Geometry · G.MG.A.1

Surface area of a sphere

A ball cannot be unrolled flat the way a box can. So how do you measure how much material covers it?

You need

Area of a circleMath 7 SpheresMath 8

Where you are 3 in Surface area

Geometry11

Solids8

Surface area3

Surface area of a sphere

Surface area of prisms and cylinders Surface area of pyramids and cones

When you reach for it

You want to know how much material covers something round: the leather on a ball, the paint on a dome, the foil around a sweet, the skin of a tank that loses heat through its surface. For a box or a prism, Surface area means unfolding the solid into flat faces and adding up their areas. A sphere has no faces to unfold. It cannot even be flattened: any piece of a sphere pressed onto a table has to stretch or tear, which is why every flat map of the Earth distorts something.

So a sphere needs a formula of its own, and it is a short one:

S = 4\pi r^2

where r is the radius of the sphere. Read it as a comparison with something you already know. Cut a sphere through its center and the cut face is a circle of radius r, called a great circle, with area \pi r^2 by the area of a circle. The surface of the sphere is exactly four of those circles.

Reach for it whenever the surface is a whole sphere, half of one, or a slice of one, and whenever a sphere sits inside a larger solid, such as a capsule or a silo with a rounded roof.

The procedure

Why it is four circles

The formula comes from comparing the sphere with the cylinder that just fits around it. That cylinder has the same radius r as the sphere, and its height is the sphere’s diameter, 2r.

Figure 1The cylinder that just contains a sphere has radius r and height 2r. Its curved side has the same area as the sphere.

The claim: the sphere’s surface has the same area as the curved side of this cylinder, leaving out its top and bottom. That side unrolls into a flat rectangle, as for any cylinder. Its length is the distance round the cylinder, the circumference 2\pi r, and its height is 2r, so its area is

2\pi r \times 2r = 4\pi r^2

Why the two are equal. Slice both solids with two horizontal planes very close together, a distance h apart. Between the planes, the cylinder has a thin band: radius r, height h, and area 2\pi r \cdot h, since it too unrolls into a rectangle.

The sphere has a band between the same planes. It is a thin ring a distance \rho from the axis (the vertical line through the center), and because the sphere’s surface slants, its width w is measured along the slant. A thin enough band is very nearly a flat strip, 2\pi\rho long and w wide, so its area is very nearly 2\pi\rho \cdot w.

The sphere’s band differs from the cylinder’s in two ways that pull in opposite directions. It is shorter around, because it sits closer to the axis: \rho < r. And it is wider, because it slants: w > h. The figure, a cut through the center, shows they cancel exactly.

Figure 2Seen from the side. The sphere’s band is closer to the axis (\rho < r) but slanted, so wider (w > h). The two effects cancel exactly.

Draw the radius from the center O to the band, length r. It meets the sphere’s surface at a right angle, because a tangent is perpendicular to the radius at the point where it touches, so the slanted piece w is perpendicular to that radius. Now compare two right triangles. The large one has the radius r as its hypotenuse and \rho as its horizontal leg. The tiny one has w as its hypotenuse and h as its vertical leg. Turn the large triangle a quarter turn: the radius turns onto the direction of w, and the horizontal turns onto the vertical. So the angle between w and the vertical equals the angle between the radius and the horizontal, the two triangles have the same angles, and they are similar. Matching sides,

\frac{h}{w} = \frac{\rho}{r} \qquad\text{so}\qquad \rho\, w = r\, h

Multiply both sides by 2\pi and the left side is the sphere’s band, the right side the cylinder’s: 2\pi\rho w = 2\pi r h. Every thin band of the sphere has the same area as the band of the cylinder level with it. Stack the bands from the bottom of the sphere to the top and the two totals agree: the sphere’s surface is 4\pi r^2.

One step above was approximate: each band of the sphere was treated as a flat strip. Thinner bands make that error smaller, and in the limit it disappears; making that precise is the work of surface area by integration. Archimedes proved the result itself, that the surface of a sphere is four times its great circle, in his book On the Sphere and Cylinder, by a careful argument of his own.

A rough experiment agrees. Peel an orange, trace its great circle onto paper four times, and the pieces of peel come close to covering the four circles. An experiment cannot prove a formula, but it does show the 4 is sensible.

