litres drawn off, 3 × 12,00036in m³ (÷ 1000)28.29base area πr², r = 3 m1.27 mdrop in level = volume ÷ base area

tank diameter = 6 mtankers = 3each tanker holds = 12,000 litres

First change the litres to cubic metres: 1 m³ = 1000 litres, so 36,000 litres = 36 m³. The liquid that leaves is a cylinder of the tank's radius and height equal to the drop, so πr² × drop = 36, and drop = 36 ÷ 28.29 = 1.27 m.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q10 (a)

A cylindrical fuel storage tank of diameter $7\text{ m}$ is full of gasoline. If four tankers, each of capacity $13\,500$ litres, draw gasoline from the storage tank, calculate, correct to two decimal places, the height reduction, in metres, of gasoline in the tank. $\left[\text{Take }\pi = \frac{22}{7}\right]$

The volume removed
$4 \times 13\,500 = 54\,000$ litres $= 54\text{ m}^3$ .
Think first. How many litres do the four tankers take? How many m³ is that?
The tank's base
$r = 3.5$ m, so the base is $\frac{22}{7} \times 3.5^2 = 38.5\text{ m}^2$ .
Think first. The diameter is 7 m. What is the area of the base?
The drop in level
$38.5 \times \text{drop} = 54$ , so the drop is $\frac{54}{38.5} \approx 1.40$ m.

Your turn

WAEC 2011 · Paper 2 · Q10 (a)

(a)
The total surface areas of two spheres are in the ratio $9 : 49$ . If the radius of the smaller sphere is $12\text{ cm}$ , find, correct to the nearest $\text{cm}^3$ , the volume of the bigger sphere. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Try it on a graph

X at the origin, Y 3 km west, then 5 km north-west to Z.

Worked solution (try it first)

(a)

Surface areas are in the ratio of the squares of the radii: $\frac{4\pi(12)^2}{4\pi R^2} = \frac{9}{49}$ .
So $\frac{12^2}{R^2} = \frac{9}{49}$ , and taking square roots, $\frac{12}{R} = \frac37$ .
So $R = 28\text{ cm}$ .
Volume of the bigger sphere $= \frac43\pi R^3$
$= \frac43 \times \frac{22}{7} \times 28^3$
$= \frac43 \times \frac{22}{7} \times 21\,952$
$\approx 91\,989\text{ cm}^3$ .

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