Spheres: Volume, Radius, and Real-World Problems | Grade 8 Math | LevelKite

A sphere is a perfectly round solid, like a ball. Every point on its surface is the same distance from one point inside it, called the center. That distance is the radius, $r$ . The radius is a straight-line length from the center to the surface, never a length measured along the curved surface. The diameter, $d$ , is the straight-line distance across the sphere through its center, so $d = 2r$ . Both are lengths, so both are positive numbers.

You already know how to find the volume of cylinders and cones. Volume is the amount of space a solid takes up, measured in cubic units such as cubic centimeters ( $\text{cm}^3$ ). In this lesson you will find the volume of a sphere, work backward from a volume to a radius, and then use both skills to solve problems about real objects.

Where does a sphere's volume come from? Think of a cylinder that a sphere of radius $r$ just fits inside, touching the top, the bottom, and the curved side. The cylinder's base is a circle with the same radius $r$ . The cylinder is exactly as tall as the sphere, so its height is the diameter, $2r$ . That height is measured at a right angle to the base, not along a slant.

From your work with cylinders, the cylinder's volume is its base area times its height: $\pi r^2 \cdot 2r = 2\pi r^3$ . Here $\pi$ is the same constant, a little more than $3.14$ , that you used for circles and cylinders.

The sphere leaves empty space around it inside the cylinder, so its volume is less than the cylinder's. Mathematicians have proved that a sphere fills exactly two-thirds of the cylinder it just fits inside. Call the sphere's volume $V$ . Then $V$ is two-thirds of $2\pi r^3$ :

V = \frac{2}{3} \cdot 2\pi r^3 = \frac{4}{3}\pi r^3

Key concept

Volume of a sphere

The volume $V$ of a sphere is four-thirds times $\pi$ times the cube of its radius $r$ .

The radius $r$ is a positive length, measured in a straight line from the center to the surface.
If you are given the diameter $d$ , find the radius first: $r = \frac{d}{2}$ , because $d = 2r$ .
Use one unit of length throughout. A radius in centimeters gives a volume in cubic centimeters, because $r^3 = r \cdot r \cdot r$ multiplies three lengths.

V = \frac{4}{3}\pi r^3

Worked example

Volume from a radius

A sphere has a radius of $3$ cm. Find its exact volume. Then approximate the volume to the nearest cubic centimeter, using $3.14$ for $\pi$ .

Write the formula for the volume of a sphere. You can use it because you know the radius, which is a positive length.
$V = \frac{4}{3}\pi r^3$
Substitute the radius, $r = 3$ . The radius is in centimeters, so the volume will be in cubic centimeters.
$V = \frac{4}{3}\pi (3)^3$
Cube the radius before you multiply, because the exponent belongs only to $r$ : $3 \cdot 3 \cdot 3 = 27$ .
$V = \frac{4}{3}\pi \cdot 27$
Multiply $\frac{4}{3}$ by $27$ : $27 \div 3 = 9$ , and $4 \cdot 9 = 36$ . Keeping the symbol $\pi$ and not rounding makes this the exact volume.
$V = 36\pi$
Replace $\pi$ with $3.14$ to approximate. Then round to the nearest whole number: the tenths digit is $0$ , so round down.
$V \approx 36 \cdot 3.14 = 113.04 \approx 113$

Answer

The volume of the sphere is exactly $36\pi \text{ cm}^3$ , which is about $113 \text{ cm}^3$ .

Sometimes you know a sphere's volume and need its radius. The table below lists the exact volumes of six spheres. Read a row from left to right and you go from a radius to a volume, as you just did. Read it from right to left and you go the other way: a sphere with a volume of $36\pi \text{ cm}^3$ has a radius of $3$ cm. No other radius works, because a longer radius always gives a larger volume.

Exact volumes of spheres with whole-number radii

Radius $r$ (cm)	Radius cubed, $r^3$	Exact volume $V$ ( $\text{cm}^3$ )
$1$	$1$	$\frac{4}{3}\pi$
$2$	$8$	$\frac{32}{3}\pi$
$3$	$27$	$36\pi$
$4$	$64$	$\frac{256}{3}\pi$
$5$	$125$	$\frac{500}{3}\pi$
$6$	$216$	$288\pi$

A table only helps for the radii it lists. To find any radius, undo the formula $V = \frac{4}{3}\pi r^3$ one operation at a time. Do the same thing to both sides each time, so the two sides stay equal.

Multiply both sides by $3$ . This clears the denominator and gives $3V = 4\pi r^3$ .
Divide both sides by $4\pi$ . You may do this because $4\pi$ is not zero. Now $r^3$ stands alone.
Take the cube root of both sides. The cube root of a number is the value whose cube is that number. For example, $\sqrt[3]{27} = 3$ because $3 \cdot 3 \cdot 3 = 27$ .

