Comparing Cylinder, Cone, and Sphere Volumes | Grade 8 Math | LevelKite

Volume is the amount of space a solid fills. You measure it in cubic units, such as cubic centimeters, written $\text{cm}^3$ .

You already know how to calculate the volume of three solids. A cylinder has two matching, parallel circular ends, called bases. A cone has one circular base and narrows to a single tip. A sphere is perfectly round, like a ball. In this lesson you will compare their volumes.

Comparing can answer two different questions:

How much more? Subtract the smaller volume from the larger one.
How many times as much? Divide one volume by the other. This kind of comparison is a ratio. A ratio of $1 : 3$ means the second amount is 3 times the first.

Either way, both volumes must be in the same cubic unit. And you cannot settle a comparison by how big a solid looks. You have to calculate.

Each volume formula needs certain lengths, so name them before you calculate.

Radius, $r$ : the distance from the center of a circle to its edge, or from the center of a sphere to its surface. For a cylinder or cone, use the radius of the circular base.
Diameter, $d$ : the distance straight across through the center. It is twice the radius, $d = 2r$ , so halve a diameter to get $r$ .
Height, $h$ : for a cylinder, the distance between its two bases. For a cone, the distance from its base to its tip. Either way, measure it perpendicular to the base, which means at a right angle to it.
Slant height: the distance along a cone's sloped side, from its tip to the edge of its base. It is not the height, and the volume formulas in this lesson do not use it.

Each of these is a positive length, so $r > 0$ , $d > 0$ , and $h > 0$ . All lengths in one comparison must be in the same unit. If they are in centimeters, each volume $V$ is in cubic centimeters, $\text{cm}^3$ .

Volume formulas for positive lengths in the same unit

Solid	Lengths you need	Volume formula
Cylinder	radius $r$ and height $h$	$V = \pi r^2 h$
Cone	radius $r$ and height $h$ (not slant height)	$V = \frac{1}{3}\pi r^2 h$
Sphere	radius $r$	$V = \frac{4}{3}\pi r^3$

Start with the cylinder and the cone. Keep the same conditions: positive lengths, all in the same unit, and heights measured perpendicular to the base.

Both formulas contain $\pi r^2 h$ . That is the area of the circular base, $\pi r^2$ , in square units, times the height $h$ , in length units. The product is in cubic units.

So if a cone and a cylinder have the same radius and the same height, then $\pi r^2 h$ is the same number for both. The cone's volume is exactly $\frac{1}{3}$ of the cylinder's, and the cylinder's volume is exactly 3 times the cone's. It would take 3 of those cones to fill that cylinder.

Now add a sphere with the same radius $r$ . A sphere is as tall as its diameter, $2r$ . So the cylinder that matches it has radius $r$ and a height equal to that diameter, $h = 2r$ . The sphere just fits inside this cylinder, so its volume must be smaller. How much smaller?

Replace $h$ with $2r$ in the cylinder formula: $\pi r^2 (2r) = 2\pi r^3$ , because $r^2 \cdot r = r^3$ . Then divide the sphere's volume by the cylinder's. Since $r > 0$ , the factor $\pi r^3$ is not zero, so you can divide the top and the bottom by it without changing the value:

\frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{\frac{4}{3}\pi r^3}{2\pi r^3} = \frac{4}{3} \div 2 = \frac{4}{3} \cdot \frac{1}{2} = \frac{4}{6} = \frac{2}{3}

Key concept

Matching dimensions give fixed ratios

These comparisons hold when all lengths are positive and in the same unit, and each height is measured perpendicular to its base, not along a slant.

A cone and a cylinder with the same radius $r$ and the same height $h$ : the cone has $\frac{1}{3}$ of the cylinder's volume.
If that shared height also equals the diameter of a sphere with the same radius, so $h = 2r$ : the sphere has $\frac{2}{3}$ of the cylinder's volume.

Split that cylinder's volume into 3 equal parts. The cone equals 1 part and the sphere equals 2 parts, so cone to sphere to cylinder is $1 : 2 : 3$ . If the dimensions do not match this way, do not count on the pattern. Calculate each volume with its own formula, then compare.

V_{\text{cone}} : V_{\text{sphere}} : V_{\text{cylinder}} = 1 : 2 : 3 \quad \text{when all share } r > 0 \text{ and } h = 2r

Worked example

Cone and cylinder with the same radius and height

A cylinder and a cone each have radius $r = 3$ cm and height $h = 8$ cm, measured perpendicular to the base. The cylinder's volume is how many times the cone's volume? How many more cubic centimeters does the cylinder hold?

Check the measurements. Both lengths are positive and in centimeters, so each volume will be in cubic centimeters. The two solids share the same radius and the same perpendicular height.
$r = 3 \text{ cm}, \quad h = 8 \text{ cm}$
Find the cylinder's volume with $V = \pi r^2 h$ . Square the radius first, $3^2 = 9$ , then multiply by the height, $9 \cdot 8 = 72$ . Keep $\pi$ as a symbol so the value stays exact.
$V_{\text{cylinder}} = \pi (3)^2 (8) = \pi (9)(8) = 72\pi \text{ cm}^3$
Find the cone's volume with $V = \frac{1}{3}\pi r^2 h$ . The radius and height are the same, so $\pi r^2 h$ is $72\pi$ again. Take one third of it: $72 \div 3 = 24$ .
$V_{\text{cone}} = \frac{1}{3}\pi (3)^2 (8) = \frac{1}{3}(72\pi) = 24\pi \text{ cm}^3$
Compare by dividing. The cone's volume is positive, so it is not zero and you can divide by it. Both volumes are in cubic centimeters and both contain the nonzero factor $\pi$ , so the units and $\pi$ divide out and leave a plain number.
$\frac{72\pi \text{ cm}^3}{24\pi \text{ cm}^3} = \frac{72}{24} = 3$
Compare by subtracting. Both volumes are in cubic centimeters, so the difference is too. For a decimal, replace $\pi$ with $3.14$ : $48 \cdot 3.14 = 150.72$ , which rounds to $151$ . That decimal is an approximation, so use $\approx$ .
$72\pi - 24\pi = 48\pi \text{ cm}^3 \approx 151 \text{ cm}^3$

Answer

The cylinder's volume, $72\pi \text{ cm}^3$ , is exactly $3$ times the cone's volume, $24\pi \text{ cm}^3$ . The cylinder holds $48\pi \text{ cm}^3$ more, which is about $151 \text{ cm}^3$ .

Worked example

A sphere with its matching cylinder and cone

A sphere has radius $r = 6$ cm. A cylinder and a cone each have radius $6$ cm and height $h = 12$ cm, measured perpendicular to the base. Compare the three volumes.

Check that the dimensions match. All lengths are positive and in centimeters, and all three solids share the radius $6$ cm. The sphere's diameter is $d = 2r$ , and it equals the shared height of the cylinder and the cone.
$d = 2(6) = 12 \text{ cm} = h$
Sphere: use $V = \frac{4}{3}\pi r^3$ . Cube the radius, $6 \cdot 6 \cdot 6 = 216$ . Then multiply by $\frac{4}{3}$ : $216 \div 3 = 72$ and $72 \cdot 4 = 288$ .
$V_{\text{sphere}} = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi \text{ cm}^3$
Cylinder: use $V = \pi r^2 h$ , with $6^2 = 36$ and $36 \cdot 12 = 432$ .
$V_{\text{cylinder}} = \pi (6)^2 (12) = \pi (36)(12) = 432\pi \text{ cm}^3$
Cone: use $V = \frac{1}{3}\pi r^2 h$ . Its radius and height equal the cylinder's, so take one third of $432\pi$ : $432 \div 3 = 144$ .
$V_{\text{cone}} = \frac{1}{3}\pi (6)^2 (12) = \frac{1}{3}(432\pi) = 144\pi \text{ cm}^3$
Write the volumes as a ratio, cone to sphere to cylinder. Then divide every part by $144\pi$ . It is positive, so it is not zero, and dividing every part of a ratio by the same nonzero number gives an equal ratio: $288 \div 144 = 2$ and $432 \div 144 = 3$ .
$144\pi : 288\pi : 432\pi = 1 : 2 : 3$

Answer

The cone has volume $144\pi \text{ cm}^3$ , the sphere $288\pi \text{ cm}^3$ , and the cylinder $432\pi \text{ cm}^3$ . The volumes are in the ratio $1 : 2 : 3$ , so the sphere has $\frac{2}{3}$ of the cylinder's volume and twice the cone's volume.

A matching radius is not enough. Suppose a sphere has radius $3$ cm. A cylinder has radius $3$ cm and height $10$ cm, measured perpendicular to its base. A tempting shortcut says: the radii match, so the sphere must be two thirds of the cylinder. The cylinder's volume is $\pi(3)^2(10) = 90\pi \text{ cm}^3$ , so the shortcut gives $\frac{2}{3}(90\pi) = 60\pi \text{ cm}^3$ for the sphere. But the sphere formula gives $\frac{4}{3}\pi(3)^3 = 36\pi \text{ cm}^3$ .

The shortcut fails because $\frac{2}{3}$ came from replacing $h$ with $2r$ . Here $2r = 6$ cm, but $h = 10$ cm. This cylinder is taller than the sphere, so the sphere takes up less than two thirds of it. Use the pattern only when the radii match and the cylinder's height equals the sphere's diameter.

Worked example

Same radius, but the height is not the diameter

A sphere has diameter $6$ inches. A cylinder and a cone each have diameter $6$ inches and height $4$ inches, measured perpendicular to the base. The cone's slant height is $5$ inches. Compare the three volumes.

Find the radius. Every formula uses $r$ , and $d = 2r$ , so halve the diameter. All three solids share this radius. Every length is positive and in inches, so each volume will be in cubic inches, $\text{in}^3$ .
$r = 6 \div 2 = 3 \text{ in.}$
Choose the height. Use the perpendicular height, $4$ inches, not the slant height, $5$ inches. Then test the shortcut: the sphere's diameter is $6$ inches, but the shared height is $4$ inches. They differ, so you cannot use the $1 : 2 : 3$ pattern. Calculate each volume.
$h = 4 \text{ in.}, \quad 2r = 6 \text{ in.}, \quad h \ne 2r$
Sphere: $V = \frac{4}{3}\pi r^3$ . Here $3^3 = 27$ , then $27 \div 3 = 9$ and $9 \cdot 4 = 36$ .
$V_{\text{sphere}} = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi (27) = 36\pi \text{ in}^3$
Cylinder: $V = \pi r^2 h$ , with $3^2 = 9$ and $9 \cdot 4 = 36$ .
$V_{\text{cylinder}} = \pi (3)^2 (4) = \pi (9)(4) = 36\pi \text{ in}^3$
Cone: $V = \frac{1}{3}\pi r^2 h$ . It shares the cylinder's radius and height, so take one third of $36\pi$ : $36 \div 3 = 12$ .
$V_{\text{cone}} = \frac{1}{3}\pi (3)^2 (4) = \frac{1}{3}(36\pi) = 12\pi \text{ in}^3$
Compare. The sphere and the cylinder have equal volumes. Divide either one by the cone's volume, which is positive and so not zero. The units and $\pi$ divide out.
$\frac{36\pi \text{ in}^3}{12\pi \text{ in}^3} = \frac{36}{12} = 3$

Answer

The sphere and the cylinder each have volume $36\pi \text{ in}^3$ , and the cone has volume $12\pi \text{ in}^3$ . The sphere's volume equals the cylinder's, and each is $3$ times the cone's. Here the ratio of cone to sphere to cylinder is $1 : 3 : 3$ , not $1 : 2 : 3$ .

Takeaway. To compare a cylinder, a cone, and a sphere, first check the lengths: each is positive, all are in the same unit, the radius is half the diameter, and the height is measured perpendicular to the base, never along a slant. Then find each volume with its own formula and compare by dividing or subtracting.

When a cone and a cylinder share a radius and a perpendicular height, the cone has exactly $\frac{1}{3}$ of the cylinder's volume. When that height also equals the diameter of a sphere with the same radius, the three volumes are in the ratio $1 : 2 : 3$ , cone to sphere to cylinder. When the dimensions do not match, let the formulas decide.