What the bands give for free

The argument proved more than the whole-sphere formula. Any band of the sphere between two parallel planes a distance h apart has the same area as the cylinder band between them:

\text{area of the band} = 2\pi r h

wherever the planes are, as long as both cut the sphere. Take the planes through the center and the top: h = r, and the upper half has area 2\pi r^2, half of 4\pi r^2 as it must. Take one plane touching the top of the sphere: the band becomes a cap, a piece cut off by a single plane, and its curved area is 2\pi r h with h the cap’s height. Example 8 uses this.

Doing the calculation

The radius, not the diameter. The formula is built on r. If a diameter is given, halve it first. A ball 22 cm across has r = 11 cm.

Square r, then multiply by 4\pi. The square in 4\pi r^2 belongs to r alone: it means 4 \times \pi \times r^2, not (4\pi r)^2. That is the order of operations, powers before products.

Exact or decimal. Leaving \pi as a symbol gives an exact answer such as 100\pi square centimeters. When a decimal is wanted, use the calculator’s \pi and round only at the end, since rounding early lets small errors grow when they are multiplied by 4 and by r^2.

Square units. A surface is an area, so a radius in centimeters gives an answer in square centimeters, \text{cm}^2.

Working backwards. Given the area, undo the formula one operation at a time: divide by 4\pi, then take the square root,

r = \sqrt{\frac{S}{4\pi}}

Only the positive square root is kept, because a radius is a length.

Scaling. Multiply the radius by a number k and r^2 is multiplied by k^2, so the surface area is too: 4\pi (kr)^2 = k^2 \cdot 4\pi r^2. Twice the radius gives four times the skin. That is the rule of scale factor and area, seen on a sphere.

Worked examples

Example 1

A sphere from its radius

Find the surface area of a sphere of radius 5 cm, exactly and to two decimal places.

Solution

The radius is given, so it goes straight into the formula. Square it first: 5^2 = 25. Then multiply by 4\pi:

S = 4\pi r^2 = 4\pi \cdot 25 = 100\pi \text{ cm}^2

As a decimal, 100\pi \approx 314.16 \text{cm}^2. As a check on size: a great circle of this sphere has area 25\pi \approx 78.5 \text{cm}^2, and four of those is about 314.

Example 2

A ball given by its diameter

A soccer is 22 cm across. How much leather covers it, if the leather follows the ball’s surface exactly? Give the answer exactly and to two decimal places.

Solution

“Across” is the diameter, the length of a line through the center from one side to the other. The formula needs the radius, which is half of it: r = 11 cm. Then

S = 4\pi \cdot 11^2 = 4\pi \cdot 121 = 484\pi \text{ cm}^2

So the leather covers 484\pi \approx 1520.53 \text{cm}^2. Putting 22 into the formula instead would give 1936\pi, four times too much, because doubling the length quadruples the area.

Example 3

Half a sphere

(a) A glass paperweight is a solid hemisphere of radius 4 cm. Find its total surface area.

(b) A dome is a hemisphere of radius 10 m with no floor. Find the area of its curved outside surface, which is to be painted.

Give each answer exactly and to two decimal places.

Figure 3A solid hemisphere has two surfaces, the curved half of the sphere and a flat circle.

Solution

(a) A solid hemisphere has two surfaces. The curved one is half the sphere:

\frac{1}{2} \cdot 4\pi r^2 = 2\pi r^2 = 2\pi \cdot 16 = 32\pi

The flat one is the face left by the cut. The cut passed through the center, so that face is a great circle, of area \pi r^2 = 16\pi. Adding the two,

S = 32\pi + 16\pi = 48\pi \text{ cm}^2 \approx 150.80 \text{ cm}^2

In general a closed hemisphere has area 2\pi r^2 + \pi r^2 = 3\pi r^2.

(b) The dome has no floor, and only the curved part is painted, so only half the sphere counts:

2\pi r^2 = 2\pi \cdot 10^2 = 200\pi \text{ m}^2 \approx 628.32 \text{ m}^2

The flat circle is left out here because it is not there. Which pieces to count is decided by the object, not by the formula.

Example 4

Finding the radius from the area

(a) A sphere has surface area 144\pi square centimeters. Find its radius.

(b) A sphere has surface area 1000 square centimeters. Find its radius to two decimal places.

Solution

(a) Set the formula equal to the area and solve for r:

4\pi r^2 = 144\pi

Divide both sides by 4\pi, which is not zero, so the equation keeps the same solutions: r^2 = 36. Then r = 6 cm, taking the positive square root since a radius is a length. Check: 4\pi \cdot 36 = 144\pi.