A volume is positive, so $r^3$ is positive. Every positive number has exactly one positive cube root, so you get exactly one radius. If $V$ is in cubic centimeters, then $r$ is in centimeters. In symbols:

r^3 = \frac{3V}{4\pi} \qquad \text{so} \qquad r = \sqrt[3]{\frac{3V}{4\pi}}

Worked example

Radius from a volume

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

Write the formula and substitute the known volume, $V = 972\pi$ . The unknown is the radius $r$ , a positive length in centimeters.
$972\pi = \frac{4}{3}\pi r^3$
Multiply both sides by $3$ to clear the denominator. Multiplying both sides by the same number keeps them equal, and $972 \cdot 3 = 2916$ .
$2916\pi = 4\pi r^3$
Divide both sides by $4\pi$ , which is not zero. The factor $\pi$ divides out, and $2916 \div 4 = 729$ .
$r^3 = \frac{2916\pi}{4\pi} = 729$
Take the cube root of both sides. Find the positive number whose cube is $729$ : $9 \cdot 9 = 81$ , and $81 \cdot 9 = 729$ .
$r = \sqrt[3]{729} = 9$
Check by substituting $r = 9$ into the formula. It returns the given volume, so the radius is correct.
$\frac{4}{3}\pi (9)^3 = \frac{4}{3}\pi \cdot 729 = 972\pi$

Answer

The radius of the sphere is $9$ cm.

Balls, bubbles, and round tanks are all close to spheres, so the same formula answers questions about them. In a word problem, sort out four things before you calculate:

What is given? A radius, a diameter, or a volume. If it is a diameter, halve it to get the radius, because $d = 2r$ .
What is asked? For a volume, substitute the radius and evaluate. For a radius or a diameter, work backward from the volume.
Which units? Keep every length in one unit. Lengths in meters give a volume in cubic meters.
Exact or rounded? Answer in the form the question asks for, and always include the unit.

Worked example

How much space does a basketball take up?

A basketball is very close to a sphere. Treat it as a sphere with a diameter of $24$ cm. How much space does it take up? Give the exact volume. Then approximate it to the nearest hundred cubic centimeters, using $3.14$ for $\pi$ .

The diameter is given and the volume is asked for. The formula needs the radius, so halve the diameter, because $d = 2r$ .
$r = \frac{24}{2} = 12$
Substitute $r = 12$ . The radius is in centimeters, so the volume will be in cubic centimeters.
$V = \frac{4}{3}\pi (12)^3$
Cube the radius: $12 \cdot 12 = 144$ , and $144 \cdot 12 = 1728$ .
$V = \frac{4}{3}\pi \cdot 1728$
Multiply by $\frac{4}{3}$ : $1728 \div 3 = 576$ , and $576 \cdot 4 = 2304$ . This is the exact volume.
$V = 2304\pi$
Use $3.14$ for $\pi$ , then round to the nearest hundred. The tens digit is $3$ , so round down.
$V \approx 2304 \cdot 3.14 = 7234.56 \approx 7200$

Answer

Treated as a sphere, the basketball takes up exactly $2304\pi \text{ cm}^3$ of space, which is about $7200 \text{ cm}^3$ .

Worked example

The diameter of a water tank

A water tower has a tank shaped like a sphere. When it is full, the tank holds exactly $\frac{500}{3}\pi \text{ m}^3$ of water. What is the inside diameter of the tank?

The volume is given and the diameter is asked for. Work backward to the radius $r$ , in meters, then double it. Start by substituting the volume into the formula.
$\frac{500}{3}\pi = \frac{4}{3}\pi r^3$
Multiply both sides by $3$ . Each side has a denominator of $3$ , so both clear.
$500\pi = 4\pi r^3$
Divide both sides by $4\pi$ , which is not zero: $500 \div 4 = 125$ .
$r^3 = \frac{500\pi}{4\pi} = 125$
Take the cube root. The positive number whose cube is $125$ is $5$ , because $5 \cdot 5 \cdot 5 = 125$ .
$r = \sqrt[3]{125} = 5$
The question asks for the diameter, and $d = 2r$ .
$d = 2 \cdot 5 = 10$

Answer

The inside diameter of the tank is $10$ m.

Takeaway: A sphere's volume depends only on its radius. For a positive radius $r$ , the volume is $V = \frac{4}{3}\pi r^3$ , in the cubic unit that matches the unit of $r$ .

Radius to volume: cube the radius, then multiply by $\frac{4}{3}\pi$ . If you are given a diameter, halve it first, because $d = 2r$ .
Volume to radius: multiply the volume by $3$ , divide by $4\pi$ , then take the cube root.
For an exact value, keep $\pi$ as a symbol and do not round. For an approximation, use a decimal for $\pi$ and write $\approx$ .

Next, you will find the volume of composite solids.