Math 8

Spheres

A circle is every point in a flat plane at one distance from a center. What shape do you get when the points may lie anywhere in space?

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What a circle isMath 6

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Volume of a sphereMath 8 Surface area of a sphereGeometry

Where you are 7 in Three-dimensional shapes

Geometry11

Solids8

Three-dimensional shapes7

▴ 2 above Naming and describing solid shapes Polyhedra

Spheres

Prisms and pyramids The five regular solids

Why this exists

A circle answers a flat question: which points on a page sit at one distance from a center? Most round things are not flat. A soap bubble, an orange and, very nearly, the Earth are round in every direction at once. To reason about them (how large a box must be to hold one, how wide the face of a cut fruit is, whether a point is inside or outside) they need a description as exact as the circle’s. It turns out to be the circle’s own, with the words “in a plane” taken away.

Definition

A sphere is the set of all points in space that are the same distance from one fixed point.

The fixed point is the center, often called O.
The fixed distance is the radius, r. The word also names any segment from the center to a point on the sphere; every such segment has length r.
A chord is a segment with both ends on the sphere. A chord through the center is a diameter. It is two radii laid end to end, so its length is

d = 2r

A great circle is the circle where a plane through the center cuts the sphere. It has the sphere’s own center and radius. A globe’s equator is one.
A hemisphere is half a sphere: either of the two pieces a great circle cuts it into.

Figure 1The parts of a sphere. The great circle has the sphere’s own center and radius, and the shaded half above it is a hemisphere.

A sphere is only the surface, the skin of points at distance r. The solid it encloses, skin and inside together, is a ball: a soap bubble is very nearly a sphere, an orange very nearly a ball. Everyday speech uses “sphere” for both, and this lesson says “ball” whenever it means the solid. The center itself is not on the sphere; it is at distance 0, not r.

Since every point of the sphere is exactly r from the center, distance from the center sorts all of space into three groups: a point closer than r is inside the sphere, a point at exactly r is on it, and a point farther than r is outside.

Intuition

Spin a circle about one of its own diameters, as a coin spun on its edge seems to do, and every point keeps its distance from the center, so together they sweep out a sphere.

A sphere looks the same from every direction, which makes its cuts easy to describe. Slice it with a flat plane, tilted any way, and the cut edge is a circle. Through the center the cut is a great circle, the largest there is. Move the plane away from the center and the circle shrinks, down to a single point where the plane just touches, and beyond that the plane misses. Slicing other solids is cross sections of prisms and pyramids.

Figure 2Every flat cut of a sphere is a circle. The cut through the center is the largest, and the circles shrink as the cut moves away from it.

Spheres turn up wherever something is pulled equally in every direction toward a center. A sphere has the least surface for its volume, so a soap film, pulling itself as tight as it can, rounds into one. A planet’s own gravity pulls it into very nearly a sphere; the Earth is slightly flattened at the poles, so a sphere is a close model of it, not an exact one.

Results

Every diameter has the same length, d = 2r.

The diameter is the longest chord. No two points of a sphere are more than 2r apart, and two points are exactly 2r apart only when they are the ends of a diameter.

Inside, on or outside. A point at distance p from the center of a sphere of radius r is inside when p < r, on the sphere when p = r, and outside when p > r.

Every flat cut is a circle. A plane at distance h from the center, with h < r, cuts the sphere in a circle centered at the point of the plane nearest the sphere’s center. Its radius a satisfies the The Pythagorean theorem:

a^2 + h^2 = r^2

At h = 0 the cut is a great circle, of radius r, the largest cut. At h = r the plane touches the sphere at one point, and when h > r it misses.

How much a ball holds. Books say “the volume of a sphere” and mean the volume of the ball inside it. For radius r it is \displaystyle V = \frac{4}{3}\pi r^3, worked out in Volume of a sphere; the area of the skin is Surface area of a sphere.

Why it is true

Diameters. A diameter PQ runs through the center O, and its two parts OP and OQ are radii. So PQ = r + r = 2r, whichever diameter it is.

The longest chord. Take any two points P and Q on the sphere. The path from P to O and on to Q has length 2r. By the Triangle inequality, the straight segment PQ is never longer than a path with a bend in it, so PQ \le 2r, with equality only when there is no bend: when O lies on PQ and PQ is a diameter.