(b) The same steps, but this time \pi does not cancel:

r^2 = \frac{1000}{4\pi} = \frac{250}{\pi}

r = \sqrt{\frac{250}{\pi}} \approx 8.92 \text{ cm}

Keep the whole calculator value of \displaystyle \frac{250}{\pi}, about 79.58, until the square root has been taken, and round only then.

Example 5

A balloon that grows

A round balloon is blown up from a radius of 10 cm to a radius of 15 cm.

(a) By what factor does its surface area grow?

(b) What radius would have made the surface area exactly double the starting area?

Solution

(a) The radius is multiplied by \displaystyle \frac{15}{10} = \frac{3}{2}, so the area is multiplied by the square of that:

\left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25

Check directly: the area goes from 4\pi \cdot 100 = 400\pi to 4\pi \cdot 225 = 900\pi, and \displaystyle \frac{900\pi}{400\pi} = \frac{9}{4}. The rubber has been stretched over two and a quarter times the area, though the radius grew by only half.

(b) Doubling the area means multiplying it by 2, so the radius must be multiplied by a number k with k^2 = 2, that is \displaystyle k = \sqrt{2}. The new radius is

10\sqrt{2} \approx 14.14 \text{ cm}

Check: \displaystyle 4\pi \left(10\sqrt{2}\right)^2 = 4\pi \cdot 200 = 800\pi, which is twice 400\pi.

Example 6

A capsule

A capsule is a cylinder 10 cm long with a half sphere on each end. The radius of the cylinder and of both half spheres is 3 cm. Find the area of the capsule’s outside surface, exactly and to two decimal places.

Figure 4The capsule in Example 6. The two half spheres together make one whole sphere of radius 3 cm.

Solution

Split the surface into pieces whose areas are known. The two half spheres have the same radius, so together they cover exactly the surface of one whole sphere of radius 3 cm:

4\pi \cdot 3^2 = 36\pi

The middle is the curved side of a cylinder, with no top or bottom, since those are covered by the half spheres. It unrolls into a rectangle 2\pi r by 10:

2\pi \cdot 3 \cdot 10 = 60\pi

Adding the pieces,

S = 36\pi + 60\pi = 96\pi \text{ cm}^2 \approx 301.59 \text{ cm}^2

The seams where the half spheres meet the cylinder are inside the surface, not extra faces, so nothing else is added.

Example 7

The sphere and its cylinder

A sphere of radius r fits exactly inside a closed cylinder, top and bottom included. What fraction of the cylinder’s total surface area is the sphere’s surface area?

Solution

The cylinder has radius r and height 2r. Its total surface is the curved side and two circular ends:

2\pi r \cdot 2r + 2\pi r^2 = 4\pi r^2 + 2\pi r^2 = 6\pi r^2

The sphere has area 4\pi r^2. So the fraction is

\frac{4\pi r^2}{6\pi r^2} = \frac{2}{3}

The r and the \pi cancel, so the answer is the same for every sphere. Archimedes found this pair fitting together in two ways. The sphere’s surface is two thirds of the cylinder’s whole surface, as just calculated (and equal to its curved side alone, as shown in the procedure), and the volume of a sphere is two thirds of the cylinder’s volume too.

Example 8

A cap cut from a sphere

A sphere of radius 10 cm is cut by a plane 6 cm from its center. Find the curved area of the smaller piece, exactly and to two decimal places, and say what fraction of the whole sphere’s surface it is.

Figure 5The sphere in Example 8, seen from the side, lengths in centimeters. The cap is 4 cm tall.

Solution

The smaller piece is a cap. Its height is the distance from the cutting plane to the top of the sphere, measured along the line through the center perpendicular to the plane. The top is 10 cm from the center and the plane is 6 cm from it, so

h = 10 - 6 = 4 \text{ cm}

A cap is a band between the cutting plane and a plane touching the sphere at the top, so its curved area is 2\pi r h:

2\pi \cdot 10 \cdot 4 = 80\pi \text{ cm}^2 \approx 251.33 \text{ cm}^2

The whole sphere has area 4\pi \cdot 100 = 400\pi, so the cap is \displaystyle \frac{80\pi}{400\pi} = \frac{1}{5} of it, exactly the same fraction as its height: \displaystyle \frac{4}{20} of the sphere’s height is also one fifth. That is the band result again: equal heights of a sphere carry equal areas. Check: the larger piece has height 16 cm and curved area 2\pi \cdot 10 \cdot 16 = 320\pi, and 80\pi + 320\pi = 400\pi.

