Volume — Definition, Formula & Examples

Volume

The total amount of space enclosed in a solid.

For the following tables,
h = height of solid	s = slant height	P = perimeter or circumference of the base
l = length of solid	B = area of the base	r = radius of sphere
w = width of solid	R = radius of the base	a = length of an edge

Figure	Volume	Lateral Surface Area	Area of the Base(s)	Total Surface Area
Box (also called rectangular parallelepiped, right rectangular prism)	lwh	2lh + 2wh	2lw	2lw + 2lh + 2wh
Prism	Bh	Ph	2B	Ph + 2B
Pyramid		-	B	-
Right Pyramid			B
Cylinder	Bh	-	2B	-
Right Cylinder	Bh	Ph	2B	Ph + 2B
Right Circular Cylinder	πR²h	2πRh	2πR²	2πRh + 2πR²
Cone		-	B	-
Right Circular Cone		πRs or	πR²	πRs + πR² or

Figure	Volume	Total Surface Area
Sphere
Regular Tetrahedron
Cube (regular hexahedron)	a³	6a²
Regular Octahedron
Regular Dodecahedron
Regular Icosahedron

For the tables above,
h = height of solid	s = slant height	P = perimeter or circumference of the base
l = length of solid	B = area of the base	r = radius of sphere
w = width of solid	R = radius of the base	a = length of an edge

Key Formula

\text{Box: } V = lwh \qquad \text{Prism/Cylinder: } V = Bh \qquad \text{Pyramid/Cone: } V = \tfrac{1}{3}Bh \qquad \text{Sphere: } V = \tfrac{4}{3}\pi r^3

Where:

$V$ = Volume of the solid
$l$ = Length of the solid (for a box)
$w$ = Width of the solid (for a box)
$h$ = Height (perpendicular distance from base to top)
$B$ = Area of the base of the solid
$r$ = Radius of the sphere
$\pi$ = The constant pi, approximately 3.14159

Worked Example

Problem: Find the volume of a right circular cylinder with a radius of 5 cm and a height of 12 cm.

Step 1: Identify the formula for the volume of a right circular cylinder.

V = \pi R^2 h

Step 2: Substitute the given values: radius R = 5 cm and height h = 12 cm.

V = \pi (5)^2 (12)

Step 3: Square the radius and multiply.

V = \pi \cdot 25 \cdot 12 = 300\pi

Step 4: Compute the numerical value.

V \approx 300 \times 3.14159 \approx 942.48 \text{ cm}^3

Answer: The volume of the cylinder is

300\pi \approx 942.48

cm³.

Another Example

This example uses a cone rather than a cylinder, highlighting the critical factor of 1/3 that appears whenever a solid tapers to a point (pyramids and cones) instead of maintaining a constant cross-section.

Problem: A cone-shaped paper cup has a radius of 6 cm and a height of 10 cm. How much water can it hold?

Step 1: Since a cone tapers to a point, its volume is one-third the volume of a cylinder with the same base and height.

V = \tfrac{1}{3}\pi R^2 h

Step 2: Substitute R = 6 cm and h = 10 cm.

V = \tfrac{1}{3}\pi (6)^2 (10)

Step 3: Simplify inside the expression first: 6² = 36, and 36 × 10 = 360.

V = \tfrac{1}{3}\pi \cdot 360 = 120\pi

Step 4: Calculate the decimal approximation.

V \approx 120 \times 3.14159 \approx 376.99 \text{ cm}^3

Answer: The cone holds

120\pi \approx 377.0

cm³ of water.

Frequently Asked Questions

What is the difference between volume and surface area?

Volume measures the three-dimensional space inside a solid and is expressed in cubic units (cm³, m³). Surface area measures the total area covering the outside of the solid and is expressed in square units (cm², m²). Think of volume as how much a container can hold, and surface area as how much wrapping paper you would need to cover it.

Why do pyramids and cones have 1/3 in their volume formulas?

A pyramid or cone tapers from its full base down to a single point, so it occupies less space than a prism or cylinder with the same base and height. Rigorous calculus proofs show the ratio is exactly one-third. An intuitive way to see this: you can fill a cone with water exactly three times to fill a cylinder of the same base and height.

How do you find the volume of an irregular shape?

For irregular shapes, you can use water displacement: submerge the object in water and measure how much the water level rises. The displaced volume equals the object's volume. In advanced math, you can also use integration (calculus) to compute volumes by summing infinitely thin cross-sectional slices.

Volume vs. Surface Area

	Volume	Surface Area
What it measures	The 3D space enclosed inside a solid	The total area of all outer faces/surfaces of a solid
Units	Cubic units (e.g., cm³, m³, ft³)	Square units (e.g., cm², m², ft²)
Box formula	V = lwh	SA = 2lw + 2lh + 2wh
Sphere formula	V = (4/3)πr³	SA = 4πr²
Real-world analogy	How much water a tank holds	How much paint to coat the tank

Why It Matters

Volume appears throughout geometry courses when you study prisms, cylinders, pyramids, cones, and spheres — it is one of the most tested topics on standardized math exams. Beyond the classroom, volume calculations are essential in science (measuring liquid capacity, gas volumes), engineering (sizing tanks and containers), cooking (converting recipe quantities), and architecture (estimating material for concrete or fill).

Common Mistakes

Mistake: Forgetting the 1/3 factor for pyramids and cones.

Correction: Any solid that tapers to a point uses V = (1/3)Bh, not V = Bh. The factor of 1/3 accounts for the tapering shape. A quick check: a cone should hold exactly one-third the volume of a cylinder with the same base and height.

Mistake: Confusing radius and diameter when computing volume.

Correction: Formulas like V = πR²h and V = (4/3)πr³ require the radius, not the diameter. If you are given the diameter, divide it by 2 before substituting. Using the diameter in place of the radius makes the answer 4 times too large (for cylinders) or 8 times too large (for spheres).

Related Terms

Solid — A 3D figure whose volume can be measured
Surface Area — Measures outside area, often studied alongside volume
Prism — A solid whose volume equals base area times height
Cylinder — Circular prism; volume uses πR²h
Pyramid — Tapered solid with volume (1/3)Bh
Cone — Circular pyramid; volume is (1/3)πR²h
Sphere — Volume is (4/3)πr³
Volume by Parallel Cross Sections — Calculus method for finding volume of irregular solids