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Cylinders: Volume and Surface Area

Find the volume and surface area of cans, tanks and pipes, and work backwards to find a cylinder's height, radius or diameter.

Practice this skill 25 min Cylinders Leer en español
Key idea

What you'll learn

A cylinder is the shape of a soup can, a water tank, a candle, a pipe or a round cake pan. In this lesson you will learn to:

  • find the volume of a cylinder (how much it holds),
  • find the surface area (how much material covers it, such as a label or sheet metal),
  • work backwards to find a missing height, radius or diameter.

Why it matters on the GED: Cylinder questions are among the most common 3-D questions on the test. Both cylinder formulas are on the GED formula sheet, but the test checks the details: Did you use the radius, not the diameter? Did you square the radius? Did you include both circular ends? Should the answer be left "in terms of \pi" or rounded with \pi \approx 3.14? Master those details and these become easy points. In daily life, the same math tells you how many gallons a tank holds or how much paper wraps around a can.

Words to know

  • Cylinder — a 3-D shape with two identical, parallel circles (the bases) joined by a curved side, like a can.
  • Base — one of the two circular ends. Its area is B = \pi r^2 (radius 3: B = 9\pi).
  • Radius (r) — the distance from the center of the circle to its edge (half of a 10\text{ cm} diameter is r = 5\text{ cm}).
  • Diameter (d) — the distance all the way across the circle through the center: d = 2r.
  • Height (h) — the straight distance between the two circular bases (how tall the can is).
  • Pi (\pi) — the number 3.14159\ldots, often rounded to 3.14. The GED formula sheet says \pi \approx 3.14.
  • Circumference (C) — the distance around a circle: C = 2\pi r = \pi d (r = 2: C = 4\pi \approx 12.57).
  • Lateral area — the area of the curved side only: 2\pi r h (the label on a can).
  • Surface area (SA) — the lateral area plus both circles: 2\pi r h + 2\pi r^2.
  • Volume (V) — the space inside, in cubic units (\text{in}^3, \text{ft}^3, \text{cm}^3).
  • "In terms of \pi" — leave \pi as a symbol in the answer, like 160\pi, instead of multiplying it out.
A cylinder with its parts labelled: two identical circular bases (top and bottom), the radius r from the centre to the edge, the diameter d = 2r across the bottom, the height h between the bases, and the curved lateral side. radius r diameter d = 2r height h base (circle) curved side base (circle)

All Math words in the glossary

Volume: a stack of circles

Cylinder with radius r drawn from the center of the top circle to its edge, and height h measured between the two circular bases. The diameter d is twice the radius. r h

A cylinder is like a stack of identical coins. Just as with a box, the volume is the area of the base times the height:

V = Bh

The base is a circle, so B = \pi r^2, and

V = \pi r^2 h
A cylinder pictured as a stack of identical circles (like coins): volume = area of one circle B = πr² times the height h. B = πr² h (number of layers) V = Bh = πr²h

Example: radius 4\text{ cm}, height 10\text{ cm}.

Cylinder with radius 4 cm and height 10 cm. Base area 16π, volume 160π ≈ 502.7 cubic cm. r = 4 cm h = 10 cm V = π(4²)(10) = 160π cm³
  1. Square the radius: r^2 = 4^2 = 16.
  2. Base area: B = 16\pi.
  3. Volume: V = 16\pi \times 10 = 160\pi\text{ cm}^3.

Exact or rounded? Read what the question wants:

Instruction in the questionWhat to write for V = 160\pi
"in terms of \pi"160\pi\text{ cm}^3
"use 3.14 for \pi"160 \times 3.14 = 502.4\text{ cm}^3
"to the nearest whole number" (use the \pi key)502.65\ldots \approx 503\text{ cm}^3

Diameter given? Halve it first. A can with diameter 6\text{ in} has r = 3\text{ in}. If you use 6 by mistake, r^2 becomes 36 instead of 9, and the volume comes out four times too big.

Top view of a can 6 inches across: the diameter is 6 in, the radius is half of it, 3 in. Using r = 3 gives r² = 9; using 6 by mistake gives 36, four times too big. d = 6 in r = 3 in r = d ÷ 2 = 3, r² = 9 (not 36)

Order of operations: only the radius is squared. \pi r^2 h means \pi \times (r \times r) \times h, not (\pi r)^2 and not \pi \times 2r \times h.

Scaling: doubling the height doubles the volume, but doubling the radius multiplies the volume by 2^2 = 4, because r is squared.

Surface area: label plus two lids

Net of a cylinder: a rectangle whose width is the circumference 2πr and whose height is the cylinder height h, with one circle of radius r attached above and one below. lateral area = 2πr · h width = circumference = 2πr h r r

Peel the label off a can and lay it flat. It is a rectangle:

  • its height is the height of the can, h;
  • its width is the distance around the can, the circumference 2\pi r.

So the curved side (the lateral area) is 2\pi r \times h = 2\pi r h.

Add the two circular lids, each \pi r^2:

SA = 2\pi r h + 2\pi r^2

Example: a closed can with r = 2\text{ in} and h = 7\text{ in}.

  1. Lateral area: 2\pi(2)(7) = 28\pi.
  2. Two lids: 2\pi(2^2) = 2\pi(4) = 8\pi.
  3. Total: SA = 28\pi + 8\pi = 36\pi \approx 113.1\text{ in}^2.
Net of a closed can with radius 2 in and height 7 in: the label unrolls to a rectangle 2π(2) = 4π ≈ 12.57 in wide and 7 in tall (area 28π), plus two circles of radius 2 (area 4π each). Total 36π ≈ 113.1 square inches. r = 2 in r = 2 in width = 2πr = 4π ≈ 12.57 in h = 7 in label: 4π × 7 = 28π SA = 28π + 2(4π) = 36π ≈ 113.1 in²

Notice: this is the same idea as SA = ph + 2B for prisms. The "perimeter" of a circle is its circumference 2\pi r, and B = \pi r^2.

Which parts are covered? Read the situation:

  • A label covers only the curved side: 2\pi r h.
  • An open tank, cup, or pipe with one end closed: 2\pi r h + \pi r^2 (one lid).
  • A pipe open at both ends: 2\pi r h only.
  • A closed can or tank: the full 2\pi r h + 2\pi r^2.

Working backwards: height, radius or diameter

Write the formula, substitute what you know, and solve.

Missing height (from volume). Divide the volume by the base area:

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

Example: V = 1{,}570\text{ ft}^3, r = 5\text{ ft}, \pi \approx 3.14. Base area = 3.14 \times 25 = 78.5, so h = 1{,}570 \div 78.5 = 20\text{ ft}.

Cylindrical tank with radius 5 ft and unknown height h. Volume 1,570 cubic feet; base area 3.14 × 25 = 78.5, so h = 20 ft. r = 5 ft h = ? V = 1,570 ft³ → h = 1,570 ÷ 78.5 = 20 ft

Missing radius (from volume). Divide by \pi h, then take the square root:

r^2 = \frac{V}{\pi h} \qquad r = \sqrt{\frac{V}{\pi h}}

Example: V = 245\pi\text{ cm}^3, h = 5\text{ cm}. Then r^2 = 245\pi \div 5\pi = 49, so r = \sqrt{49} = 7\text{ cm}, and the diameter is 14\text{ cm}. Do not stop at 49, and check whether the question wants r or d.

Cylinder with unknown radius r and height 5 cm. Volume 245π cubic cm, so r² = 49, r = 7 cm and the diameter is 14 cm. r = ? h = 5 cm V = 245π cm³ → r² = 49, r = 7 cm

Missing height (from surface area). Subtract the two lids first, then divide by 2\pi r.

Example: SA = 170\pi, r = 5. Lids: 2\pi(25) = 50\pi. Then 10\pi h = 170\pi - 50\pi = 120\pi, so h = 12.

Closed cylinder with radius 5 and unknown height h. Surface area 170π: the two lids are 50π, so 10πh = 120π and h = 12. r = 5 h = ? SA = 170π → h = 12

Tip for \pi answers: if both sides contain \pi, divide it out. 10\pi h = 120\pi becomes 10h = 120.

Formula

The formulas (on the GED formula sheet)

Both formulas are printed on the GED formula sheet under cylinder:

V = \pi r^2 h \qquad\qquad SA = 2\pi r h + 2\pi r^2
  • r = radius of the circular base (half the diameter).
  • h = height, the distance between the two bases.
  • \pi \approx 3.14 (the formula sheet gives this value; on the calculator you can use the π key).

Also on the sheet: circle area A = \pi r^2 and circumference C = 2\pi r (or C = \pi d).

Memorize: d = 2r and r = d \div 2. The sheet uses r, but questions often give d. Unit conversions such as gallons per cubic foot will be given in the question.

Worked example

Example 1 (basic): volume in terms of π

A cylinder has radius 5\text{ cm} and height 8\text{ cm}. Find its volume in terms of \pi, then to the nearest tenth.

Cylinder with radius 5 cm and height 8 cm. Its volume is V = πr²h. r = 5 cm h = 8 cm V = πr²h
  1. r^2 = 5^2 = 25.
  2. V = \pi r^2 h = \pi(25)(8) = 200\pi\text{ cm}^3.
  3. With the π key: 200\pi \approx 628.318\ldots \approx 628.3\text{ cm}^3.

Answer: 200\pi \approx 628.3\text{ cm}^3

Check: 628.3 \div (\pi \times 25) \approx 8, the height ✓.

Worked example

Example 2: diameter given, find V and SA

A coffee can has a diameter of 8\text{ in} and a height of 10\text{ in}. Find its volume and its total surface area (closed can). Use \pi \approx 3.14.

Coffee can with diameter 8 inches (radius 4 inches) and height 10 inches. d = 8 in h = 10 in r = 8 ÷ 2 = 4 in
  1. Radius: r = 8 \div 2 = 4\text{ in}.
  2. Volume: V = \pi(4^2)(10) = 160\pi \approx 160 \times 3.14 = 502.4\text{ in}^3.
  3. Lateral area: 2\pi r h = 2\pi(4)(10) = 80\pi.
  4. Two lids: 2\pi r^2 = 2\pi(16) = 32\pi.
  5. SA = 80\pi + 32\pi = 112\pi \approx 112 \times 3.14 = 351.68\text{ in}^2.
Net of the closed coffee can, radius 4 in and height 10 in: a rectangle 2π(4) = 8π ≈ 25.1 in wide and 10 in tall (lateral area 80π) plus two lids of radius 4 (16π each). Total 112π ≈ 351.68 square inches. r = 4 in r = 4 in width = 2πr = 8π ≈ 25.1 in h = 10 in label: 8π × 10 = 80π SA = 80π + 2(16π) = 112π ≈ 351.68 in²

Answer: V \approx 502.4\text{ in}^3, SA \approx 351.68\text{ in}^2

Check: with r = 8 by mistake, V would be 640\pi, exactly 4 times 160\pi, which is why halving the diameter matters ✓.

Worked example

Example 3 (real world): gallons in a tank

A cylindrical rain barrel has radius 2\text{ ft} and height 3\text{ ft}. One cubic foot holds about 7.5 gallons. About how many gallons does the barrel hold?

Rain barrel shaped like a cylinder with radius 2 ft and height 3 ft. Volume 12π ≈ 37.7 cubic feet, about 283 gallons. r = 2 ft h = 3 ft V = π(2²)(3) = 12π ≈ 37.7 ft³
  1. V = \pi r^2 h = \pi(2^2)(3) = 12\pi\text{ ft}^3.
  2. With the π key: 12\pi \approx 37.70\text{ ft}^3.
  3. Gallons: 37.70 \times 7.5 \approx 282.7.

Answer: about 283 gallons

Check: estimate: \pi \approx 3, so V \approx 36\text{ ft}^3 and 36 \times 7.5 = 270 gallons, close to 283 ✓.

Worked example

Example 4 (work backwards): missing height

A cylindrical vase holds 942\text{ in}^3 of water when full. Its radius is 5\text{ in}. How tall is it? Use \pi \approx 3.14.

Cylindrical vase with radius 5 in and unknown height h. It holds 942 cubic inches. r = 5 in h = ? V = 942 in³
  1. Formula: V = \pi r^2 h.
  2. Substitute: 942 = 3.14 \times 5^2 \times h = 3.14 \times 25 \times h = 78.5h.
  3. Divide: h = 942 \div 78.5 = 12.

Answer: 12\text{ in}

Check: 3.14 \times 25 \times 12 = 78.5 \times 12 = 942 ✓.

Worked example

Example 5 (GED-hard): missing diameter

A grain silo is a cylinder 9\text{ m} tall with a volume of 576\pi\text{ m}^3. What is the diameter of the silo?

Grain silo: a cylinder 9 m tall with volume 576π cubic metres and unknown diameter d. d = ? h = 9 m V = 576π m³
  1. Formula: V = \pi r^2 h, so 576\pi = \pi r^2 (9).
  2. Divide both sides by 9\pi: r^2 = 576 \div 9 = 64.
  3. Square root: r = \sqrt{64} = 8\text{ m}.
  4. The question asks for the diameter: d = 2r = 16.

Answer: 16\text{ m}

Check: r = 16 \div 2 = 8; \pi(8^2)(9) = \pi(64)(9) = 576\pi ✓.

Calculator tip

On the TI-30XS MultiView

  • The π key enters \pi. For V = \pi r^2 h with r = 4, h = 10: π × 4 x² × 10 enter → 160\pi; press toggle → 502.6548….
  • Use the x² key right after the radius so you never forget to square it.
  • "Use 3.14" questions: type 3.14 instead of pressing π. The two answers differ slightly (502.65 vs. 502.4), and the answer choices are built from whichever the question says.
  • Square root (finding r): 2nd x² gives \sqrt{\;}. For r^2 = 49: 2nd x² 49 enter → 7.
  • Working backwards in one line: for h = V \div (\pi r^2), use parentheses: 1570 ÷ ( 3.14 × 5 x² ) enter → 20. Without the parentheses the calculator divides by 3.14 only and then multiplies by 25.

When not to use it: "in terms of \pi" answers (160\pi) are usually faster by hand: multiply the numbers and keep \pi as a symbol.

Common mistake

Common mistakes

  • Diameter used as radius. ✗ d = 6, so V = \pi(6^2)(5) = 180\pi. ✓ r = 3: V = \pi(9)(5) = 45\pi. Using d makes the volume 4 times too big.
  • Forgetting to square r. ✗ \pi(4)(10) = 40\pi. ✓ \pi(4^2)(10) = 160\pi.
  • Only one lid, or none. ✗ SA = 2\pi r h + \pi r^2 for a closed can. ✓ A closed can has two circles: +2\pi r^2. (But a label really is just 2\pi r h.)
  • Stopping at r^2. ✗ "r^2 = 49, so r = 49." ✓ Take the square root: r = 7. Then check whether the question wants the diameter, 14.
  • Mixing up volume and surface area. 2\pi r h is an area (label). \pi r^2 h is a volume (contents). Look for the words holds/fills versus covers/wraps.
  • Dropping \pi when working backwards. ✗ 1{,}570 \div 25 = 62.8. ✓ Divide by the whole base area 3.14 \times 25 = 78.5, giving 20.
Test strategy

GED test strategy

  • Circle the word radius or diameter in the question or on the figure before you calculate.
  • Match the answer form. If the choices contain \pi, keep \pi as a symbol. If they are decimals, check whether the question says "use 3.14."
  • Spot the distractor family. Choices that are 4 times too big (diameter error), or that look like 2\pi r h when the question asked for volume, are built from real mistakes. Identify which one is yours.
  • Estimate with \pi \approx 3. For r = 3, h = 10: 3 \times 9 \times 10 = 270, so the answer should be a bit more than 270 (it is 282.7).
  • Plug answers back in on work-backwards questions: try each choice as h or r in \pi r^2 h and see which gives the stated volume.
  • Watch units. Volume is cubic, area is square; convert to gallons only at the end, using the rate given.

Check yourself

Cylinders

What is the volume of the cylinder shown, in terms of \pi?

Cylinder with radius 4 centimeters and height 10 centimeters. 4 cm 10 cm
Answer choices
Hint

The formula sheet gives V = \pi r^2 h. Which number is the radius, and which is the height?

Check yourself

CylindersFill in

A can of tomatoes has the dimensions shown.

Can shaped like a cylinder with diameter 6 inches across the top and height 5 inches. 6 in 5 in

What is the volume of the can, to the nearest tenth of a cubic inch?

\text{in}^3

Type your answer in the box. Fractions like 3/4 are fine.

Hint

The 6\text{ in} goes all the way across the top. Is that the radius or the diameter?

Summary

  • Volume of a cylinder: V = \pi r^2 h (base area times height), in cubic units.
  • Surface area: SA = 2\pi r h + 2\pi r^2 — the label rectangle (2\pi r \times h) plus two circular lids. Labels, open tanks and pipes use only some of these parts.
  • Both formulas are on the GED formula sheet; \pi \approx 3.14.
  • Always use the radius: if you are given the diameter, divide by 2 first. Square only the radius.
  • Work backwards: h = \dfrac{V}{\pi r^2}; r = \sqrt{\dfrac{V}{\pi h}}; then double r if the diameter is asked.
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