Boundaries

A sphere, and only a sphere. 4\pi r^2 needs every point of the surface to be the same distance r from the center. An egg, a rugby ball or a squashed ball has no single radius, and its surface area is not 4\pi r^2 for any choice of r; such shapes need methods beyond this one.

The surface, not the solid. The formula measures the skin and nothing inside it. A hollow ball with thick walls has two surfaces, an outer one and an inner one, with different radii.

Only the curved part. Cutting a sphere makes new flat faces, and 4\pi r^2 (or 2\pi r^2 for a half) never counts them. Whether they belong in the answer depends on the object, as Example 3 shows.

The band rule needs parallel planes. The area 2\pi r h holds for the piece between two parallel planes that both cut or touch the sphere, with h the perpendicular distance between the planes. Two planes that are not parallel cut out a different shape, and the rule says nothing about it.

Where this goes next

The surface leads straight to the inside. The volume of a sphere is \displaystyle \frac{4}{3}\pi r^3, and the two formulas fit together: a thin skin of thickness t has volume very nearly 4\pi r^2 \cdot t, so a ball can be built out of thin skins, one inside the next. Why the volume formulas hold links the two another way, through \displaystyle V = \frac{1}{3}Sr. In calculus the link runs the other way: the rate at which \displaystyle \frac{4}{3}\pi r^3 grows as r grows is exactly 4\pi r^2. That is why a balloon’s volume and surface appear together in related rates.

Capsules, silos and domes are composite solids: split into a sphere or hemisphere plus pieces whose areas are already known, as in Example 6. And the band argument, made exact with integrals, becomes the general method of surface area of a solid of revolution, where any curve spun about an axis is cut into thin bands and their areas added.

Exercises

Find the surface area

Give each answer exactly and to two decimal places.

A sphere of radius 7 inches.
A sphere of radius 0.5 meters.
A sphere of diameter 18 cm.
A ball of radius 2.5 feet.
A marble of diameter 4.2 cm.

Hemispheres

Give each answer exactly and to two decimal places.

The total surface area of a solid hemisphere of radius 6 cm.
The outside of an open bowl that is a hemisphere of radius 9 cm, with no lid.
A dome is a hemisphere of diameter 30 meters. Find the area of its curved surface.

Working backwards

A sphere has surface area 64\pi square units. Find its radius.
A ball has surface area 196\pi \text{cm}^2. Find its diameter.
A sphere has surface area 500 \text{cm}^2. Find its radius to two decimal places.
A solid hemisphere, flat face included, has total surface area 27\pi square units. Find its radius.

Scaling

The radius of a sphere is doubled. By what factor is its surface area multiplied?
Two spheres have radii 2 cm and 5 cm. Find the ratio of the smaller one’s surface area to the larger one’s.
The surface area of a sphere is multiplied by 9. By what factor was its radius multiplied?
A ball has surface area 100 \text{cm}^2. Find the surface area of a ball whose radius is 1.5 times as large.

Problems

A capsule is a cylinder 5 cm long with a half sphere on each end, all of radius 2 cm. Find its outside surface area, exactly and to two decimal places.
A silo is a cylinder of radius 4 m and height 12 m with a hemisphere for a roof. Find the area of its curved wall and roof together, leaving out the floor, exactly and to two decimal places.
A factory paints 50 balls, each of radius 3.5 cm. Find the total area painted, exactly and to two decimal places.
A sphere of radius 5 cm is cut by a plane 3 cm from its center. Find the curved area of each of the two pieces, exactly and to two decimal places.
A cube with edges 2 cm long and a sphere have the same surface area. Find the radius of the sphere to two decimal places.

Reasoning

Explain why the total surface area of a solid hemisphere is 3\pi r^2 and not 2\pi r^2.
Use the formula to explain why tripling the radius of a sphere multiplies its surface area by 9.
Near the equator a thin band of a sphere stands almost upright; near a pole a band of the same height lies almost flat. Use the band argument to explain why the two bands still have the same area.
A sphere of radius r is sliced into two pieces by a plane. Explain why the total surface area of the two pieces is more than 4\pi r^2, and find how much more when the plane passes through the center.

