Surface Area of a Sphere and a Hemisphere

Find the surface area of a sphere, $4\pi r^{2}$ , the curved surface area of a hemisphere, $2\pi r^{2}$ , and the total surface area of a solid hemisphere, $3\pi r^{2}$ , and work back from an area to the radius.

Before you start

A circle of radius $r$ has area $\pi r^{2}$ and circumference $2\pi r$ .
A hemisphere is half a sphere, cut through its centre, with a flat circular face of radius $r$ (Volume of a Sphere and a Hemisphere).

The Surface Area of a Sphere

A sphere cannot be opened out flat into a net, as a pyramid or a cone can. Its area was found by Archimedes, who showed that the surface of a sphere is exactly as large as the curved surface of the cylinder it fits in: that cylinder has radius $r$ and height $2r$ , so its curved surface is $2\pi r \times 2r = 4\pi r^{2}$ .

You can see it with an orange: the peel of an orange of radius $r$ , torn into small pieces, just covers four circles of radius $r$ . This shows the result; it does not prove it.

Surface area of a sphere

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

The radius is squared: an area is two lengths multiplied. (The volume, $\frac{4}{3}\pi r^{3}$ , has $r$ cubed.)

Example 1: A sphere

Find the surface area of a sphere of radius $5$ cm.

Solution

$4 \times \pi \times 5^{2} = 100\pi = 314.15\ldots = 314$ (3 s.f.)

The surface area is $314 \text{ cm}^{2}$ .

Example 2: From the diameter

A ball has a diameter of $12$ cm. Find its surface area.

Solution

$r = 12 \div 2 = 6$ cm.

$4 \times \pi \times 6^{2} = 144\pi = 452.38\ldots = 452$ (3 s.f.)

The surface area is $452 \text{ cm}^{2}$ .

Example 3: Taking π = 3.142, and in terms of π

(a) Taking $\pi = 3.142$ , find the surface area of a sphere of radius $7$ cm. (b) Find the surface area of a sphere of radius $2.5$ cm in terms of $\pi$ .

Solution

(a) $4 \times 3.142 \times 7^{2} = 4 \times 3.142 \times 49 = 615.832 = 616$ (3 s.f.)

(b) $4 \times \pi \times 2.5^{2} = 4 \times 6.25\pi = 25\pi$

(a) $616 \text{ cm}^{2}$ (b) $25\pi \text{ cm}^{2}$

Hemispheres: Solid or Hollow

A hemisphere's curved surface area is half of the sphere's: $2\pi r^{2}$ . What else to count depends on the object.

A solid hemisphere (a paperweight, a dome-shaped kueh) has a flat circular face as well, of area $\pi r^{2}$ , so its total surface area is $2\pi r^{2} + \pi r^{2} = 3\pi r^{2}\text{.}$ A hollow hemisphere open at the top (a thin bowl) has no flat face: only the curved surface, outside, inside, or both, as the question asks. Always say which you mean: "the total surface area of a hemisphere" on its own is not clear.

Surface area of a hemisphere

Curved surface area of a hemisphere $= 2\pi r^{2}$

Total surface area of a solid hemisphere $= 2\pi r^{2} + \pi r^{2} = 3\pi r^{2}$

A hollow, thin hemisphere: $2\pi r^{2}$ for one side of it, $4\pi r^{2}$ for inside and outside together.

All the formulas of the sphere and the hemisphere side by side. Each area has $r^{2}$ and each volume $r^{3}$ :

Solid	Surface area	Volume
Sphere	$4\pi r^{2}$	$\frac{4}{3}\pi r^{3}$
Solid hemisphere	$3\pi r^{2}$ (curved $2\pi r^{2}$ )	$\frac{2}{3}\pi r^{3}$
Hollow hemisphere	$2\pi r^{2}$ a side	(holds $\frac{2}{3}\pi r^{3}$ )

Example 4: A solid hemisphere

A solid hemisphere has a radius of $4$ cm. Find (a) its curved surface area, (b) its total surface area.

Solution

(a) $2 \times \pi \times 4^{2} = 32\pi = 100.53\ldots = 101$ (3 s.f.)

(b) Add the flat face, $\pi \times 4^{2} = 16\pi$ : total $= 32\pi + 16\pi = 48\pi = 150.79\ldots = 151$ (3 s.f.)

(a) $101 \text{ cm}^{2}$ (b) $151 \text{ cm}^{2}$

Example 5: A hollow bowl

A thin hemispherical bowl, open at the top, has a radius of $10$ cm. Find the area to be painted if (a) only the outside is painted, (b) both the inside and the outside are painted.

Solution

(a) Outside only: $2 \times \pi \times 10^{2} = 200\pi = 628.31\ldots = 628$ (3 s.f.)

(b) The inside is the same size: $2 \times 200\pi = 400\pi = 1256.6\ldots = 1260$ (3 s.f.)

There is no flat face to add: the top of the bowl is open.

(a) $628 \text{ cm}^{2}$ (b) $1260 \text{ cm}^{2}$

Finding the Radius from an Area

Put the area into the formula, make $r^{2}$ the subject and take the square root (an area has $r$ squared, so here it is the square root, not the cube root).

Example 6: A sphere

A sphere has a surface area of $200 \text{ cm}^{2}$ . Find its radius.

Solution

$4\pi r^{2} = 200$ , so $r^{2} = \frac{200}{4\pi} = 15.915\ldots$

$r = \sqrt{15.915\ldots} = 3.989\ldots = 3.99$ (3 s.f.)

The radius is $3.99$ cm.

Example 7: A solid hemisphere

The total surface area of a solid hemisphere is $75\pi \text{ cm}^{2}$ . Find its radius.

Solution

$3\pi r^{2} = 75\pi$ , so $r^{2} = 25$ and $r = 5$ .

The radius is $5$ cm.

Example 8: From the area to the volume

A sphere has a surface area of $256\pi \text{ cm}^{2}$ . Find its volume.

Solution

First the radius: $4\pi r^{2} = 256\pi$ , so $r^{2} = 64$ and $r = 8$ cm.

Then the volume: $\frac{4}{3} \times \pi \times 8^{3} = \frac{2048}{3}\pi = 2144.6\ldots = 2140$ (3 s.f.)

The volume is $2140 \text{ cm}^{3}$ .

Leather and Paint

Example 9: Leather for a football

A football has a diameter of $22$ cm. Ignoring the seams, what area of leather covers it?

Solution

$r = 11$ cm. $4 \times \pi \times 11^{2} = 484\pi = 1520.5\ldots = 1520$ (3 s.f.)

About $1520 \text{ cm}^{2}$ of leather

Example 10: Painting a dome

The roof of a planetarium is a hemispherical dome of radius $15$ m. One litre of paint covers $10 \text{ m}^{2}$ . How many litres of paint are needed to paint the outside of the dome once? Give a whole number of litres.

Solution

Only the curved surface is painted (the flat face is the floor): $2 \times \pi \times 15^{2} = 450\pi = 1413.7\ldots \text{ m}^{2}\text{.}$

$1413.7\ldots \div 10 = 141.37\ldots$ litres; $141$ litres would not be quite enough, so round up.

$142$ litres

Common Mistakes

Cubing the radius for an area: $4\pi r^{2}$ , not $4\pi r^{3}$ or $\frac{4}{3}\pi r^{3}$ .
Using the diameter. A diameter of $12$ cm used as $r$ gives four times the area.
Forgetting the flat face of a solid hemisphere ( $3\pi r^{2}$ ), or adding one to an open bowl ( $2\pi r^{2}$ ).
Taking the cube root to find $r$ from an area: an area needs the square root.
Rounding $r^{2}$ before taking the square 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 surface area of a sphere of radius $3$ cm.

ShowHide solution

$4 \times \pi \times 9 = 36\pi = 113.09\ldots = 113$ (3 s.f.)

$113 \text{ cm}^{2}$

Question 2

Find the surface area of a sphere of diameter $20$ cm.

ShowHide solution

$r = 10$ cm. $4 \times \pi \times 100 = 400\pi = 1256.6\ldots = 1260$ (3 s.f.)

$1260 \text{ cm}^{2}$

Question 3

Find the total surface area of a solid hemisphere of radius $7$ cm, in terms of $\pi$ .

ShowHide solution

$3 \times \pi \times 7^{2} = 147\pi$

$147\pi \text{ cm}^{2}$

Question 4

Taking $\pi = 3.142$ , find the curved surface area of a hemisphere of radius $5$ cm.

ShowHide solution

$2 \times 3.142 \times 25 = 157.1 = 157$ (3 s.f.)

$157 \text{ cm}^{2}$

Question 5

A sphere has a surface area of $196\pi \text{ cm}^{2}$ . Find its radius.

ShowHide solution

$4\pi r^{2} = 196\pi$ , so $r^{2} = 49$ and $r = 7$ .

$7$ cm

Question 6

A sphere has a surface area of $1000 \text{ cm}^{2}$ . Find its radius.

ShowHide solution

$r^{2} = \frac{1000}{4\pi} = 79.57\ldots\text{,}$ so $r = \sqrt{79.57\ldots} = 8.920\ldots = 8.92$ (3 s.f.)

$8.92$ cm

Question 7

A solid hemisphere has a total surface area of $300 \text{ cm}^{2}$ . Find its radius.

ShowHide solution

$3\pi r^{2} = 300$ , so $r^{2} = \frac{300}{3\pi} = 31.83\ldots$ and $r = 5.641\ldots = 5.64$ (3 s.f.)

$5.64$ cm

Question 8

A sphere has a surface area of $64\pi \text{ cm}^{2}$ . Find its volume.

ShowHide solution

$4\pi r^{2} = 64\pi$ , so $r^{2} = 16$ and $r = 4$ cm.

$\frac{4}{3} \times \pi \times 4^{3} = \frac{256}{3}\pi = 268.08\ldots = 268$ (3 s.f.)

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

Question 9

A thin hemispherical bowl of radius $9$ cm, open at the top, is glazed inside and outside. Find the area glazed.

ShowHide solution

Inside and outside, no flat face: $2 \times 2\pi \times 9^{2} = 324\pi = 1017.8\ldots = 1020$ (3 s.f.)

$1020 \text{ cm}^{2}$

Question 10

A hemispherical glass dome over a garden has a radius of $30$ m. Find the area of glass needed to cover its curved surface.

ShowHide solution

$2 \times \pi \times 30^{2} = 1800\pi = 5654.8\ldots = 5650$ (3 s.f.)

About $5650 \text{ m}^{2}$

Question 11

Wei Jie writes the total surface area of a solid hemisphere of radius $6$ cm as $2\pi \times 6^{2} = 72\pi \text{ cm}^{2}\text{.}$ Explain his mistake and find the correct answer.

ShowHide solution

He found only the curved surface. A solid hemisphere also has its flat circular face, $\pi \times 6^{2} = 36\pi$ .

$72\pi + 36\pi = 108\pi = 339.29\ldots = 339$ (3 s.f.)

He left out the flat face; the total is $339 \text{ cm}^{2}$ .

Question 12

A cube and a sphere each have a surface area of $600 \text{ cm}^{2}$ . Which has the greater volume, and by how much?

ShowHide solution

Cube: $6 \times \text{edge}^{2} = 600$ , so the edge is $10$ cm and the volume $1000 \text{ cm}^{3}$ .

Sphere: $r^{2} = \frac{600}{4\pi} = 47.74\ldots\text{,}$ so $r = 6.909\ldots$ cm and the volume is $\frac{4}{3}\pi \times 6.909\ldots^{3} = 1381.9\ldots \text{ cm}^{3}\text{.}$

Difference: $1381.9\ldots - 1000 = 381.9\ldots = 382$ (3 s.f.)

The sphere, by about $382 \text{ cm}^{3}$ .

Key Points

Surface area of a sphere $= 4\pi r^{2}$ , the area of four circles of radius $r$ .
Curved surface area of a hemisphere $= 2\pi r^{2}$ ; total surface area of a solid hemisphere $= 3\pi r^{2}$ .
Say whether a hemisphere is solid (count the flat face) or hollow (no flat face; one side or both).
To find $r$ from an area, make $r^{2}$ the subject and take the square root.
Areas have $r^{2}$ ; volumes have $r^{3}$ . Round only at the end.