Applications of surface areas and volumes

A water tank needs to be filled, and its outer wall needs to be painted. Both jobs use the same cylinder, yet they call for two completely different formulas — one measures a skin (surface area), the other measures a stuffing (volume). Confusing the two is the single most common mistake students make in this chapter, and it is entirely avoidable once you learn to read the verb in the question. This article gathers the surface area and volume formulas for the six solids of Class 9 — cube, cuboid, cylinder, cone, sphere, hemisphere — and focuses on the real skill the exam tests: translating a real-world sentence about painting, filling, wrapping, or melting into the correct formula, then carrying the arithmetic through cleanly, including unit conversions between cubic metres and litres.

The core insight

Surface-area formulas answer how much of a solid's skin must be covered, painted, or made from sheet material, while volume formulas answer how much a solid can hold or how much matter it contains — and "melt and recast" problems always conserve volume, never surface area.

The core idea

Every problem in this chapter reduces to picking the right formula from a short list, then plugging in numbers correctly.

The six solids and their formulas. For a cube of side $a$ , a cuboid of length $l$ , breadth $b$ , height $h$ , a cylinder of radius $r$ and height $h$ , a cone of radius $r$ , height $h$ and slant height $\ell$ , a sphere of radius $r$ , and a hemisphere of radius $r$ :

\text{Cube:}\ \text{TSA}=6a^2,\ V=a^3 \qquad \text{Cuboid:}\ \text{TSA}=2(lb+bh+hl),\ V=lbh

\text{Cylinder:}\ \text{CSA}=2\pi rh,\ \text{TSA}=2\pi r(r+h),\ V=\pi r^2h

\text{Cone:}\ \ell=\sqrt{r^2+h^2},\ \text{CSA}=\pi r\ell,\ \text{TSA}=\pi r\ell+\pi r^2,\ V=\tfrac13\pi r^2h

\text{Sphere:}\ \text{TSA}=4\pi r^2,\ V=\tfrac43\pi r^3 \qquad \text{Hemisphere:}\ \text{CSA}=2\pi r^2,\ \text{TSA}=3\pi r^2,\ V=\tfrac23\pi r^3

where: TSA is total surface area (every face counted), CSA is curved surface area (only the curved part, excluding flat circular ends), and $V$ is volume, all in consistent length units cubed for $V$ and squared for area.

Reading the question. Three verb families point to three different quantities, and spotting the family is most of the battle.

Painting, whitewashing, varnishing, wrapping, or making a solid from a flat sheet of material — these all measure a surface, so reach for CSA (curved part only) or TSA (every face), depending on which part of the solid the question actually covers.
Filling, holding, capacity, "how much can it contain," or the amount of raw material used to make a solid object — these all measure a volume.
Cost questions almost always multiply an already-identified area or volume by a rate given per unit (₹ per m², ₹ per litre), so the first job is still to decide area or volume before any multiplication happens.

Melt and recast. When a solid metal object is melted down and recast into a different shape, its volume is conserved exactly — the same amount of metal simply takes a new form — but its surface area is not conserved, since a sphere and twenty-four small cones do not expose the same amount of outer skin even though they contain identical amounts of metal.

Figure

A cylindrical tank showing that curved surface area governs the paint needed for the outer wall while volume governs the litres of water the tank holds

Derivation ANumber of cones cast from a melted sphere

The goal is to show, using nothing beyond the volume formulas already stated above and the physical fact that melting and recasting a solid metal object neither creates nor destroys metal, that the number of smaller solids obtained from a melted-down solid equals the total volume of the original divided by the volume of one smaller piece, and to carry this through with concrete numbers so the reasoning is checkable at every step.

Step 1. State the conservation principle. When a solid is melted and poured into new moulds, its mass and hence its volume stay the same, because melting only changes the arrangement of the material, not its quantity; so $V_{\text{original}} = n\times V_{\text{one small piece}}$ , where $n$ is the number of pieces obtained.

Step 2. Compute the volume being melted. Take a solid metal sphere of radius $6$ cm. Its volume is $V_{\text{sphere}} = \tfrac43\pi r^3 = \tfrac43\pi(6)^3 = \tfrac43\pi\times216 = 288\pi$ cm³.

Step 3. Compute the volume of one recast piece. Each small solid cone has base radius $3$ cm and height $4$ cm, so $V_{\text{cone}} = \tfrac13\pi r^2h = \tfrac13\pi(3)^2(4) = \tfrac13\pi\times9\times4 = 12\pi$ cm³.

