Surface Areas and Volumes

A truck carrying water down a highway is not a cylinder, and a test tube on a lab bench is not a cone — yet both are built entirely from shapes you already know. The truck's tank is a cylinder capped by two hemispherical bowls; the test tube is a cylinder with one hemispherical bottom. Real objects are almost never pure cuboids, cylinders, cones, spheres or hemispheres in isolation — they are combinations of solids, and this chapter gives you the one idea that lets you handle every one of them: break the unfamiliar shape into the familiar pieces you have already mastered, and add. But "add" means two different things depending on what you are adding. For volume, the rule is exactly what intuition suggests — the volume of the whole equals the sum of the volumes of the parts, because volume measures the space enclosed, and joining two solids simply pools their interiors together. For surface area, intuition misleads you. When a hemisphere is glued onto a cylinder, the flat circular face where they meet vanishes from view — it is no longer part of the outer skin of the new solid. So the total surface area of a combination is built from the curved surface areas of the pieces, not their total surface areas, and getting this distinction right is the single skill this entire chapter tests. This article walks through both rules with the exact worked examples NCERT uses — a colourful spinning top, a decorative block, a toy rocket, a bird-bath, a kulfi glass with a false bottom, and a solid toy circumscribed by a cylinder — and turns each into a reusable method. By the end you should be able to look at any composite solid, identify its constituent cuboids, cylinders, cones, spheres and hemispheres, and correctly assemble either its surface area or its volume without ever needing a new formula beyond the five you already know.

Key takeaways

A combination of solids is any object built by joining two or more basic solids — cuboid, cylinder, cone, sphere, hemisphere — face to face.
Volume of a combination = sum of the volumes of its parts. Nothing is added or removed at the joint, so volumes simply add.
Surface area of a combination is NOT the sum of the total surface areas (TSA) of its parts. The flat face where two solids are joined is hidden inside the new solid and must be dropped from both pieces.
The safe rule: TSA of a combination = sum of the curved surface areas (CSA) of the parts that are actually exposed, plus any flat area that remains genuinely on the outside.
When a hemisphere sits on a cone or cylinder of the same radius, only the CSA of each contributes — the flat circular faces that meet cancel exactly.
When a hemisphere is scooped out of a flat face, surface area changes by (CSA of hemisphere) minus (circular base area removed) — a net gain of $2\pi r^2$ for a cube-cavity.
For a cone of base radius $r$ and height $h$ , the slant height $l = \sqrt{r^2+h^2}$ is needed for both CSA and TSA — always compute it before using $\pi r l$ .
Capacity problems (glasses, vessels) use volume; apparent capacity (as if the vessel were a pure cylinder or cone) minus the volume of any raised inner solid gives the actual capacity.
When comparing a solid to a cylinder that exactly circumscribes it, the two share the same radius, and the circumscribing cylinder's height equals the sum of the heights of the solid's parts.
Key idea: for a combination of basic solids, volumes simply add, but surface area is the sum of only the curved surface areas that remain exposed on the outside, because the flat faces where solids are joined disappear from view.

The core insight

When two or more basic solids — cuboid, cylinder, cone, sphere, hemisphere — are joined to form one composite object, its volume is the plain sum of the parts' volumes, but its total surface area is the sum of only the parts' curved surface areas (the joined flat faces are hidden and must be excluded, sometimes with a base-area adjustment where a solid is scooped out rather than stacked on).

What this chapter covers

This chapter has exactly two working ideas, built from the same five basic solids you met in Class IX: the cuboid (and its special case, the cube), the cylinder, the cone, the sphere, and the hemisphere. Section 12.2, Surface Area of a Combination of Solids, sets up the central caution — joining solids hides flat faces, so total surface area must be built from curved surface areas — through examples such as a cone-and-hemisphere spinning top, a cube-and-hemisphere decorative block, and a cone-on-cylinder toy rocket. Section 12.3, Volume of a Combination of Solids, is the friendlier half: volumes simply add, with no hidden-face correction, illustrated by a cuboid-and-half-cylinder shed, a glass with a false hemispherical bottom, and a hemisphere-and-cone toy compared against its circumscribing cylinder. Master the CSA-versus-TSA distinction and the straightforward addition, and you have the whole chapter.

The core ideas

Recap: the five basic solids and their formulas

Every combination in this chapter is built from exactly five shapes, so before combining anything you must have their individual curved surface area (CSA), total surface area (TSA) and volume formulas at instant recall. For a cuboid of length $l$ , breadth $b$ and height $h$ (and its special case the cube of edge $a$ ), and for a right circular cylinder of radius $r$ and height $h$ :

\text{Cuboid: TSA} = 2(lb+bh+hl), \ \ V = lbh; \qquad \text{Cube: TSA} = 6a^2, \ \ V = a^3; \qquad \text{Cylinder: CSA} = 2\pi rh, \ \ \text{TSA} = 2\pi r(h+r), \ \ V = \pi r^2 h. \qquad (1)

where: $l, b, h$ are the three edges of a cuboid, $a$ is the edge of a cube, and $r, h$ are the radius and height of a cylinder. The cylinder's TSA splits naturally into its curved part $2\pi rh$ (the label wrapped around it) plus its two circular ends $2\pi r^2$ — a split you will use constantly once cylinders start losing one or both of those ends inside a combination. For a right circular cone of base radius $r$ , height $h$ and slant height $l$ , for a sphere of radius $r$ , and for a hemisphere of radius $r$ :

\text{Cone: } l = \sqrt{r^2+h^2}, \ \ \text{CSA} = \pi rl, \ \ \text{TSA} = \pi r(l+r), \ \ V = \tfrac13 \pi r^2 h; \qquad \text{Sphere: surface area} = 4\pi r^2, \ \ V = \tfrac43 \pi r^3; \qquad \text{Hemisphere: CSA} = 2\pi r^2, \ \ \text{TSA} = 3\pi r^2, \ \ V = \tfrac23 \pi r^3. \qquad (2)

where: the slant height $l$ of a cone is not given directly in most problems — it must first be computed from the Pythagoras relation $l^2 = r^2+h^2$ , using the cone's own radius and height as the two legs of a right triangle. A hemisphere's TSA ( $3\pi r^2$ ) is its curved part ( $2\pi r^2$ ) plus one flat circular base ( $\pi r^2$ ); once a hemisphere is glued flat-face-down onto another solid, only its curved $2\pi r^2$ survives as exposed surface, and its volume is exactly half a sphere's, $\tfrac23\pi r^3 = \tfrac12 \times \tfrac43\pi r^3$ .

Figure

A cone mounted on a hemisphere of the same radius forming a spinning top with the radius, cone height and slant height labelled, showing that only the two curved surfaces remain visible

Surface area of a combination of solids

Consider joining two basic solids flat-face to flat-face, such as a hemisphere glued onto the top of a cylinder, or a cone stacked on a cylinder of the same radius. The seam where they meet is a flat circle that sits entirely inside the new solid — it is no longer part of the boundary you would paint. So the total surface area of the combination is built only from the surfaces that remain genuinely on the outside:

\text{TSA of the combined solid} = \text{CSA of piece one} + \text{CSA of piece two} + (\text{any flat faces that remain exposed}). \qquad (3)

where: "flat faces that remain exposed" covers cases where a solid is placed on top of another without covering it entirely — for instance a hemisphere fixed on top of a cube leaves most of the cube's top face still visible, so that leftover ring of flat area (cube's top face area minus the hemisphere's circular base) must be added back in. The rule is never "add the TSAs of the parts" — that double-counts nothing at the seam but also keeps the two flat faces that have actually disappeared, over-counting the true surface by exactly those two hidden circles.

Figure

A capsule shape made from a cylinder with a hemisphere attached at each end showing the cylinder height and shared radius, illustrating that the two end hemispheres combine to one full sphere for volume while only their curved surfaces count for area

Volume of a combination of solids

Volume behaves the way intuition already expects. When two solids are joined, the space each one encloses simply pools together into the space enclosed by the whole — nothing about the interior is hidden or removed by the act of joining, unlike the exterior skin. So:

V_{\text{combination}} = V_1 + V_2 + \cdots + V_n. \qquad (4)

where: $V_1, V_2, \ldots, V_n$ are the volumes of the constituent basic solids, computed with the ordinary formulas of equations (1) and (2), using whatever radius and height each piece actually has (a hemisphere glued onto a cone, for example, shares the cone's radius but has its own independent contribution to the total volume). This is the one place in the chapter where you do not need to hunt for hidden or exposed surfaces — you only need to correctly identify each constituent piece and its dimensions. The only extra care needed is in scooped-out or cavity problems, where a smaller solid's volume is subtracted rather than added, because that material has been removed rather than joined on.

Derivations

Derivation AWhy total surface area of a combination uses curved surface areas, not total surface areas

The goal is to justify, from first principles, the central rule of Section 12.2: when a solid $A$ and a solid $B$ are joined so that a flat face of $A$ coincides exactly with a flat face of $B$ of the same shape and size, the total surface area of the combined solid equals (TSA of $A$ minus the joined face's area) plus (TSA of $B$ minus the joined face's area) — which, when the joined faces are the only flat faces of $A$ and $B$ , reduces to simply the sum of their curved surface areas. This is not a new formula to memorise; it is an argument about which parts of a boundary remain a boundary after gluing.

Step 1. Define the surface area of any solid as the total area of the boundary separating its interior from the surrounding air — the points you could physically touch or paint from outside. Before joining, both $A$ and $B$ contribute their full TSA.

Step 2. Glue $A$ 's flat face (area $F$ ) onto $B$ 's matching flat face (also area $F$ ). Every point on $A$ 's glued face now touches a point on $B$ 's glued face; neither face is touchable any more — both become purely internal, sealed away from the air.

Step 3. The combined solid's surface area is the original total minus the two faces that became internal: $\text{TSA}_{\text{new}} = (\text{TSA}_A + \text{TSA}_B) - F - F = (\text{TSA}_A - F) + (\text{TSA}_B - F)$ . Every other part of $A$ 's and $B$ 's boundary is still exposed exactly as before.

Step 4. Specialise to hemisphere-on-cone (or hemisphere-on-cylinder). A hemisphere's only flat face is its circular base, area $\pi r^2$ , and $\text{TSA}_{\text{hemisphere}} - \pi r^2 = 3\pi r^2 - \pi r^2 = 2\pi r^2$ , exactly its CSA. If the cone or cylinder shares the same radius $r$ , its one flat circular end is likewise removed. So $\text{TSA}_{\text{new}} = \text{CSA}_{\text{hemisphere}} + \text{CSA}_{\text{other solid}}$ — the Section 12.2 rule — and this also shows exactly when it needs modification: whenever the joined face does not consume an entire flat face of one solid, a leftover ring of exposed flat area (as with the cube-and-hemisphere block) must be added back.

Go deeper · Derivation B · 1 minGeneral surface area and volume of a capsule shaped like a cylinder with a hemisphere at each end

Derivation BGeneral surface area and volume of a capsule shaped like a cylinder with a hemisphere at each end

The goal is to derive reusable general formulas for the total surface area and volume of the "capsule" family of solids — a cylinder of radius $r$ and length $h$ with a hemisphere of the same radius $r$ fixed at each end — of which NCERT's medicine-capsule problem is the standard instance. Deriving the general formula in $r$ and $h$ turns every future capsule-shaped question (pills, rounded test tubes, Exercise 12.2's Question 2 with cones instead) into a single substitution.

Step 1. Identify the three pieces: one cylinder of radius $r$ , height $h$ (the straight middle), plus two hemispheres of radius $r$ (the rounded ends). Since both hemispheres share the cylinder's radius, each flat circular face is fully consumed by its join — only curved surfaces survive anywhere on the capsule.

Step 2. Add the curved surfaces: $\text{TSA} = 2\pi rh + 2\pi r^2 + 2\pi r^2 = 2\pi rh + 4\pi r^2 = 2\pi r(h+2r)$ .

Step 3. Add the three volumes directly, since volume needs no hidden-face correction: $V = \pi r^2 h + \tfrac23\pi r^3 + \tfrac23\pi r^3 = \pi r^2 h + \tfrac43\pi r^3$ . The two hemisphere volumes combine into exactly one full sphere's volume — a useful shortcut whenever a solid has two same-radius hemispherical ends.

Step 4. Check against NCERT's own capsule (radius $2.5\ \text{mm}$ , cylindrical length $14-2(2.5)=9\ \text{mm}$ , $\pi=22/7$ ): $\text{TSA} = 2\times\tfrac{22}{7}\times2.5\times(9+5) = 220\ \text{mm}^2$ , matching the textbook's answer and confirming the formula.

Go deeper · Derivation C · 1 minVolume and surface area when a smaller solid is scooped out of a larger one

Derivation CVolume and surface area when a smaller solid is scooped out of a larger one

The goal is to extend the addition rules of Sections 12.2 and 12.3 to the reverse situation — a cavity, where material is removed rather than added, as when a hemispherical depression is cut into a cube, a cone is hollowed from a cylinder, or lead shots are dropped into a cone of water. Volume subtracts exactly as it adds, but surface area is more interesting: the flat mouth-area removed is lost, while a new curved surface is exposed inside, so the two effects must be tracked separately rather than assumed to cancel.

Step 1. If solid $S$ has volume $V_S$ , and a smaller solid $C$ of volume $V_C$ is scooped out (leaving an empty hollow), the remaining volume is $V_{\text{remaining}} = V_S - V_C$ — the same additive logic as Section 12.3, run in reverse.

Step 2. Set up the surface area rule for a cavity as a two-part accounting. First, the flat face of $S$ where the cavity's mouth sits loses exactly the area of that mouth (a circle of area $\pi r^2$ for a hemisphere or cone of base radius $r$ ) — now a hole, no longer part of the boundary. Second, a new surface is exposed: the curved inner wall of the cavity, equal to the cavity's own CSA (hemispherical dent: $2\pi r^2$ ; conical dent: $\pi rl$ ).

Step 3. Combine both effects: $\text{TSA}_{\text{remaining}} = \text{TSA}_S - (\text{mouth area}) + (\text{CSA of the cavity}).$ For a hemispherical cavity of radius $r$ , this gives $\text{TSA}_S - \pi r^2 + 2\pi r^2 = \text{TSA}_S + \pi r^2$ — surface area increases, since the exposed bowl ( $2\pi r^2$ ) exceeds the flat circle it replaced ( $\pi r^2$ ). This is easy to get backwards if you assume "removed material means less area."

Step 4. The same logic runs in reverse for displacement: lead shots dropped into a cone of water displace a volume equal to their own total volume, $n \times \tfrac43\pi r_{\text{shot}}^3 = V_{\text{displaced water}}$ , which is solved for the count $n$ . Cavities, scoops and displacement are all one idea viewed from different angles.

Worked examples

Board level - the spinning top: cone surmounted by a hemisphere (NCERT Example 1)

Rasheed got a playing top (lattu) which is a cone surmounted by a hemisphere. The entire top is $5\ \text{cm}$ in height and the diameter of the top is $3.5\ \text{cm}$ . Find the area he has to colour. (Take $\pi = \tfrac{22}{7}$ .)

Step 1. Read off the shared radius: diameter $3.5\ \text{cm}$ gives $r = 1.75\ \text{cm}$ . TSA of the top means CSA of hemisphere plus CSA of cone (a flush join of equal radii): $2\pi r^2 = 2 \times \tfrac{22}{7} \times 1.75 \times 1.75 = 19.25\ \text{cm}^2$ .

Step 2. Find the cone's own height by subtracting the hemisphere's radius from the total height: $h = 5 - 1.75 = 3.25\ \text{cm}$ — skipping this and using the full $5\ \text{cm}$ is the commonest mistake here.

Step 3. Slant height $l = \sqrt{r^2+h^2} = \sqrt{1.75^2 + 3.25^2} = \sqrt{13.625} \approx 3.7\ \text{cm}$ , so cone's CSA $= \pi rl = \tfrac{22}{7} \times 1.75 \times 3.7 \approx 20.35\ \text{cm}^2$ .

Step 4. Add: $19.25 + 20.35 = 39.6\ \text{cm}^2$ . Answer: the area to be coloured is approximately $\mathbf{39.6\ \text{cm}^2}$ .

Board level - the kulfi glass: apparent versus actual capacity (NCERT Example 6)

A juice seller used glasses with an inner diameter $5\ \text{cm}$ and height $10\ \text{cm}$ , but the base had a hemispherical raised portion that reduced the capacity. Find the apparent capacity of the glass and its actual capacity. (Use $\pi = 3.14$ .)

Step 1. Compute the apparent capacity as if the glass were a plain cylinder of radius $r = 2.5\ \text{cm}$ , height $10\ \text{cm}$ : $V_{\text{apparent}} = \pi r^2 h = 3.14 \times 2.5 \times 2.5 \times 10 = 196.25\ \text{cm}^3$ .

Step 2. The raised hemispherical bump at the base displaces liquid, so its volume is subtracted, not added: $V_{\text{hemisphere}} = \tfrac23\pi r^3 = \tfrac23 \times 3.14 \times 15.625 \approx 32.71\ \text{cm}^3$ .

Step 3. Subtract to get the true drinkable volume: $V_{\text{actual}} = 196.25 - 32.71 = 163.54\ \text{cm}^3$ .

Step 4. Sanity-check: the bump occupies space liquid would otherwise fill, so actual capacity must be less than apparent, and $163.54 < 196.25$ confirms it. Answer: apparent capacity $\mathbf{196.25\ \text{cm}^3}$ , actual capacity $\mathbf{163.54\ \text{cm}^3}$ .

Board HOTS / foundation JEE-NEET - hemisphere-and-cone toy inside a circumscribing cylinder (NCERT Example 7)

A solid toy is a hemisphere surmounted by a right circular cone. The cone's height is $2\ \text{cm}$ and the base diameter is $4\ \text{cm}$ . Find the volume of the toy. If a right circular cylinder exactly circumscribes the toy, find the difference between the volumes of the cylinder and the toy. (Take $\pi = 3.14$ .)

Step 1. The shared radius is half the diameter: $r = 2\ \text{cm}$ , and the cone's height is given directly as $h = 2\ \text{cm}$ — no subtraction needed here, unlike Example 1, since the total height was not given as one combined number.

Step 2. Add the two volumes directly (no hidden-face correction for volume): $V_{\text{toy}} = \tfrac23\pi r^3 + \tfrac13\pi r^2 h = 16.75 + 8.37 = 25.12\ \text{cm}^3$ .

Step 3. The circumscribing cylinder shares the toy's radius $r = 2\ \text{cm}$ , and its height spans the toy's full axial extent — hemisphere's radius plus cone's height, $H = r+h = 4\ \text{cm}$ . Its volume: $V_{\text{cyl}} = \pi r^2 H = 3.14 \times 4 \times 4 = 50.24\ \text{cm}^3$ .

Step 4. Subtract to find the empty space between toy and cylinder: $50.24 - 25.12 = 25.12\ \text{cm}^3$ . Answer: volume of the toy is $\mathbf{25.12\ \text{cm}^3}$ , and the difference from the circumscribing cylinder is $\mathbf{25.12\ \text{cm}^3}$ . Timing note: aim for under 6 minutes on a two-part circumscribing-solid problem like this. The reusable pattern is: find the shared radius first (it locks every other dimension), add the toy's own volumes as in Step 2, then separately work out the circumscribing cylinder's height as (hemisphere radius) plus (cone height) — never assume it equals the toy's own "visual height" without checking which parts stack along the axis. Save the subtraction for the very last step; doing it earlier by mistake is the most common slip on this exact question type.

Common traps

The single most common error is adding the total surface areas of the constituent solids instead of their curved surface areas, which silently double-counts the hidden joined face. A close second is using the total height of a composite solid as the height of just one part (Worked Example 1): whenever only a combined height is given, subtract the hemisphere's radius (or the other part's height) before finding the missing part's own height, and recompute the slant height $l=\sqrt{r^2+h^2}$ from that corrected height, not the total one. In cavity problems, a frequent sign error is subtracting the cavity's curved surface area instead of adding it — Derivation 3 shows a scooped surface typically increases, not decreases, which surprises most students. Finally, in capacity problems, mixing up "apparent" and "actual" capacity — forgetting a raised inner solid reduces liquid volume while a scooped cavity increases it — is a common source of backwards subtraction; always sanity-check whether the answer should be larger or smaller than the base figure.

Common trapTotal surface area of a combination is the sum of curved surface areas only, never the sum of total surface areas

The chapter's single biggest misconception is treating "find the total surface area of the combined solid" as "add up the TSA of each piece." Do that for the spinning top of Worked Example 1 and you would wrongly include the hemisphere's flat circular base (area $\pi r^2$ ) and the cone's own flat circular base (also area $\pi r^2$ ) — but neither of those two circles is actually part of the outer boundary of the assembled top; they are pressed together, sealed inside the solid, invisible and unpaintable. The correct method, proved in the first derivation above, is to start from each solid's curved surface area only, and add back any flat area that is not consumed by the join (as with the cube-and-hemisphere block, where most of the cube's top face remains genuinely exposed around the hemisphere's base). Volume has no such trap — volumes of the parts simply add, always — so this warning applies specifically and only to surface area questions.

You should be able to…

State the CSA, TSA and volume formulas for a cuboid, cube, cylinder, cone, sphere and hemisphere from memory, including the Pythagoras relation $l=\sqrt{r^2+h^2}$ for a cone's slant height.
Identify the constituent basic solids that make up a given composite figure (top, rocket, capsule, bird bath, decorative block) and their shared or independent radii.
Explain, using the hidden-face argument, why total surface area of a combination sums curved surface areas rather than total surface areas.
Correctly compute the height of one constituent solid when only the combined total height of a composite figure is given.
Compute the total surface area of a combination of two solids joined flat-face to flat-face, including cases where a flat face is only partially covered (leaving an exposed ring).
Compute the volume of a combination of solids by straightforward addition of the individual volumes.
Handle scooped-out or cavity problems by subtracting the removed solid's volume and correctly adjusting the surface area (subtract the mouth's flat area, add the cavity's curved surface area).
Solve apparent-versus-actual capacity problems and displacement (lead-shot) problems by correctly identifying which volume adds and which subtracts.
Set up and solve problems involving a solid circumscribed by a cylinder or other container of matching radius.

Ask the tutor: Want to work through a combination-of-solids problem from your own textbook, step by step, together?

Why does surface area not simply add up the way volume does?

Volume measures the space enclosed inside a solid, and when two solids are joined, their interiors simply pool together — nothing about the inside is hidden or lost, so volumes add without any correction. Surface area, in contrast, measures the outer boundary — the part you could touch or paint. When two flat faces are glued together, both of them stop being part of the outer boundary; they become sealed inside the new solid. So surface area must be built from only the curved parts (and any flat parts that remain genuinely exposed), never from a plain sum of the parts' total surface areas. This is proved carefully in the first derivation of this article, and it is the one rule this entire chapter is really testing.

How do I know whether to add or subtract volumes in a combination problem?

Ask whether the second solid is stacked onto the first (added material) or scooped out of the first (removed material, leaving a hollow). Toys, tops, rockets and capsules are almost always additions — every piece contributes its full volume. Cavities, depressions, and "a solid is scooped out from" phrasing signal subtraction — the scooped solid's volume comes out of the original. Reread the question for verbs like "surmounted," "mounted" (add) versus "scooped out," "a depression is cut" (subtract) before writing any formula.

When a problem gives only the total height of a composite solid, how do I find the height of just the cone or cylinder part?

Identify which other part occupies height along the same axis, and subtract its contribution from the total. For a cone-on-hemisphere top, the hemisphere's own "height" is exactly its radius $r$ (its flat face sits at the equator, its highest point is $r$ above that), so the cone's own height is (total height) minus $r$ . For a cone-on-cylinder rocket, the cylinder's height is usually given separately, so the cone's height is (total height) minus (cylinder's height). Always draw a quick vertical sketch marking which segment belongs to which solid before subtracting.

Is a frustum or the conversion of one solid into another part of this chapter?

No, not in the current NCERT syllabus for this chapter — the two sections here are strictly about joining basic solids together (or scooping one out of another) and computing surface area or volume for the resulting shape. Melting-and-recasting problems (reshaping one solid into another of equal volume) and the frustum of a cone (a cone with its top sliced off) belong to a different, separate topic and are not covered by this chapter in its present form — treat them as a related but distinct extension if your board's paper includes them.