Answers

196\pi \approx 615.75 square inches. 4\pi \cdot 7^2 = 4\pi \cdot 49.
\pi \approx 3.14 \text{m}^2. 4\pi \cdot 0.5^2 = 4\pi \cdot 0.25.
324\pi \approx 1017.88 \text{cm}^2. The radius is 9, and 4\pi \cdot 81 = 324\pi.
25\pi \approx 78.54 square feet. 4\pi \cdot 2.5^2 = 4\pi \cdot 6.25.
17.64\pi \approx 55.42 \text{cm}^2. The radius is 2.1, and 4\pi \cdot 4.41 = 17.64\pi.
108\pi \approx 339.29 \text{cm}^2. Curved part 2\pi \cdot 36 = 72\pi, flat face \pi \cdot 36 = 36\pi.
162\pi \approx 508.94 \text{cm}^2. Half a sphere, 2\pi \cdot 81, with no flat face since the bowl is open.
450\pi \approx 1413.72 \text{m}^2. The radius is 15, and 2\pi \cdot 225 = 450\pi.
r = 4. 4\pi r^2 = 64\pi gives r^2 = 16.
Diameter 14 cm. 4\pi r^2 = 196\pi gives r^2 = 49, so r = 7.
r \approx 6.31 cm. \displaystyle r^2 = \frac{500}{4\pi} = \frac{125}{\pi} \approx 39.79, then take the square root.
r = 3. 3\pi r^2 = 27\pi gives r^2 = 9.
By 4. 4\pi (2r)^2 = 4 \cdot 4\pi r^2.
4 : 25. The areas are 16\pi and 100\pi, or directly \displaystyle \left(\frac{2}{5}\right)^2 = \frac{4}{25}.
By 3. The area is multiplied by the square of the radius factor, and 3^2 = 9.
225 \text{cm}^2. The area is multiplied by 1.5^2 = 2.25, and 100 \times 2.25 = 225.
36\pi \approx 113.10 \text{cm}^2. The ends make one sphere, 4\pi \cdot 4 = 16\pi; the side is 2\pi \cdot 2 \cdot 5 = 20\pi.
128\pi \approx 402.12 \text{m}^2. Roof 2\pi \cdot 16 = 32\pi; wall 2\pi \cdot 4 \cdot 12 = 96\pi.
2450\pi \approx 7696.90 \text{cm}^2. Each ball is 4\pi \cdot 3.5^2 = 49\pi, and 50 \times 49\pi = 2450\pi.
20\pi \approx 62.83 \text{cm}^2 and 80\pi \approx 251.33 \text{cm}^2. The caps have heights 5 - 3 = 2 and 5 + 3 = 8, so 2\pi \cdot 5 \cdot 2 and 2\pi \cdot 5 \cdot 8. Together they make 100\pi = 4\pi \cdot 25.
r \approx 1.38 cm. The cube’s area is 6 \cdot 2^2 = 24, so 4\pi r^2 = 24, \displaystyle r^2 = \frac{6}{\pi}, and \displaystyle r = \sqrt{\frac{6}{\pi}}.
The curved part is half the sphere, \displaystyle \frac{1}{2} \cdot 4\pi r^2 = 2\pi r^2. A solid hemisphere also has the flat face made by the cut, and the cut passes through the center, so that face is a great circle of area \pi r^2. The total is 2\pi r^2 + \pi r^2 = 3\pi r^2. The value 2\pi r^2 is right only for an open shell, such as a bowl or a dome with no floor.
Replacing r by 3r gives 4\pi (3r)^2 = 4\pi \cdot 9r^2 = 9 \cdot 4\pi r^2. The radius appears squared, once for each of the two directions an area spreads in, so the factor 3 is used twice.
Each band has area 2\pi\rho w, where \rho is its distance from the axis and w its slanted width. Near the equator \rho is close to r and the band is nearly upright, so w is close to h. Near a pole \rho is small, but the band lies almost flat, so w is much larger than h. The similar triangles give \rho w = r h in every position, so both bands have area 2\pi r h.
The curved surfaces of the two pieces together are the whole sphere, 4\pi r^2. The cut also makes two new flat faces, one on each piece, and each is a circle with positive area, so the total is larger. Through the center each new face is a great circle, \pi r^2, so the total is 4\pi r^2 + 2\pi r^2 = 6\pi r^2, which is 2\pi r^2 more.