How to Find the Volume of Cylinders, Cones, and Spheres? (+FREE Worksheet!)

Three-dimensional shapes are everywhere—cans in your kitchen, traffic cones on the road, and soccer balls on the field. In 8th-grade math you learn the formulas for calculating the volume of three essential solids: cylinders, cones, and spheres. Once you memorize the three formulas and understand how they relate to each other, you can solve any volume problem that comes your way.

This guide walks you through each formula, explains where it comes from, provides worked examples, and gives you plenty of practice with full solutions.

The Three Volume Formulas

Bonus — Hemisphere: \(V = \dfrac{2}{3}\pi r^{3}\) (exactly half of a sphere).

Understanding the Relationships

Notice how the formulas connect:

• A cone is exactly \(\frac{1}{3}\) the volume of a cylinder with the same radius and height. It takes 3 cones to fill the cylinder.
• A sphere with radius \(r\) has the same volume as a cylinder with radius \(r\) and height \(\frac{4}{3}r\). Alternatively, a sphere fits inside a cylinder of height \(2r\), and the sphere fills exactly \(\frac{2}{3}\) of that cylinder.

Step-by-Step Guide

1. Identify the shape (cylinder, cone, or sphere).
2. Write the correct formula.
3. Identify \(r\) and \(h\) from the problem. If you are given the diameter, divide by 2 to get the radius.
4. Substitute the values into the formula.
5. Compute. Leave the answer in terms of \(\pi\) or use \(\pi \approx 3.14\) as directed.
6. Include units (cubic inches, cubic cm, etc.).

Worked Examples

Example 1 — Cylinder

Find the volume of a cylinder with radius 5 cm and height 10 cm.

\(V = \pi r^{2} h = \pi (5)^{2}(10) = \pi (25)(10) = 250\pi \approx 785 \text{ cm}^{3}\)

Example 2 — Cone

Find the volume of a cone with radius 3 in and height 7 in.

\(V = \dfrac{1}{3}\pi r^{2} h = \dfrac{1}{3}\pi (3)^{2}(7) = \dfrac{1}{3}\pi (9)(7) = \dfrac{63\pi}{3} = 21\pi \approx 65.97 \text{ in}^{3}\)

Example 3 — Sphere

Find the volume of a sphere with radius 6 m.

\(V = \dfrac{4}{3}\pi r^{3} = \dfrac{4}{3}\pi (6)^{3} = \dfrac{4}{3}\pi (216) = 288\pi \approx 904.78 \text{ m}^{3}\)

Example 4 — Hemisphere

Find the volume of a hemisphere with diameter 12 ft.

Radius \(= 6\) ft.

\(V = \dfrac{2}{3}\pi r^{3} = \dfrac{2}{3}\pi (6)^{3} = \dfrac{2}{3}\pi (216) = 144\pi \approx 452.39 \text{ ft}^{3}\)

Example 5 — Given Diameter

A cylindrical can has a diameter of 8 cm and height of 12 cm. Find the volume.

Radius \(= \frac{8}{2} = 4\) cm.

\(V = \pi (4)^{2}(12) = 192\pi \approx 603.19 \text{ cm}^{3}\)

Video Lesson

Watch this video for additional examples and a step-by-step walkthrough:

Solving for a Missing Dimension

Sometimes the volume is given and you need to find the radius or height.

Example 6 — Find the height of a cylinder

A cylinder has volume \(100\pi\) cm³ and radius 5 cm. Find \(h\).

\(100\pi = \pi (5)^{2} h \Rightarrow 100\pi = 25\pi h \Rightarrow h = 4 \text{ cm}\)

Example 7 — Find the radius of a sphere

A sphere has volume \(36\pi\) in³. Find \(r\).

\(36\pi = \dfrac{4}{3}\pi r^{3} \Rightarrow 36 = \dfrac{4}{3}r^{3} \Rightarrow 27 = r^{3} \Rightarrow r = 3 \text{ in}\)

Practice Problems

1. Find the volume of a cylinder: \(r = 4\) cm, \(h = 9\) cm.
2. Find the volume of a cone: \(r = 6\) in, \(h = 10\) in.
3. Find the volume of a sphere: \(r = 3\) m.
4. A hemisphere has radius 8 cm. Find its volume.
5. A cylinder has diameter 10 ft and height 7 ft. Find the volume.
6. A cone has diameter 12 cm and height 15 cm. Find the volume.
7. A sphere has diameter 14 in. Find its volume.
8. A cylinder has volume \(200\pi\) cm³ and radius 5 cm. Find the height.
9. A cone has volume \(48\pi\) in³ and height 9 in. Find the radius.
10. Compare: a cone and a cylinder both have \(r = 4\) and \(h = 6\). How many times larger is the cylinder?
11. A tennis ball has a radius of about 3.3 cm. What is the volume of one ball?
12. An ice cream cone (cone shape) has \(r = 2\) in and \(h = 5\) in, topped with a hemisphere of ice cream (\(r = 2\) in). What is the total volume?

Solutions

1. \(V = \pi(4)^{2}(9) = 144\pi \approx 452.39\) cm³
2. \(V = \frac{1}{3}\pi(6)^{2}(10) = 120\pi \approx 376.99\) in³
3. \(V = \frac{4}{3}\pi(3)^{3} = 36\pi \approx 113.10\) m³
4. \(V = \frac{2}{3}\pi(8)^{3} = \frac{1024\pi}{3} \approx 1072.33\) cm³
5. \(r = 5\). \(V = \pi(5)^{2}(7) = 175\pi \approx 549.78\) ft³
6. \(r = 6\). \(V = \frac{1}{3}\pi(6)^{2}(15) = 180\pi \approx 565.49\) cm³
7. \(r = 7\). \(V = \frac{4}{3}\pi(7)^{3} = \frac{1372\pi}{3} \approx 1436.76\) in³
8. \(200\pi = \pi(25)h \Rightarrow h = 8\) cm
9. \(48\pi = \frac{1}{3}\pi r^{2}(9) = 3\pi r^{2} \Rightarrow r^{2} = 16 \Rightarrow r = 4\) in
10. Cylinder: \(\pi(16)(6) = 96\pi\). Cone: \(\frac{1}{3}(96\pi) = 32\pi\). The cylinder is 3 times larger.
11. \(V = \frac{4}{3}\pi(3.3)^{3} \approx \frac{4}{3}\pi(35.937) \approx 150.53\) cm³
12. Cone: \(\frac{1}{3}\pi(4)(5) = \frac{20\pi}{3}\). Hemisphere: \(\frac{2}{3}\pi(8) = \frac{16\pi}{3}\). Total: \(\frac{36\pi}{3} = 12\pi \approx 37.70\) in³

Real-World Applications

• Packaging: Companies calculate cylinder volume to determine how much soup, soda, or paint fits in a can.
• Construction: Traffic cones, grain silos (cylinders), and dome roofs (hemispheres) all require volume calculations for materials and capacity.
• Sports: Knowing the volume of a basketball (\(V = \frac{4}{3}\pi r^{3}\)) helps manufacturers determine how much air it holds.

Common Mistakes to Avoid

• Using diameter instead of radius. Always divide the diameter by 2 before substituting into a formula.
• Forgetting \(\frac{1}{3}\) for the cone. A cone is one-third of a cylinder, not the same.
• Cubing vs. squaring the radius. Cylinder and cone formulas use \(r^{2}\); the sphere formula uses \(r^{3}\). Don’t mix them up.
• Dropping the units. Volume is always in cubic units (cm³, in³, m³).

Frequently Asked Questions

What if the shape is a combination of solids?

Break it into individual shapes, calculate each volume separately, then add (or subtract, if one is hollow) the results.

When should I leave the answer in terms of \(\pi\)?

Leave it in terms of \(\pi\) unless the problem says “use \(\pi \approx 3.14\)” or “round to the nearest tenth.” Exact answers in terms of \(\pi\) are more precise.