Math Core

A waffle cone, a funnel and a traffic cone all come to a point. A cone has one circular base and a curved side that narrows to a single point, called the vertex. Its volume turns out to be closely tied to the cylinder you just studied.

A cone is one third of a cylinder

Picture a cone and a cylinder with the same base and the same height. Fill the cone with sand and pour it into the cylinder. It takes exactly three full cones to fill the cylinder.

So the cone holds one third as much:

V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} = \frac{1}{3}\pi r^2 h

This makes sense if you look at the shapes. The cylinder is just as wide at the top as at the bottom. The cone keeps getting narrower, so it leaves a lot of empty space near the top.

A cone with radius r and height h. The height goes straight down from the vertex to the center of the base.

Volume of a cone

V = \frac{1}{3}\pi r^2 h

where $r$ is the radius of the base and $h$ is the height, measured straight down from the vertex to the center of the base.

Worked example: A basic cone

A cone has a radius of 3 m and a height of 5 m. Find its volume in terms of $\pi$ and to the nearest tenth.

\begin{aligned} V &= \tfrac{1}{3}\pi r^2 h \\ &= \tfrac{1}{3} \cdot \pi \cdot 3^2 \cdot 5 \\ &= \tfrac{1}{3} \cdot 45\pi \\ &= 15\pi \approx 47.1 \end{aligned}

The volume is $15\pi$ , or about $47.1$ cubic meters. Notice that a cylinder with the same radius and height holds $45\pi$ , which is three times as much.

Worked example: Starting from the diameter

A paper cup shaped like a cone is 12 cm across the top and 7 cm deep. Find its volume in terms of $\pi$ .

The radius is $12 \div 2 = 6$ cm.

V = \tfrac{1}{3} \cdot \pi \cdot 6^2 \cdot 7 = \tfrac{1}{3} \cdot 252\pi = 84\pi

The cup holds $84\pi$ cubic centimeters.

Tip

Multiply in whichever order is easiest. In the last example, $\tfrac{1}{3} \cdot 36 = 12$ first, then $12 \cdot 7 = 84$ .

Height versus slant height

The slant height runs along the side of the cone, from the vertex to the edge of the base. It is not the height you use in the formula.

Look at the figure again. The height, the radius and the slant height form a right triangle. The slant height is the hypotenuse. So if a problem gives the slant height, use the Pythagorean theorem to find the height first:

r^2 + h^2 = (\text{slant height})^2

Worked example: Using the Pythagorean theorem

A cone has a radius of 5 inches and a slant height of 13 inches. Find its volume in terms of $\pi$ .

Step 1: find the height.

\begin{aligned} 5^2 + h^2 &= 13^2 \\ 25 + h^2 &= 169 \\ h^2 &= 144 \\ h &= 12 \end{aligned}

Step 2: find the volume.

V = \tfrac{1}{3} \cdot \pi \cdot 5^2 \cdot 12 = \tfrac{1}{3} \cdot 300\pi = 100\pi

The volume is $100\pi$ cubic inches.

Common mistake

Don't put the slant height into the volume formula. In the last example, using 13 instead of 12 gives $\tfrac{1}{3} \cdot 25 \cdot 13\,\pi \approx 108.3\pi$ , which is wrong. The height must go straight down, making a right angle with the base.

Working backward

Worked example: Finding a missing height

A cone has a volume of $48\pi$ cubic feet and a radius of 4 feet. What is its height?

\begin{aligned} \tfrac{1}{3} \cdot \pi \cdot 4^2 \cdot h &= 48\pi \\ \tfrac{16}{3}\pi h &= 48\pi \\ h &= 48 \cdot \tfrac{3}{16} = 9 \end{aligned}

The height is 9 feet. Check: $\tfrac{1}{3} \cdot 16 \cdot 9 = 48$ . ✓

Practice

Practice 1

A cone has a radius of 3 cm and a height of 4 cm. Find its volume in cubic centimeters, in terms of $\pi$ .