Cylinder: Volume and Surface Area

Base area times height: V = πr²·h. First compute the circle area πr² of the base, then multiply by the height. Example: a can with r = 3 cm and h = 12 cm holds V = π·9·12 = 108π ≈ 339.3 cm³, about 0.34 litres (1000 cm³ = 1 l). Most common mistake: the diameter is inserted as the radius; always halve d first, otherwise the result is four times too large. The formula holds for every right circular cylinder, lying or standing; what matters is that r belongs to the circular cross-section and h to the axis perpendicular to the base.

The lateral surface M = 2πrh is only the side wall, intuitively the label of a can: unrolled it is a rectangle with the circumference 2πr as width and h as height. The total surface O = 2πr² + 2πrh = 2πr(r + h) additionally includes base and lid, i.e. two circle areas. Example with r = 4 cm and h = 10 cm: M = 80π ≈ 251.3 cm², O = 112π ≈ 351.9 cm². In word problems the context decides: painting a closed barrel needs O, a label or an open pipe needs only M, a cup without a lid needs M plus one circle.

Rearrange V = πr²h for h: h = V/(πr²). So you divide the volume by the base area. Example: a vessel should hold V = 500 cm³ and has r = 5 cm; then h = 500/(π·25) = 500/78.54 ≈ 6.37 cm. In the same way you find the radius when V and h are given: r = √(V/(πh)), here with a square root, because r enters quadratically. Make sure to convert all data to the same unit before rearranging (convert litres to cm³ first). Such rearrangements are standard in filling tasks: the filling amount is given, the matching vessel dimension is required.

The two quantities act with different strength, because r enters quadratically and h only linearly in V = πr²h. Doubling the height doubles the volume. Doubling the radius quadruples it, since (2r)² = 4r². Both together multiply the volume by eight. Example: r = 3 cm, h = 12 cm gives 339.3 cm³; with r = 6 cm it becomes 1357.2 cm³. This asymmetry explains why wide vessels hold so much more than one estimates, and it is the core of many exam questions: a price comparison task with a twice-as-wide cup is almost always a scaling question in disguise.

Through a famous ratio due to Archimedes. Take a cylinder with radius r and height h = 2r, the inscribed cone (same base, same height) and the inscribed sphere with radius r. Then: cone 2/3·πr³, sphere 4/3·πr³, cylinder 2πr³, i.e. the ratio 1 : 2 : 3. The cone holds one third of the cylinder, the sphere two thirds. This relation is a strong checking tool in exams: once you have computed two of the three solids, you can immediately verify the third. Archimedes was so proud of this connection that he had it immortalized on his tombstone.