Right circular cylinder

A garden roller, a tin of paint, a length of pipe — all are right circular cylinders, two identical circular ends joined by a curved wall that rises straight up at right angles to the base. ICSE Class 10 mensuration asks you to find three things about this shape: how much wraps around it (curved surface area), how much would cover it entirely including the two circular ends (total surface area), and how much it can hold (volume). The trick that unlocks all three at once is a single mental picture — peel the curved wall off the cylinder and unroll it flat, and it becomes an ordinary rectangle whose formulas you already know. This article rebuilds that picture from scratch, extends it to hollow pipes, and works through both a solid tank and a hollow pipe problem so you can handle either style of question with confidence.

The core insight

Unrolling a cylinder's curved surface gives a rectangle of length $2\pi r$ (the base circumference) and breadth $h$ , and from that one picture the formulas for curved surface area, total surface area, and volume all follow directly.

The core idea

Where it fits. Mensuration in Class X is entirely about solids with a circular cross-section — the cylinder, the cone, and the sphere/hemisphere. Surface area and volume of the cube and cuboid were Class IX work, assumed known here. For the cylinder specifically, you must be able to compute CSA, TSA and volume for solid cylinders, extend the same ideas to hollow cylinders (pipes), and apply both to cost and melting-and-recasting problems.

Generating the shape. A right circular cylinder is what you get by rotating a rectangle about one of its sides — the fixed side becomes the axis, and the far side sweeps out the curved surface. Because the base is a circle of radius $r$ , the curved surface has a very simple structure: if you cut it along one vertical line and flatten it, it opens into a rectangle. The rectangle's breadth is just the cylinder's height $h$ , and its length is the distance once around the circular base, the circumference $2\pi r$ .

Figure

A cylinder next to its unrolled curved surface shown as a rectangle of length two pi r and breadth h

Solid cylinder formulas. Because a rectangle's area is length times breadth, the flattened curved surface immediately gives the curved surface area. Adding the two circular ends gives the total surface area, and the volume is simply base area times height.

\text{CSA} = 2\pi r h

\text{TSA} = 2\pi r h + 2\pi r^2 = 2\pi r(r+h)

V = \pi r^2 h

where: $r$ is the radius of the circular base, $h$ is the perpendicular height, and $\pi \approx \dfrac{22}{7}$ (use this when $r$ or $h$ is a multiple of 7) or $3.14$ otherwise.

Hollow cylinders. A pipe has two radii — an outer radius $R$ and an inner radius $r$ — and its wall is the annulus (ring) between them. Its volume is only the material of the wall, and its surface consists of the outer curved surface, the inner curved surface, and the two flat rims at the ends.

\text{Volume of material} = \pi h (R^2 - r^2)

\text{Total surface area} = 2\pi R h + 2\pi r h + 2\pi (R^2 - r^2)

where: the three terms of the TSA are, in order, the outer CSA, the inner CSA, and the combined area of the two flat rims (each rim is an annulus of area $\pi(R^2-r^2)$ ).

Derivation ADeriving the cylinder's surface area and volume formulas

Show, starting only from "a right circular cylinder is generated by rotating a rectangle about one side" and the standard facts that a circle of radius $r$ has circumference $2\pi r$ and area $\pi r^2$ , how the flattening argument produces $\text{CSA}=2\pi rh$ , $\text{TSA}=2\pi r(r+h)$ and $V=\pi r^2h$ without any calculus, and why the same reasoning extends cleanly to a hollow pipe.

Step 1. Picture the cylinder as a stack of the base circle repeated at every height from $0$ to $h$ . The base circle has area $\pi r^2$ ; since every cross-section at every height is the identical circle, the solid's volume is just that base area multiplied by how far it is stacked — this is the general "volume = base area times height" rule for any prism-like solid with a constant cross-section, giving $V = \pi r^2 h$ .

Step 2. Now handle the curved wall alone. Make one straight vertical cut in the curved surface, from the bottom rim to the top rim, and unroll the surface flat onto a table. Because the cut is straight and the surface doesn't stretch, it opens into a flat rectangle: one pair of sides has length $h$ (the height you cut along), and the other pair has length equal to the full distance around the base circle, the circumference $2\pi r$ .

Step 3. The area of a rectangle is length times breadth, so the area of this unrolled piece — which is exactly the curved surface of the cylinder, just laid flat — is $2\pi r \times h = 2\pi rh$ . This gives the curved surface area formula directly, with no separate proof needed beyond the rectangle-area fact you already know.

Step 4. To close the solid into a total surface, glue back the two circular ends that were cut away, each of area $\pi r^2$ . Adding these to the curved surface gives $\text{TSA} = 2\pi rh + 2\pi r^2 = 2\pi r(r+h)$ , the factored form being convenient for calculation since $r+h$ is computed once and multiplied by $2\pi r$ .

Step 5. For a hollow pipe, apply Step 1's logic twice and subtract: the volume enclosed by the outer radius $R$ is $\pi R^2 h$ , the volume of the inner hollow (removed) is $\pi r^2 h$ , so the material actually present is the difference, $\pi h(R^2-r^2)$ . The surface, by the same unrolling idea, now has an outer rectangle of area $2\pi Rh$ , an inner rectangle of area $2\pi rh$ (the inside wall you'd see looking into the pipe), and two flat rims, each an annulus of area $\pi(R^2-r^2)$ , giving $\text{TSA} = 2\pi Rh + 2\pi rh + 2\pi(R^2-r^2)$ exactly as stated above.

Worked example

A closed cylindrical tank has radius $1.4\text{ m}$ and height $3\text{ m}$ . Find its CSA, TSA, and volume, taking $\pi = \dfrac{22}{7}$ .

Step 1. Curved surface area: $\text{CSA} = 2\pi rh = 2 \times \dfrac{22}{7} \times 1.4 \times 3 = 2 \times 22 \times 0.2 \times 3 = 26.4\text{ m}^2$ .

Step 2. Total surface area: $\text{TSA} = 2\pi r(r+h) = 2 \times \dfrac{22}{7} \times 1.4 \times (1.4+3) = 2 \times 22 \times 0.2 \times 4.4 = 38.72\text{ m}^2$ .

Step 3. Volume: $V = \pi r^2 h = \dfrac{22}{7} \times 1.4 \times 1.4 \times 3 = 22 \times 0.2 \times 1.4 \times 3 = 18.48\text{ m}^3$ .

The tank's curved surface is $26.4\text{ m}^2$ , its total surface is $38.72\text{ m}^2$ , and it holds $18.48\text{ m}^3$ .

Competitive stretch

A hollow metal pipe has external diameter $10\text{ cm}$ , internal diameter $6\text{ cm}$ , and length $21\text{ cm}$ . Find the volume of metal and the total surface area, taking $\pi = \dfrac{22}{7}$ .

Step 1. Halve the diameters: $R = 5\text{ cm}$ , $r = 3\text{ cm}$ , $h = 21\text{ cm}$ . Volume of metal: $V = \pi h(R^2-r^2) = \dfrac{22}{7} \times 21 \times (25-9) = 22 \times 3 \times 16 = 1056\text{ cm}^3$ .

Step 2. Outer CSA $= 2\pi Rh = 2 \times \dfrac{22}{7} \times 5 \times 21 = 660\text{ cm}^2$ ; inner CSA $= 2\pi rh = 2 \times \dfrac{22}{7} \times 3 \times 21 = 396\text{ cm}^2$ ; rim area $= 2\pi(R^2-r^2) = 2 \times \dfrac{22}{7} \times 16 \approx 100.57\text{ cm}^2$ .

Step 3. Adding: $\text{TSA} = 660 + 396 + 100.57 = 1156.57\text{ cm}^2$ . The pipe uses $1056\text{ cm}^3$ of metal and has total surface area about $1156.57\text{ cm}^2$ .

Common traps

Confusing height with slant height — a right circular cylinder has no slant height at all; that term belongs only to the cone.
Using the diameter instead of the radius in a formula — always halve a given diameter first.
For a hollow pipe open at both ends, forgetting the rim area $2\pi(R^2-r^2)$ , or wrongly treating it as having two full solid circular ends.
In melting-and-recasting problems, equating the surface areas of the two solids instead of their volumes — melted material keeps the same volume, not the same surface area.

Ask the tutor: Stuck on a hollow-pipe question or a recasting sum that won't balance — ask the AI tutor to walk through it with you step by step.

Why does the curved surface unroll into exactly a rectangle and not some other shape?

Because the cut you make is a single straight vertical line from bottom rim to top rim, and the surface itself does not stretch or tear when flattened. A straight cut on a surface with a constant circular cross-section always opens into a flat shape with straight edges on all four sides, and since every point on the curved wall is the same distance $h$ from the base, two opposite edges are equal and parallel — that is precisely a rectangle, with breadth $h$ and length equal to the circumference you cut open, $2\pi r$ .

Do I need to memorise TSA as two separate terms or as the factored form?

Either works, but the factored form $2\pi r(r+h)$ is faster to compute and less error-prone, because you add $r+h$ once and then do a single multiplication by $2\pi r$ , instead of calculating $2\pi rh$ and $2\pi r^2$ separately and then adding. Keep the unfactored form $2\pi rh + 2\pi r^2$ in mind too, since some questions ask you to identify the curved and end-cap contributions separately, particularly in hollow-cylinder problems where you build TSA up piece by piece.

How do I know which value of pi to use in a problem?

Use $\pi = \dfrac{22}{7}$ whenever the radius or diameter given is a multiple of 7 (like 7, 14, 21, 3.5), since the sevens cancel cleanly and you avoid long decimals. Use $\pi = 3.14$ when the given values are not multiples of 7, or when the question explicitly tells you which value to use — always follow an explicit instruction over this general rule.

Right circular cylinder

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads