Volume of 3-D Objects

Volume is the amount of space an object takes up, and capacity is how much a container can hold. You need them to know how much juice is in a box, how much water fills a fish tank, or how much concrete to order for a project. Best of all, one big idea covers prisms and cylinders, and a second one (“one third”) covers pyramids and cones.

Key ideas

Prisms and cylinders: base area times height

A prism or a cylinder has the same cross-section all the way through, like a stack of identical slices. So its volume is the area of the base times the height:

V = (\text{area of base}) \times (\text{height}) = Bh

Object	Base	Volume
Rectangular prism, $l$ by $w$ by $h$	rectangle, $B = lw$	$V = lwh$
Triangular prism, length $L$	triangle, $B = \frac{1}{2}bh_{\triangle}$	$V = \frac{1}{2}bh_{\triangle} \times L$
Cylinder, radius $r$ , height $h$	circle, $B = \pi r^2$	$V = \pi r^2 h$

For a triangular prism, be careful: there are two “heights”. $h_{\triangle}$ is the height of the triangle, and $L$ is the length of the prism.

Pyramids and cones: one third

Picture a hollow pyramid and a hollow prism with the same base and the same height. If you fill the pyramid with water or sand and pour it into the prism, it takes exactly three pyramids to fill the prism. The same is true for a cone and a cylinder with the same base and height.

A pyramid holds one third as much as a prism with the same base and height. A cone holds one third as much as the matching cylinder.

V_{\text{pyramid}} = \frac{1}{3} B h \qquad\qquad V_{\text{cone}} = \frac{1}{3}\pi r^2 h

For a square-based pyramid with base side $b$ , $B = b^2$ , so $V = \dfrac{1}{3} b^2 h$ .

The $h$ here is the height, measured straight down from the apex to the base, not the slant height. If you’re given the slant height, use the Pythagorean theorem to find the height first.

Spheres (extension)

A sphere with radius $r$ has volume

V = \frac{4}{3}\pi r^3

Where this comes from. The Greek mathematician Archimedes (around 250 BCE) discovered that a sphere that fits snugly inside a cylinder has exactly two thirds of the cylinder’s volume. He is said to have been so proud of it that he asked for a sphere and cylinder to be carved on his tomb. You can check his result in Practice question 8. Much earlier, Egyptian and Babylonian scribes were already calculating the volumes of pyramid-shaped and other solid shapes for building and grain storage.

Units and capacity

Volume is measured in cubic units, such as $\text{cm}^3$ or $\text{m}^3$ . To change to capacity units, use

1 \text{ cm}^3 = 1 \text{ mL} \qquad 1000 \text{ cm}^3 = 1 \text{ L} \qquad 1 \text{ m}^3 = 1000 \text{ L}

(See measurement conversions for more.) Make sure all the lengths are in the same unit before you multiply.

Composite objects

For an object made of simpler pieces, find the volume of each piece and add them. For an object with a piece cut out (like a hole), find the volume of the whole thing and subtract the missing piece.

On the SAT

The SAT reference sheet lists the volume formulas for rectangular prisms, cylinders, spheres, cones, and pyramids, so you don’t need to memorize them, only to pick the right one and identify the radius and height. Keep $\pi$ in your answer when the choices are in terms of $\pi$ , and use Desmos (type pi) only for decimal choices. A common SAT question is about scaling: if every dimension is multiplied by $k$ , the volume is multiplied by $k^3$ (so doubling every dimension multiplies the volume by $8$ ). See using Desmos on the SAT.

Worked examples

Example 1: Prisms

(a) A juice box is $6$ cm long, $4$ cm wide and $10$ cm tall. How many millilitres does it hold?
(b) A chocolate bar comes in a triangular prism box. Each triangular end has a base of $4$ cm and a height of $3.5$ cm, and the box is $20$ cm long. Find its volume.

Solution.

(a)

V = lwh = 6 \times 4 \times 10 = 240 \text{ cm}^3

Since $1 \text{ cm}^3 = 1 \text{ mL}$ , it holds $240$ mL.

(b) First the area of the triangular base, then times the length:

\begin{aligned} B &= \frac{1}{2} \times 4 \times 3.5 = 7 \text{ cm}^2 \\ V &= B \times L = 7 \times 20 = 140 \text{ cm}^3 \end{aligned}

Example 2: A cylinder

A can of soup has a radius of $3.5$ cm and a height of $10$ cm. Find its volume, and its capacity to the nearest millilitre.

Solution.

\begin{aligned} V &= \pi r^2 h \\ &= \pi (3.5)^2 (10) \\ &= 122.5\pi \\ &\approx 384.8 \text{ cm}^3 \end{aligned}

The can holds about $385$ mL.

Example 3: The one-third relationship

(a) A cylinder and a cone both have a radius of $3$ cm and a height of $10$ cm. Find both volumes, to one decimal place. How many cones of water would fill the cylinder?
(b) A square-based pyramid has a base side of $6$ m and a height of $4$ m. Find its volume.

Solution.

(a)

\begin{aligned} V_{\text{cylinder}} &= \pi(3)^2(10) = 90\pi \approx 282.7 \text{ cm}^3 \\ V_{\text{cone}} &= \frac{1}{3}\pi(3)^2(10) = 30\pi \approx 94.2 \text{ cm}^3 \end{aligned}

Since $90\pi \div 30\pi = 3$ , it takes $3$ cones of water to fill the cylinder.

(b) The base is a square, so $B = 6^2 = 36 \text{ m}^2$ :

V = \frac{1}{3} B h = \frac{1}{3} \times 36 \times 4 = 48 \text{ m}^3

Check: a prism with the same base and height would be $36 \times 4 = 144 \text{ m}^3$ , and $144 \div 3 = 48$ . ✓

Example 4: A composite object

A grain silo is a cylinder with radius $2$ m and height $6$ m, topped by a cone with the same radius and a height of $1.5$ m. Find the total volume, and the capacity in litres.

Solution. Find each piece, then add.

\begin{aligned} V_{\text{cylinder}} &= \pi(2)^2(6) = 24\pi \text{ m}^3 \\ V_{\text{cone}} &= \frac{1}{3}\pi(2)^2(1.5) = 2\pi \text{ m}^3 \\ V_{\text{total}} &= 24\pi + 2\pi = 26\pi \approx 81.68 \text{ m}^3 \end{aligned}

Since $1 \text{ m}^3 = 1000 \text{ L}$ , the silo holds about $81.68 \times 1000 = 81\,680$ L (roughly $81\,700$ L).

Common mistakes

Forgetting the one third. Pyramids and cones hold one third as much as the matching prism or cylinder. If you leave out the $\dfrac{1}{3}$ , your answer will be three times too big.

Using the slant height as the height. The volume formulas use the height, measured straight down from the apex to the base. If you’re given the slant height, find the height with the Pythagorean theorem first (Practice question 9 does this).

Using the diameter in place of the radius. If a cup is $7$ cm across, $r = 3.5$ cm. Using $7$ in $\pi r^2 h$ makes the answer four times too big.

Squaring the wrong thing. In $\pi r^2 h$ , only $r$ is squared. Work it out in order: square the radius, then multiply by $\pi$ and by the height.

Mixing units. Change all the lengths to the same unit before you multiply. Then remember $1 \text{ cm}^3 = 1 \text{ mL}$ and $1 \text{ m}^3 = 1000 \text{ L}$ (not $1$ L).

Writing square units. Volume is in cubic units, like $\text{cm}^3$ or $\text{m}^3$ , because you multiplied three lengths.

Practice

1. (Warm-up) Find the volume of a cube with $5$ cm edges. How many millilitres would it hold?

Solution

V = 5^3 = 125 \text{ cm}^3

That’s $125$ mL.

2. (Warm-up) A rectangular prism is $12$ cm by $5$ cm by $4$ cm.

(a) Find its volume.
(b) Find the volume of a pyramid with the same base and height.

Solution

(a) $V = 12 \times 5 \times 4 = 240 \text{ cm}^3$ .

(b) The pyramid is one third of the prism: $240 \div 3 = 80 \text{ cm}^3$ .

3. (Core) A cylinder has a radius of $5$ cm and a height of $8$ cm. Find its volume and the volume of a cone with the same radius and height, both to one decimal place.

Solution

V_{\text{cylinder}} = \pi(5)^2(8) = 200\pi \approx 628.3 \text{ cm}^3

V_{\text{cone}} = \frac{1}{3} \times 200\pi \approx 209.4 \text{ cm}^3

4. (Core) A fish tank is $80$ cm long, $35$ cm wide and $40$ cm tall. How many litres of water does it hold when full?

Solution

V = 80 \times 35 \times 40 = 112\,000 \text{ cm}^3

That’s $112\,000$ mL. Divide by $1000$ : $112$ L.

5. (Core) A paper water cup is a cone $7$ cm across the top and $9$ cm deep. How much water does it hold, to the nearest millilitre?

Solution

The radius is half the width: $r = 3.5$ cm.

\begin{aligned} V &= \frac{1}{3}\pi r^2 h \\ &= \frac{1}{3}\pi(3.5)^2(9) \\ &= 36.75\pi \\ &\approx 115.45 \text{ cm}^3 \end{aligned}

The cup holds about $115$ mL.

6. (Core) A company wants a cylindrical can that holds exactly $1$ L, with a radius of $5$ cm. How tall must it be, to the nearest tenth of a centimetre?

Solution

$1 \text{ L} = 1000 \text{ cm}^3$ , so

\begin{aligned} \pi(5)^2 h &= 1000 \\ 25\pi h &= 1000 \\ h &= \frac{1000}{25\pi} \approx 12.7 \text{ cm} \end{aligned}

Check: $\pi(25)(12.73) \approx 1000$ . ✓

7. (Core) The Great Pyramid of Giza in Egypt was built with a square base about $230$ m on each side and a height of about $146$ m. Estimate its volume, to the nearest hundred thousand cubic metres.

Solution

\begin{aligned} V &= \frac{1}{3} b^2 h \\ &= \frac{1}{3} \times 230^2 \times 146 \\ &= \frac{1}{3} \times 52\,900 \times 146 \\ &\approx 2\,574\,467 \text{ m}^3 \end{aligned}

That’s about $2\,600\,000 \text{ m}^3$ (roughly $2.6$ million cubic metres).

8. (Challenge) A ball with a radius of $12$ cm fits snugly inside a cylindrical can: the can’s radius is $12$ cm and its height is $24$ cm (the ball’s diameter).

(a) Find the volume of the ball and of the can, to one decimal place.
(b) What fraction of the can does the ball fill?

Solution

(a)

V_{\text{ball}} = \frac{4}{3}\pi(12)^3 = \frac{4}{3}\pi(1728) = 2304\pi \approx 7238.2 \text{ cm}^3

V_{\text{can}} = \pi(12)^2(24) = 3456\pi \approx 10\,857.3 \text{ cm}^3

(b)

\frac{2304\pi}{3456\pi} = \frac{2}{3}

The ball fills two thirds of the can, which is exactly what Archimedes discovered.

9. (Challenge) A pile of gravel is shaped like a cone. Its base has a radius of $3$ m, and the slant height (from the top of the pile straight down the side to the ground) is $5$ m.

(a) Find the height of the pile.
(b) Find the volume of gravel, to one decimal place.
(c) A truck carries $8 \text{ m}^3$ per load. How many loads are needed to move the whole pile?

Solution

(a) The height, radius and slant height form a right triangle, with the slant height as the hypotenuse:

h^2 = 5^2 - 3^2 = 25 - 9 = 16 \quad\Rightarrow\quad h = 4 \text{ m}

(b)

V = \frac{1}{3}\pi(3)^2(4) = 12\pi \approx 37.7 \text{ m}^3