Step 4. Divide total volume by the volume of one piece to isolate $n$ from the Step 1 equation: $n = V_{\text{sphere}}/V_{\text{cone}} = 288\pi/12\pi = 24$ . The $\pi$ cancels because both solids are measured in the same units, confirming the division is dimensionally sound and the answer is a pure count. The sphere yields exactly $24$ cones, and this same four-step pattern — conserve volume, compute each side, divide — solves every melt-and-recast problem in the chapter regardless of which two solids are involved.

Worked example

Problem. A cylindrical water tank has radius $1.4$ m and height $3$ m. Find (a) its capacity in litres, and (b) the cost of painting its curved outer surface at ₹15 per m². Use $\pi = \tfrac{22}{7}$ .

Step 1. Identify what each part needs. Part (a) asks "how many litres it can hold," which is a capacity question, so it needs volume. Part (b) asks for the cost of painting the curved outer surface, which needs curved surface area (CSA), not total surface area, since only the curved wall is being painted.

Step 2. Compute the volume. $V = \pi r^2h = \tfrac{22}{7}\times(1.4)^2\times3 = \tfrac{22}{7}\times1.96\times3$ . Since $\tfrac{22}{7}\times1.96 = 22\times0.28 = 6.16$ , we get $V = 6.16\times3 = 18.48$ m³.

Step 3. Convert to litres. Since $1$ m³ $= 1000$ litres, the capacity is $18.48\times1000 = 18{,}480$ litres.

Step 4. Compute the curved surface area. $\text{CSA} = 2\pi rh = 2\times\tfrac{22}{7}\times1.4\times3 = 2\times22\times0.2\times3 = 26.4$ m².

Step 5. Compute the painting cost. $\text{Cost} = 26.4\times15 = ₹396$ . The tank holds 18,480 litres of water, and painting its curved outer wall costs ₹396.

Competitive stretch

Suppose the same tank also needs its two circular ends covered with a metal sheet costing ₹80 per m². The area of both circles is $2\pi r^2 = 2\times\tfrac{22}{7}\times1.96 = 12.32$ m², so the extra cost is $12.32\times80 = ₹985.60$ , bringing the total covering cost (curved wall plus both ends, i.e. the full TSA-based job) to $₹396+₹985.60 = ₹1381.60$ . Notice this only makes sense if the question actually asks for both ends to be covered — many real tanks are open at the top, so read carefully before adding this extra term.

Common traps

Reaching for volume when the question is about a surface, or vice versa — always check the verb: "paint," "cover," "wrap," "sheet" mean area; "hold," "fill," "melt," "cast," "made of" mean volume.
Using total surface area when only the curved part is being painted (an open-top tank has no top circle to paint), or using curved surface area when every face genuinely needs covering.
Forgetting that melting and recasting conserves volume, not surface area — the recast pieces can have a very different total surface area from the original solid even though the metal content is identical.
Mixing units mid-problem, such as radius in centimetres with height in metres, or forgetting to convert cubic metres to litres when a capacity answer is expected.

Ask the tutor: Not sure whether your word problem wants surface area or volume, or want more practice on melt-and-recast and cost problems? Ask the AI tutor to walk through a few more applied examples with you.

How do I decide between curved surface area and total surface area for a cylinder or cone

Ask whether the flat circular parts are physically present and relevant to the question. A closed can that is fully painted needs TSA. An open-top tank being painted only on the inside wall, or a pillar wrapped in a poster only around its side, needs CSA. If the question mentions painting or covering "only the curved surface," it is telling you directly to use CSA; if it says "entirely covered" or "painted on all sides," use TSA.

Why does volume stay the same when a shape is melted and recast but surface area does not

Volume measures the amount of material, and melting simply rearranges that material into a new mould without adding or removing any of it, so the total amount — the volume — cannot change. Surface area, by contrast, measures how much outer skin the shape exposes, and a fixed amount of material can be shaped into many different forms with very different amounts of exposed skin, for example one big sphere versus two dozen small cones from the same metal.

I keep getting the wrong number of litres — what am I doing wrong

This is almost always a units slip. Compute the volume first in whatever units the radius and height are given, typically cm³ or m³, and only convert to litres at the very last step, using $1$ m³ $=1000$ litres or $1$ cm³ $= 1$ millilitre so $1000$ cm³ $= 1$ litre. Converting too early, or mixing a radius in one unit with a height in another before computing volume, is the most common source of a wrong final answer in capacity problems.

Applications of surface areas and volumes

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads