Volume & Surface Area of Cylinders - Standard Maths

The circle underneath

Every cylinder rule starts from the circle at its end. A circle of radius \(r\) has area \(\pi r^2\) and circumference \(2\pi r\). Keep both handy: the area feeds the volume, and the circumference feeds the curved surface.

Radius: The distance from the centre of a circle to its edge. The diameter is twice the radius, so if you are given a diameter, halve it first.

Volume of a cylinder

A cylinder is a prism whose cross-section is a circle, so volume is again cross-sectional area times length. Here the area is \(\pi r^2\) and the length is the height \(h\):

\[ V = \pi r^2 h \]

Square the radius before anything else

In \(\pi r^2 h\) only the radius is squared, not the height. Work out \(r^2\) first, then multiply by \(h\), then by \(\pi\). Doing it in that order keeps the common slip of squaring the wrong quantity out of your working.

Surface area of a cylinder

Unroll a closed cylinder and you get three pieces: two identical circles for the top and bottom, and one rectangle for the curved wall. The rectangle has height \(h\) and width equal to the circumference \(2\pi r\), so its area is \(2\pi r h\). Adding the two circles gives:

\[ \text{Surface area} = 2\pi r^2 + 2\pi r h \]

The \(2\pi r^2\) is the two ends; the \(2\pi r h\) is the curved wall on its own, which is called the curved surface area.

Putting it together

Worked example

A closed cylinder has radius 4 cm and height 7 cm. Find its volume and its total surface area, each to 1 decimal place.

Volume: \(V = \pi r^2 h = \pi \times 4^2 \times 7 = \pi \times 16 \times 7 = 112\pi\).

Evaluate: \(112\pi = 351.85\ldots\), so \(V \approx 351.9\) cm\(^3\).

Two ends: \(2\pi r^2 = 2\pi \times 16 = 32\pi\). Curved wall: \(2\pi r h = 2\pi \times 4 \times 7 = 56\pi\).

Add: \(32\pi + 56\pi = 88\pi = 276.46\ldots\)

\(V \approx 351.9\) cm\(^3\), surface area \(\approx 276.5\) cm\(^2\)

Where this is assessed

Selecting and substituting into the right formula is Criterion A (Knowing and understanding). Writing each stage clearly, keeping answers exact as multiples of \(\pi\) until the final rounding, is Criterion C (Communicating). Designing a tin or a pipe to hold a set amount is Criterion D (Applying maths in context).

Check yourself

1. Find the volume of a cylinder with radius 2 cm and height 10 cm (to 1 d.p.). +

\(V = \pi r^2 h = \pi \times 2^2 \times 10 = \pi \times 4 \times 10 = 40\pi = 125.66\ldots\) So \(V \approx \mathbf{125.7}\) cm\(^3\).

2. Find the total surface area of a closed cylinder with radius 5 cm and height 8 cm (to 1 d.p.). +

Two ends: \(2\pi \times 5^2 = 50\pi\). Curved wall: \(2\pi \times 5 \times 8 = 80\pi\). Total \(= 130\pi = 408.40\ldots \approx \mathbf{408.4}\) cm\(^2\).

3. Find just the curved surface area of a cylinder with radius 3 cm and height 10 cm (to 1 d.p.). +

Curved surface area \(= 2\pi r h = 2\pi \times 3 \times 10 = 60\pi = 188.49\ldots \approx \mathbf{188.5}\) cm\(^2\).

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