How much water fits in a cylindrical tank? How much ice cream sits in a cone? How much air is inside a ball? All of these questions ask for a volume. In this chapter you will learn three powerful formulas, see how they connect, and use them to solve real problems about everyday objects.
1. What volume measures
The volume of a solid is the amount of space inside it. We measure it by counting how many unit cubes fit inside. That is why volume is written in cubic units: cubic centimeters (\(\text{cm}^3\)), cubic meters (\(\text{m}^3\)), cubic inches (\(\text{in}^3\)) or cubic feet (\(\text{ft}^3\)).
You already know the volume of a rectangular prism: \(V = \text{(area of the base)} \times \text{height}\). A cylinder works the same way, and the cone and the sphere are connected to the cylinder in a surprising way.
2. Volume of a cylinder
A cylinder has two identical circular bases. If you stack many thin circular slices, each with the same area, you build the whole solid. So the volume is the area of the circular base times the height.
For a cylinder with base radius \(r\) and height \(h\):
\[ V = \pi r^2 h \]
The base area is \(B = \pi r^2\), so \(V = B \times h\).
Find the volume of the cylinder above: \(r = 4\) cm and \(h = 9\) cm.
\(V = \pi \times 4^2 \times 9 = \pi \times 16 \times 9 = 144\pi\). The exact volume is \(144\pi \text{ cm}^3\). With \(\pi \approx 3.14\), \(V \approx 452.16 \text{ cm}^3\).
The formulas need the radius, which is half of the diameter. If a problem gives you a diameter of 10 inches, use \(r = 5\) inches. Forgetting this step is the most common mistake in this chapter.
3. Volume of a cone
A cone has one circular base and comes to a point called the apex. Its height \(h\) is measured straight down from the apex to the base, not along the slanted side.
For a cone with base radius \(r\) and height \(h\):
\[ V = \dfrac{1}{3}\,\pi r^2 h \]
A paper cone has a base diameter of 10 inches and a height of 12 inches. The radius is \(r = 10 \div 2 = 5\) inches.
\(V = \dfrac{1}{3} \times \pi \times 5^2 \times 12 = \dfrac{1}{3} \times 300\pi = 100\pi\). So \(V = 100\pi \text{ in}^3 \approx 314 \text{ in}^3\) using \(\pi \approx 3.14\).
4. Cones and cylinders are related
Picture a cone and a cylinder that have the same base and the same height. If you fill the cone with sand and pour it into the cylinder, you will need to do it exactly three times to fill the cylinder.
\(V_{\text{cone}} = \dfrac{1}{3}\,V_{\text{cylinder}}\), or equivalently \(V_{\text{cylinder}} = 3 \times V_{\text{cone}}\).
This explains the \(\dfrac{1}{3}\) in the cone formula. It also gives you a quick check: a cone must always have a smaller volume than the cylinder that surrounds it.
5. Volume of a sphere
A sphere is the set of all points in space at the same distance \(r\) from a center. Its volume depends only on the radius.
For a sphere of radius \(r\):
\[ V = \dfrac{4}{3}\,\pi r^3 \]
A hemisphere is half a sphere, so \(V_{\text{hemisphere}} = \dfrac{2}{3}\,\pi r^3\).
A spherical water tank has a radius of 6 meters.
\(V = \dfrac{4}{3} \times \pi \times 6^3 = \dfrac{4}{3} \times 216\pi = 288\pi\). The volume is \(288\pi \text{ m}^3 \approx 904.32 \text{ m}^3\).
In the sphere formula the radius is cubed (\(r^3 = r \times r \times r\)), not squared and not multiplied by 3. Compute \(r^3\) first, then multiply.
6. Finding a missing dimension
Sometimes the volume is known and a radius or a height is missing. Then you write the formula, substitute what you know, and solve the equation.
- Write the correct formula for the solid.
- Substitute every known value, including the volume.
- Simplify the numbers, then isolate the unknown (divide, or multiply by 3 or \(\tfrac{3}{4}\), then take a square root or cube root if needed).
- Check your answer by putting it back in the formula, and write the unit.
A cylinder has volume \(360\pi \text{ cm}^3\) and radius 6 cm. Find its height.
\(\pi \times 6^2 \times h = 360\pi\), so \(36h = 360\) and \(h = 10\) cm. Check: \(36 \times 10 = 360\). The height is 10 cm.
On my planet we always cancel \(\pi\) first when it appears on both sides of the equation. It makes the arithmetic much lighter, and the exact answer is often a whole number!
7. Composite solids
A composite solid is built from several simple solids. To find its volume, split it into cylinders, cones and spheres (or parts of them), compute each volume, then add. If a part is hollowed out, subtract it.
A silo is a cylinder with radius 3 m and height 8 m, topped by a hemisphere of the same radius.
Cylinder: \(\pi \times 3^2 \times 8 = 72\pi\). Hemisphere: \(\dfrac{2}{3} \times \pi \times 3^3 = 18\pi\). Total: \(72\pi + 18\pi = 90\pi \approx 282.6 \text{ m}^3\).
8. Volume in the real world
Real problems often ask for a capacity in liters or gallons, or for the number of trips needed to move a material. Keep these conversions in mind:
- \(1 \text{ mL} = 1 \text{ cm}^3\) and \(1 \text{ L} = 1{,}000 \text{ cm}^3\).
- \(1 \text{ m}^3 = 1{,}000 \text{ L}\).
- \(1 \text{ gallon} = 231 \text{ in}^3\), and \(1 \text{ ft}^3 \approx 7.48\) gallons.
Always make sure all lengths are in the same unit before you apply a formula, and decide whether the question needs an exact answer in terms of \(\pi\) or a rounded decimal. When you round answers about counting objects (trucks, bottles, buckets), think about whether you must round up.
Key takeaways
- Volume is measured in cubic units.
- Cylinder: \(V = \pi r^2 h\). Cone: \(V = \dfrac{1}{3}\pi r^2 h\). Sphere: \(V = \dfrac{4}{3}\pi r^3\).
- A cone with the same base and height as a cylinder has exactly one third of its volume.
- Use the radius, never the diameter, in every formula.
- To find a missing dimension, substitute the known values and solve the equation.
- For a composite solid, add the volumes of the parts, and subtract any hollow part.
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