Cylinder: Volume and Surface Area

Base area times height: $V = \pi{}r^{2}\cdot{}h$. First compute the circle area $\pi{}r^{2}$ of the base, then multiply by the height. Example: a can with $r = 3 cm$ and $h = 12 cm$ holds $V = \pi{}\cdot{}9\cdot{}12 = 108\pi{} \approx{} 339.3 cm^{3}$, about 0.34 litres $(1000 cm^{3} = 1 l)$. 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 $h = 10 cm$: $M = 80\pi{} \approx{} 251.3 cm^{2}$, $O = 112\pi{} \approx{} 351.9 cm^{2}$. 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 $V = \pi{}r^{2}h$ for h: $h =\frac{ V}{(\pi{}r^{2})}$. So you divide the volume by the base area. Example: a vessel should hold $V = 500 cm^{3}$ and has $r = 5 cm$; then $h =\frac{ 500}{(\pi{}\cdot{}25)} =\frac{ 500}{78.54} \approx{} 6.37 cm$. In the same way you find the radius when V and h are given: $r = \sqrt{(\frac{V}{(\pi{}h)})}$, here with a square root, because r enters quadratically. Make sure to convert all data to the same unit before rearranging (convert litres to $cm^{3}$ 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 $V = \pi{}r^{2}h$. Doubling the height doubles the volume. Doubling the radius quadruples it, since $(2r)^{2} = 4r^{2}$. Both together multiply the volume by eight. Example: $r = 3 cm$, $h = 12 cm$ gives $339.3 cm^{3}$; with $r = 6 cm$ it becomes $1357.2 cm^{3}$. 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 $\frac{2}{3}\cdot{}\pi{}r^{3}$, sphere $\frac{4}{3}\cdot{}\pi{}r^{3}$, cylinder $2\pi{}r^{3}$, 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.