Chapter 10.3 — Exercise 10.2 — Cylinder
Surface area and volume of cylinder. This is Lesson 3 of 5 in Chapter 10: Surface Areas and Volumes.
The First Curved Solid
A right circular cylinder is built from one curved surface wrapped around two congruent circular ends, with the line joining the two centers running perpendicular to both. Every one of these ten problems is really just a different combination of which two of its three key quantities are already known, and which one still needs recovering.
Lateral surface area = 2πrh
Total surface area = 2πr(h + r)
Volume = πr²hReading the Formulas Off a Real Tank
A closed cylindrical tank, height 1.4 m, base radius 0.56 m, needs its total metal-sheet requirement calculated precisely. Substituting directly: total surface area = 2πr(h + r) = 2 × 3.14 × 0.56 × (1.4 + 0.56) = 6.89 m². Every quantity a closed tank needs — the curved wall plus both circular ends — is already folded into this single total-surface-area formula, so no separate calculation for the ends is needed once "closed tank" signals that the total, not the lateral, formula applies. Notice how naturally the total-surface-area formula splits into two recognizable pieces once it's expanded: 2πrh, exactly the lateral (curved-side) surface area on its own, plus 2πr², the combined area of both circular ends together — the "+r" tucked inside the (h + r) bracket is really standing in for one extra radius contributed by each of the two flat ends.
Recovering a Radius From a Given Volume
A cylinder has volume 308 cm³ and height 8 cm; find both its lateral surface area and its total surface area. Since πr²h = 308, (22/7)r²(8) = 308, giving r² = 12.25, so r = 3.5 cm. Then lateral surface area = 2πrh = 2(22/7)(3.5)(8) = 176 cm², and total surface area = 2πr(h + r) = 2(22/7)(3.5)(11.5) = 253 cm². This is the same two-step pattern already seen with cuboids: volume alone can't be substituted directly into a surface-area formula, so the radius has to be recovered first, then reused in both remaining formulas. This is the exact same two-step shape seen throughout the cube-and-cuboid exercise before it — a linear dimension buried inside a squared or cubed formula almost always needs isolating through a square root or similar step before it can be reused anywhere else, and a cylinder's radius is no exception to that general pattern.
Melting One Solid, Casting Another
A metal cuboid, 22 × 15 × 7.5 cm, is melted and recast into a cylinder of height 14 cm; find its radius. Since melting preserves the total amount of metal, volume of cuboid = volume of cylinder: 22 × 15 × 7.5 = (22/7)r²(14). Solving this out, r² = 56.25 exactly, so r = 7.5 cm. The physical fact making this solvable — that melting and recasting changes a solid's shape but never its volume — is really the entire problem; once "volume stays constant" is recognized as the governing equation, the rest is ordinary algebra swapping one volume formula for another. This same melting-and-recasting idea reappears throughout this chapter in different disguises — a shape changes entirely, its surface area and every one of its individual dimensions along with it, but the one quantity that never budges is the volume, since melting neither creates material out of nothing nor destroys any that already existed. Spotting that a problem is secretly a "volume stays fixed" problem, disguised behind a completely different physical description, is the real skill being tested here rather than any of the individual formulas themselves.
Two Radii, One Pipe
A metal pipe, 77 cm long, has inner diameter 4 cm and outer diameter 4.4 cm; find its inner curved surface area, outer curved surface area, and total surface area.
| Surface | Formula | Value |
|---|---|---|
| Inner curved surface | 2πrh | 968 cm² |
| Outer curved surface | 2πRh | 1064.8 cm² |
| Two flat ring ends | 2π(R² − r²) | 5.28 cm² |
| Total surface area | sum of all three | 2038.08 cm² |
A hollow pipe is really and truly just two separate cylinders, inner and outer, sharing the exact same height but genuinely different radii — which is exactly why this problem needs two separate applications of the curved-surface formula rather than one, plus the flat ring area that a completely solid cylinder would never need at all. The ring area itself, π(R² − r²), is worth noticing as its own small pattern: it's simply the outer circle's full area minus the inner circle's full area, the same "big shape minus small shape" logic used constantly for shaded or hollow regions in plane geometry, just applied here to the two flat annular ends of a genuinely three-dimensional pipe instead of to a flat page.
From Curved Surface to Real Cost
16 cylindrical pillars, each 56 cm in diameter and 35 m tall, need their curved surfaces painted at ₹5.50 per m². Base radius = 28 cm = 0.28 m. Lateral surface area of one pillar = 2πrh = 2(22/7)(0.28)(35) = 61.6 m². Cost for one single pillar = 61.6 × 5.50 = ₹338.80, and for all 16 pillars together, 338.80 × 16 = ₹5420.80. A separate problem follows the identical structure with a roller instead of a pillar: a roller of diameter 84 cm and length 120 cm covers, in one revolution, exactly its own lateral surface area, 31680 cm² — so 500 revolutions cover 500 times that, converting to 1584 m² of playground leveled. Both problems share the same underlying insight, even though one talks about paint and the other about ground leveled: a cylinder rolling or standing upright contributes exactly its own lateral surface area to whatever it touches per pillar painted or per single revolution rolled, so multiplying that one lateral-surface-area value by a plain count — 16 pillars, 500 revolutions — is enough to finish either problem, with no further geometry needed beyond the initial formula.
Reversing the Formula: Two Facts, One Unknown
A cylinder's curved surface area is 1760 cm² and its volume is 12320 cm³; find its height. Rather than solving for r and h entirely separately from scratch, dividing one formula directly by the other cancels cleanly: (πr²h) / (2πrh) = 12320 / 1760, which simplifies to r/2 = 7, so r = 14 cm. Substituting back into 2πrh = 1760 gives h = 1760 / (2 × 22/7 × 14) = 20 cm.
This division trick is genuinely worth remembering as its own separate technique, distinct from every other problem worked through in this exercise so far: whenever both a curved-surface-area fact and a volume fact are given together, dividing volume by curved surface area always leaves r/2 behind, since the shared πh factor present in both formulas at once cancels out entirely, leaving nothing behind but a plain ratio — a shortcut that avoids setting up and solving two full separate equations in two unknowns from scratch, cutting the whole problem down to a single quick division followed by one simple back-substitution.
A Second Curved Solid, With One More Measurement
A cylinder needed just radius and height. Exercise 10.3 introduces the cone, which reuses both of those but adds a third, the slant height, connected to the other two by its own hidden right-triangle relationship.