Volume of a Sphere and a Hemisphere

Find the volume of a sphere with $V = \frac{4}{3}\pi r^{3}$ and of a hemisphere, find the radius from a volume with a cube root, and solve melting and recasting problems.

Before you start

The volume of a cylinder is $\pi r^{2} h$ .
The cube root undoes cubing: $\sqrt[3]{64} = 4$ because $4^{3} = 64$ . On a calculator, use the $\sqrt[3]{\phantom{x}}$ key.
$1000 \text{ cm}^{3} = 1$ litre.

Spheres and Hemispheres

A sphere is a perfectly round ball: every point of its surface is the same distance, the radius $r$ , from its centre $O$ . A straight line through the centre from one side to the other is a diameter, $2r$ long.

Cut a sphere in half through its centre and each half is a hemisphere: a curved surface and a flat circular face of radius $r$ .

Where the formula comes from. Over two thousand years ago Archimedes showed that a sphere fitting exactly inside a cylinder (the cylinder's radius $r$ and height $2r$ ) fills two thirds of it. The cylinder holds $\pi r^{2} \times 2r = 2\pi r^{3}$ , and two thirds of that is $\frac{4}{3}\pi r^{3}$ . This is his result, quoted here without its proof.

Volume of a sphere and of a hemisphere

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

Hemisphere: $V = \frac{1}{2} \times \frac{4}{3}\pi r^{3} = \frac{2}{3}\pi r^{3}$

The radius is cubed: a volume is three lengths multiplied.

Example 1: A sphere

Find the volume of a sphere of radius $6$ cm.

Solution

$V = \frac{4}{3} \times \pi \times 6^{3} = \frac{4}{3} \times 216\pi = 288\pi = 904.77\ldots = 905$ (3 s.f.)

The volume is $905 \text{ cm}^{3}$ .

Example 2: From the diameter, taking π = 3.142

A ball has a diameter of $10$ cm. Taking $\pi = 3.142$ , find its volume.

Solution

$r = 10 \div 2 = 5$ cm.

$V = \frac{4}{3} \times 3.142 \times 5^{3} = \frac{4}{3} \times 3.142 \times 125 = 523.66\ldots = 524$ (3 s.f.)

The volume is $524 \text{ cm}^{3}$ .

Example 3: In terms of π

Find the volume of a sphere of radius $3$ cm, leaving your answer in terms of $\pi$ .

Solution

$V = \frac{4}{3} \times \pi \times 27 = 36\pi$

$36\pi \text{ cm}^{3}$

Example 4: A hemisphere

Find the volume of a solid hemisphere of radius $9$ cm.

Solution

$V = \frac{2}{3} \times \pi \times 9^{3} = \frac{2}{3} \times 729\pi = 486\pi = 1526.8\ldots = 1530$ (3 s.f.)

The volume is $1530 \text{ cm}^{3}$ .

Example 5: A hemispherical bowl

A mixing bowl is a hemisphere with an internal radius of $12$ cm. How many litres does it hold when full to the brim?

Solution

Use the internal radius: the bowl holds the space inside it.

$V = \frac{2}{3} \times \pi \times 12^{3} = 1152\pi = 3619.1\ldots \text{ cm}^{3}$

$3619.1\ldots \div 1000 = 3.619\ldots = 3.62$ litres (3 s.f.)

It holds $3.62$ litres.

Example 6: A cone, a hemisphere and a cylinder

A cone, a hemisphere and a cylinder each have a radius of $6$ cm, and the cone and the cylinder each have a height of $6$ cm, the same as the radius. Find each volume in terms of $\pi$ , and compare them.

Solution

Cone: $\frac{1}{3} \times \pi \times 6^{2} \times 6 = 72\pi\text{.}$ Hemisphere: $\frac{2}{3} \times \pi \times 6^{3} = 144\pi\text{.}$ Cylinder: $\pi \times 6^{2} \times 6 = 216\pi$ .

Solid	Volume ( $r = 6$ )
Cone	$\frac{1}{3}\pi r^{3} = 72\pi$
Hemisphere	$\frac{2}{3}\pi r^{3} = 144\pi$
Cylinder	$\pi r^{3} = 216\pi$

The volumes are in the ratio $72 : 144 : 216 = 1 : 2 : 3\text{,}$ for any radius: the hemisphere holds twice the cone and two thirds of the cylinder.

$72\pi$ , $144\pi$ and $216\pi \text{ cm}^{3}$ , in the ratio $1 : 2 : 3$ .

Example 7: A ball in a box

A ball of radius $5$ cm fits exactly into a cube-shaped box of edge $10$ cm. (a) What volume of the box is empty? (b) What percentage of the box does the ball fill?

Solution

Box: $10^{3} = 1000 \text{ cm}^{3}$ . Ball: $\frac{4}{3}\pi \times 5^{3} = 523.59\ldots \text{ cm}^{3}\text{.}$

(a) $1000 - 523.59\ldots = 476.40\ldots = 476$ (3 s.f.)

(b) $\frac{523.59\ldots}{1000} \times 100\% = 52.35\ldots\% = 52.4\%$ (3 s.f.)

(a) $476 \text{ cm}^{3}$ (b) about $52.4\%$

Finding the Radius from the Volume

Put the volume into the formula and make $r^{3}$ the subject; then take the cube root, not the square root. Rearranged, a sphere has $r = \sqrt[3]{\frac{3V}{4\pi}}$ .

Example 8: A sphere

A sphere has a volume of $500 \text{ cm}^{3}$ . Find its radius.

Solution

$\frac{4}{3}\pi r^{3} = 500$ , so $4\pi r^{3} = 1500$ .

$r^{3} = \frac{1500}{4\pi} = 119.36\ldots$

$r = \sqrt[3]{119.36\ldots} = 4.923\ldots = 4.92$ (3 s.f.)

Check: $4.92^{3} = 119.1$ , close to $119.36$ , as it should be.

The radius is $4.92$ cm.

Example 9: A hemisphere

A hemispherical dome has a volume of $2000 \text{ m}^{3}$ . Find its radius.

Solution

$\frac{2}{3}\pi r^{3} = 2000$ , so $2\pi r^{3} = 6000$ .

$r^{3} = \frac{6000}{2\pi} = 954.92\ldots$

$r = \sqrt[3]{954.92\ldots} = 9.847\ldots = 9.85$ (3 s.f.)

The radius is $9.85$ m.

Melting and Recasting

When a solid is melted down and made into new solids, none of the material is lost: the total volume stays the same. Write the old volume equal to the new, then solve.

Example 10: Many small spheres

A solid metal sphere of radius $6$ cm is melted down and recast into small spheres of radius $2$ cm. How many small spheres are made?

Solution

Large sphere: $\frac{4}{3}\pi \times 6^{3} = 288\pi \text{ cm}^{3}\text{.}$ Small sphere: $\frac{4}{3}\pi \times 2^{3} = \frac{32}{3}\pi \text{ cm}^{3}\text{.}$

Number $= 288\pi \div \frac{32}{3}\pi = 288 \times \frac{3}{32} = 27$

The radius is $3$ times as long, and $3^{3} = 27$ : the $\pi$ and the $\frac{4}{3}$ cancel.

$27$ small spheres

Example 11: From a cylinder to a sphere

A cylinder of modelling clay has a radius of $3$ cm and a height of $16$ cm. It is rolled into one sphere. Find the radius of the sphere.

Solution

Cylinder: $\pi \times 3^{2} \times 16 = 144\pi \text{ cm}^{3}\text{.}$

$\frac{4}{3}\pi r^{3} = 144\pi$ . Divide by $\pi$ : $\frac{4}{3} r^{3} = 144$ , so $r^{3} = 108$ .

$r = \sqrt[3]{108} = 4.762\ldots = 4.76$ (3 s.f.)

The radius is $4.76$ cm.

Example 12: A rising water level

A cylindrical jar of radius $6$ cm holds water. A solid metal ball of radius $3$ cm is lowered in and is completely covered. By how much does the water level rise?

Solution

The ball pushes up water equal to its own volume: $\frac{4}{3}\pi \times 3^{3} = 36\pi \text{ cm}^{3}\text{.}$

That water forms a layer, a cylinder of radius $6$ cm: rise $\times \pi \times 6^{2} = 36\pi$ , so $36\pi \times$ rise $= 36\pi$ .

Rise $= 1$

The water rises by $1$ cm.

Common Mistakes

Squaring instead of cubing. $\frac{4}{3}\pi r^{2}$ is not a volume: the radius is cubed.
Using the diameter. A diameter of $10$ cm used as $r$ gives $8$ times the volume.
The square root for the cube root. From $r^{3} = 108$ , $r = \sqrt[3]{108} = 4.76\text{,}$ not $\sqrt{108}$ .
Forgetting to halve for a hemisphere: $\frac{2}{3}\pi r^{3}$ , not $\frac{4}{3}\pi r^{3}$ .
Rounding $r^{3}$ before taking the cube root.

Practice Questions

Try each question before you open its solution. Use the calculator's $\pi$ unless the question says otherwise, and give answers correct to 3 significant figures where they are not exact.

Question 1

Find the volume of a sphere of radius $4$ cm.

ShowHide solution

$\frac{4}{3} \times \pi \times 64 = 268.08\ldots = 268$ (3 s.f.)

$268 \text{ cm}^{3}$

Question 2

Find the volume of a sphere of diameter $14$ cm.

ShowHide solution

$r = 7$ cm. $\frac{4}{3} \times \pi \times 343 = 1436.7\ldots = 1440$ (3 s.f.)

$1440 \text{ cm}^{3}$

Question 3

Find the volume of a solid hemisphere of radius $5$ cm, in terms of $\pi$ .

ShowHide solution

$\frac{2}{3} \times \pi \times 5^{3} = \frac{2}{3} \times \pi \times 125 = \frac{250}{3}\pi$

$\frac{250}{3}\pi \text{ cm}^{3}$

Question 4

Taking $\pi = 3.142$ , find the volume of a hemisphere of radius $10$ cm.

ShowHide solution

$\frac{2}{3} \times 3.142 \times 1000 = 2094.66\ldots = 2090$ (3 s.f.)

$2090 \text{ cm}^{3}$

Question 5

A football has a diameter of $22$ cm. Find the volume of air inside it, taking its thickness as negligible.

ShowHide solution

$r = 11$ cm. $\frac{4}{3} \times \pi \times 1331 = 5575.2\ldots = 5580$ (3 s.f.)

$5580 \text{ cm}^{3}$

Question 6

A sphere has a volume of $972\pi \text{ cm}^{3}$ . Find its radius.

ShowHide solution

$\frac{4}{3}\pi r^{3} = 972\pi$ , so $r^{3} = 972 \times \frac{3}{4} = 729$ and $r = \sqrt[3]{729} = 9$ .

$9$ cm

Question 7

A spherical flask holds exactly $1$ litre. Find its internal radius.

ShowHide solution

$\frac{4}{3}\pi r^{3} = 1000$ , so $r^{3} = \frac{3000}{4\pi} = 238.73\ldots$

$r = \sqrt[3]{238.73\ldots} = 6.203\ldots = 6.20$ (3 s.f.)

$6.20$ cm

Question 8

A hemispherical pot has an internal radius of $15$ cm. Can it hold $7$ litres of soup?

ShowHide solution

$\frac{2}{3} \times \pi \times 15^{3} = 2250\pi = 7068.5\ldots \text{ cm}^{3} = 7.068\ldots$ litres

$7.07$ litres is more than $7$ litres.

Yes: it holds $7.07$ litres.

Question 9

A solid metal cube of edge $10$ cm is melted down and made into ball bearings, each a sphere of radius $1.5$ cm. How many whole ball bearings can be made?

ShowHide solution

Cube: $10^{3} = 1000 \text{ cm}^{3}$ . One ball: $\frac{4}{3}\pi \times 1.5^{3} = 4.5\pi = 14.137\ldots \text{ cm}^{3}\text{.}$

$1000 \div 14.137\ldots = 70.7\ldots\text{,}$ so $70$ whole ball bearings (some metal is left over).

$70$

Question 10

A cylindrical glass of radius $4$ cm holds some water. Twenty marbles, each a sphere of radius $0.5$ cm, are dropped in and sink below the surface. By how much does the water level rise?

ShowHide solution

Twenty marbles: $20 \times \frac{4}{3}\pi \times 0.5^{3} = \frac{10}{3}\pi \text{ cm}^{3}\text{.}$ The water rises by this volume.

Rise $\times \pi \times 4^{2} = \frac{10}{3}\pi\text{,}$ so rise $= \frac{10}{3} \div 16 = 0.2083\ldots = 0.208$ (3 s.f.)

$0.208$ cm

Question 11

A hemispherical bowl holds $5$ litres when full. Find its internal radius.

ShowHide solution

$5$ litres $= 5000 \text{ cm}^{3}$ .

$\frac{2}{3}\pi r^{3} = 5000$ , so $r^{3} = \frac{3 \times 5000}{2\pi} = \frac{7500}{\pi} = 2387.3\ldots$

$r = \sqrt[3]{2387.3\ldots} = 13.36\ldots = 13.4$ (3 s.f.)

$13.4$ cm

Question 12

A student finds the volume of a sphere of diameter $9$ cm as $\frac{4}{3}\pi \times 9^{3} = 3054 \text{ cm}^{3}\text{.}$ Explain the mistake and find the correct volume.

ShowHide solution

The student used the diameter as the radius. The radius is $9 \div 2 = 4.5$ cm.

$\frac{4}{3}\pi \times 4.5^{3} = \frac{4}{3}\pi \times 91.125 = 381.7\ldots = 382$ (3 s.f.)

The wrong answer is $2^{3} = 8$ times too big.

The diameter was used as the radius; the volume is $382 \text{ cm}^{3}$ .

Key Points

Volume of a sphere $= \frac{4}{3}\pi r^{3}$ ; of a hemisphere $= \frac{2}{3}\pi r^{3}$ .
A sphere fills two thirds of the cylinder it fits in exactly (Archimedes).
To find the radius, make $r^{3}$ the subject and take the cube root.
Melting and recasting keeps the total volume the same.
Use the radius, not the diameter, and round only at the end.