The cut is a circle. Let a plane lie at distance h < r from O, and let F be its point nearest O, so that OF is perpendicular to the plane and OF = h. Take any point P on both the sphere and the plane. The segment FP lies in the plane, so triangle OFP has its right angle at F and hypotenuse OP = r. By the Pythagorean theorem

FP^2 = r^2 - h^2

Figure 3Seen from the side, a cut makes a right triangle with hypotenuse r, so a^2 + h^2 = r^2.

That is the same number for every such P, so all of them lie in the plane at one distance, \displaystyle a = \sqrt{r^2 - h^2}, from F: on a circle with center F. Backwards, any point of the plane at distance a from F has OP^2 = h^2 + a^2 = r^2, so it is on the sphere, and the cut is the whole circle. The radius a is largest when h = 0, giving a = r. When h = r only F is left. When h > r, every point of the plane has OP^2 = h^2 + FP^2 > r^2, so none is on the sphere.

Worked examples

Example 1

The parts of a globe

A classroom globe measures 30 cm straight through its center, from the North Pole to the South Pole. Treat it as a sphere. (a) Find its radius. (b) The equator lies in the plane through the center at right angles to the pole-to-pole line. Explain why it is a great circle, and find its radius and length.

Solution

(a) The measurement passes through the center with both ends on the sphere, so it is a diameter, and r = 30 \div 2 = 15 cm.

(b) The equator is cut out by a plane through the center, which is what a great circle is. So its radius is the globe’s, 15 cm, and it splits the globe into the northern and southern hemispheres. Its length is its Circumference, 2\pi r:

2\pi \cdot 15 = 30\pi \approx 94.25 \text{ cm}

Example 2

Inside, on or outside the glass

Hedda’s snow globe is a glass ball whose inside measures 12 cm across through its center; treat the inside of the glass as a sphere. A snowflake is 4.8 cm from the center, the tip of a little star on a tree is 6 cm from it, and a point on the wooden base is 6.9 cm from it. (a) Is each point inside, on or outside the sphere? (b) How far is the snowflake from the nearest point of the glass, and from the farthest?

Solution

(a) The radius is 12 \div 2 = 6 cm. The snowflake is inside, since 4.8 < 6. The star’s tip is on the sphere, touching the glass, since 6 = 6. The point on the base is outside, since 6.9 > 6.

(b) Call the center O and the snowflake S. The ray from O through S meets the glass 6 - 4.8 = 1.2 cm beyond S, and the same line continued past O meets it 4.8 + 6 = 10.8 cm from S. No point Q of the glass does better, since the triangle inequality gives OQ - OS \le SQ \le SO + OQ, that is 1.2 \le SQ \le 10.8. The nearest and farthest points are the ends of the diameter through the snowflake.

Example 3

A ball in a cube box

Figure 4A ball that fits a cube box snugly is as wide as the box. Balls stacked in a tube each take up one diameter of its height.

Freya-Lou packs a glass ornament, a ball of radius 4.5 cm, in a cube box that it fits snugly, touching all six faces. (a) How long is each edge? (b) Her other box is a cube with edges 10 cm long. How much room does the ornament have in it?

Solution

(a) Stand the box on a table. The ball touches the bottom face at the point directly below the center (no point of the sphere is more than r below the center, and that one is exactly r below), and the top face at the point directly above it. These two points are the ends of a vertical diameter, so the height of the box is

2r = 9 \text{ cm}

The same holds left to right and front to back, which is why the box is a cube: the ball has the same width in every direction. Each edge is 9 cm.

(b) Each direction has 10 - 9 = 1 cm to spare. With the ornament in the middle, that is a gap of 0.5 cm between the ball and each face.

Example 4

Juggling balls in a tube

Three juggling balls, each 6 cm across, are packed snugly in a cylindrical tube: each touches the side, they touch one another, the bottom one touches the floor and the top one the lid. (a) Find the inside width and height of the tube. (b) Gaspard-Ray says the tube is taller than the distance around it. Is he right?

Solution

(a) The side touches each ball all the way round its widest circle, a great circle, so the tube is one diameter wide: 6 cm. Each ball reaches one diameter higher than the one beneath it, so the height is

3 \times 6 = 18 \text{ cm}

(b) The distance around is the circumference of a circle 6 cm across:

6\pi \approx 18.85 \text{ cm}

That beats the height of 18 cm, so Gaspard-Ray is wrong, and for any size of ball: the height is 3d, the distance around is \pi d, and \pi > 3.

Example 5

A melon cut two ways

Ilario has two small melons, each a ball of radius 7.5 cm. He cuts the first straight through its center, and cuts a slice off the second with a flat cut 4.5 cm from its center. Find the radius of the cut face each time.

Solution

The first cut passes through the center, so its edge is a great circle and the face has the melon’s own radius, 7.5 cm.

For the second, let O be the center, F the center of the cut face and P a point on its edge. Triangle OFP has its right angle at F, with OF = 4.5 and OP = 7.5. With a = FP, the Pythagorean theorem gives

\begin{aligned} a^2 + 4.5^2 &= 7.5^2 \\ a^2 + 20.25 &= 56.25 \\ a^2 &= 36 \\ a &= 6 \end{aligned}

The cut face has radius 6 cm. Check: 36 + 20.25 = 56.25. Moving the knife 4.5 cm off center shrank the face only from 7.5 cm to 6 cm; cuts shrink slowly near the center and fast near the skin.

Example 6

A line of latitude, worked backwards

On the globe of Example 1, radius 15 cm, Delphine draws a line of latitude, a circle in a plane parallel to the equator’s. It has radius 12 cm. How far is its plane from the center, and how many such circles are there?

Solution

The pole-to-pole line is at right angles to the equator’s plane, so it is at right angles to every parallel plane, and it meets Delphine’s plane at the point nearest the center: the center of her circle. With h the distance from the globe’s center to that point,

\begin{aligned} h^2 + 12^2 &= 15^2 \\ h^2 + 144 &= 225 \\ h^2 &= 81 \\ h &= 9 \end{aligned}

The plane is 9 cm from the center. There are two such circles, one north of the equator and one south. Check: 81 + 144 = 225.

Boundaries

It needs space. “All points at distance r from a center” gives a sphere only when the points may lie anywhere in space. In a plane the same words give a circle; on a line, just two points. The radius must also be positive: with r = 0 only the center is left.

A sphere is a surface, not a solid. It has no thickness of its own; “the volume of a sphere” means the volume of the ball it encloses.

A measurement across is a diameter only through the center. A straight measurement between two points of a sphere that misses the center is a chord, shorter than 2r.

The cut formula needs the perpendicular distance. In a^2 + h^2 = r^2, h is measured at right angles to the plane, and must be at most r.

Real objects are only nearly spheres. The results hold exactly for the ideal shape and only approximately for a bumpy orange, a wobbling bubble or the slightly flattened Earth.

Where this goes next

Volume of a sphere finds how much a ball holds by comparing it with the cylinder that just contains it, the same snug fit as the ball in the box, and Surface area of a sphere finds that the skin has four times the area of a great circle. In coordinates, a sphere centered at the origin is every point (x, y, z) with x^2 + y^2 + z^2 = r^2, the Pythagorean theorem in three dimensions applied to the radius.

Exercises

Radius, diameter and great circles

A sphere has radius 7.5 cm. Find its diameter.
A soap bubble measures 9.4 cm straight across through its center. Find its radius.
A great circle of a sphere has diameter 26 cm. Find the radius of the sphere.
Half an orange, cut through its center, lies flat face down. The flat face is a disc of radius 4.2 cm. Treating the orange as a ball, how wide was the whole orange?
Treat the Earth as a sphere of radius 6400 km. Its equator is a great circle. Find the length of the equator exactly in terms of \pi, and to the nearest kilometer.

Inside, on or outside

A sphere has diameter 18 cm. Points P, Q and R are 8.6 cm, 9 cm and 9.4 cm from its center. Say whether each is inside, on or outside the sphere.
A snow globe’s glass is a sphere 8 cm across on the inside. A snowflake is 3.8 cm from the center. How far is it from the nearest point of the glass, and from the farthest?
Two points lie on a sphere of radius 5.5 cm. What is the greatest possible distance between them, and when does it happen?
A point P is on a sphere of radius 10 cm with center O. Point M is the midpoint of OP, and point T lies on the ray from O through P with OT = 2 \cdot OP. Find OM and OT, and say whether each of M and T is inside, on or outside the sphere.

Boxes and tubes

A ball fits snugly in a cube box with edges 12 cm long, touching all six faces. Find the ball’s radius.
Four balls, each 6.5 cm across, are stacked snugly in a tube. Find the inside height and width of the tube.
A tube is 21 cm tall inside and holds three identical balls snugly, touching the floor, each other and the lid. Find the radius of each ball.
A box is 24 cm long, 16 cm wide and 8 cm deep. How many balls 8 cm across fit in it when they are packed in straight rows and columns, each touching its neighbors?

Cutting a sphere

These use the The Pythagorean theorem.

A sphere of radius 17 cm is cut by a plane 8 cm from its center. Find the radius of the cut.
A ball of modeling clay of radius 10 cm is sliced flat by a wire 6 cm from its center. Find the radius and the diameter of the flat face.
A globe has radius 25 cm. A line of latitude on it is a circle of radius 24 cm. How far is the plane of that circle from the center of the globe?
A sphere has radius 9 cm. Describe what a plane cuts from it when the plane is (a) 0 cm, (b) 9 cm and (c) 10 cm from the center.

Reasoning

Erasmus measures straight across a ball between two points on its surface and gets 11 cm. Hedda measures straight across the same ball between two other points and gets 13 cm. Both measured correctly. What can you say about the ball’s diameter, and which of the two measurements could be a diameter?
Explain why the cut through the center is the largest circle that can be cut from a sphere.
Explain why two different great circles of the same sphere always cross at exactly two points, and why those two points are the ends of a diameter.

Answers

15 cm, since 2 \times 7.5 = 15.
4.7 cm. The measurement through the center is a diameter, and 9.4 \div 2 = 4.7.
13 cm. A great circle has the sphere’s own radius, half of 26.
8.4 cm. The cut face is a great circle, with the orange’s radius 4.2 cm, so the orange was 2 \times 4.2 = 8.4 cm across.
12800\pi km, about 40212 km: 2\pi \cdot 6400 = 12800\pi \approx 40212.4.
The radius is 18 \div 2 = 9 cm. P is inside (8.6 < 9), Q is on the sphere (9 = 9) and R is outside (9.4 > 9).
0.2 cm and 7.8 cm. The radius is 4 cm, so the nearest point is 4 - 3.8 = 0.2 cm away and the farthest, on the same line through the center, 3.8 + 4 = 7.8 cm away.
11 cm, when the two points are the ends of a diameter, the longest chord: 2 \times 5.5 = 11.
OM = 5 cm, so M is inside, since 5 < 10. OT = 20 cm, so T is outside, since 20 > 10.
6 cm. The ball’s diameter equals the edge, 12 cm.
Height 26 cm and width 6.5 cm. Each ball takes one diameter of height: 4 \times 6.5 = 26.
3.5 cm. Each ball takes 21 \div 3 = 7 cm of height, its diameter.
6 balls: 24 \div 8 = 3 along the length, 16 \div 8 = 2 across, one layer deep, and 3 \times 2 = 6.
15 cm. a^2 = 17^2 - 8^2 = 289 - 64 = 225, so a = 15.
Radius 8 cm, diameter 16 cm. a^2 = 10^2 - 6^2 = 100 - 36 = 64, so a = 8.
7 cm. h^2 = 25^2 - 24^2 = 625 - 576 = 49, so h = 7.
(a) A great circle of radius 9 cm, the largest cut. (b) A single point: the plane touches the sphere. (c) Nothing: the plane is farther from the center than the radius.
The diameter is at least 13 cm, because it is the longest chord and a chord of 13 cm exists. So Erasmus’s 11 cm cannot be a diameter. Hedda’s 13 cm could be one, but need not be; it is a diameter only if no chord of the sphere is longer.
A plane at distance h from the center cuts a circle of radius a with a^2 = r^2 - h^2. Any plane not through the center has h > 0, so a^2 < r^2 and a < r. The plane through the center has h = 0 and a = r, the largest circle.
Each great circle lies in a plane through the center O. Two different planes that share the point O meet in a line through O. A point lies on both great circles exactly when it is on the sphere and on that line. A line through the center meets the sphere at exactly two points, one at distance r on each side of O, and those two points are the ends of a diameter.