Combination of solids

A medicine capsule is a cylinder with two hemispherical caps. An ice-cream cone with a scoop on top is a cone plus a hemisphere. A wooden toy is often a cone glued onto a cylinder. None of these is a single named solid, yet every Class 10 mensuration problem about them can be solved using only the cylinder, cone and sphere or hemisphere formulas you already know — cylinder $V=\pi r^2h$ , cone $V=\tfrac13\pi r^2h$ , sphere or hemisphere volume and curved surface area (CSA). The entire skill of "combination of solids" is learning two short rules for how to combine those known formulas correctly: volumes simply add or subtract, but surface areas do not, because the surface at the join is hidden inside the solid. This article compresses that skill into the rules, one full worked derivation, and the traps that cost the most exam marks.

The core insight

Volumes of joined or hollowed-out solids add or subtract directly, but total surface area must drop the flat circular face(s) buried at every join, keeping only the curved surfaces plus any base that stays exposed.

The core idea

Composite solids are built from three known pieces. Every combination question in this syllabus assembles the right circular cylinder, the right circular cone, and the sphere/hemisphere — never a frustum, which is outside the ICSE Class 10 syllabus. Recognising which two (or three) of these three shapes make up the object, and which dimension (radius, height, slant height) is shared at the join, is the first and most error-prone step.

Rule 1 — volume is always additive or subtractive. If the solid is built up by joining two pieces (a cone sitting on a cylinder), the combined volume is simply the sum of the separate volumes. If the solid is formed by removing one shape from another (a conical cavity drilled out of a cylinder), the combined volume is the difference.

V_{\text{combined}} = V_1 + V_2 \ \text{(joined)} \qquad \text{or} \qquad V_{\text{combined}} = V_1 - V_2 \ \text{(hollowed)}

where: $V_1$ and $V_2$ are the volumes of the individual solids computed with their own standard formulas, using whichever radius and height belong to that piece.

Rule 2 — surface area is never a plain sum. Wherever two solids meet, a flat circular face becomes internal and must be dropped from both pieces before adding what remains exposed. For a cone of base radius $r$ and slant height $l$ sitting on a cylinder of the same radius $r$ and height $h$ :

\text{TSA}_{\text{combined}} = \pi r l + 2\pi r h + \pi r^2

where: $\pi r l$ is the cone's curved surface only (its flat base is buried), $2\pi rh$ is the cylinder's curved surface, and the single $\pi r^2$ is the cylinder's bottom base — the only flat circular face still on the outside. A capsule (cylinder with a hemispherical cap at each end) is the cleanest case of all: it has no flat circular face anywhere, so its TSA is just the cylinder's CSA plus two hemispherical CSAs, $2\pi rh + 2(2\pi r^2)$ .

Figure

A cone sitting on top of a cylinder with the shared circular join marked as hidden and only the slanted cone surface and curved cylinder wall shown as the exposed outer surface

The core idea continued

Common combinations to recognise on sight. Cone-on-hemisphere (a toy top), hemisphere-on-cylinder (a pillar with a domed cap), cylinder-with-conical-cavity (a wooden block drilled out), and the capsule (cylinder with two hemispherical ends). In every case, first list which radius and heights are shared, then decide whether the problem wants volume (add/subtract) or TSA (drop the hidden circle).

Derivation ATSA and volume of a cone mounted on a hemisphere of equal radius

Find the general total surface area and volume formulas for a solid made of a right circular cone of base radius $r$ and height $h$ sitting exactly on top of a hemisphere of the same radius $r$ , given that the two rules above (additive volume, join-excluding surface area) apply, and show algebraically why the flat circle where cone meets hemisphere never appears in the final TSA expression.

Step 1. Write the volume of each piece separately using the standard formulas, keeping their shared radius $r$ as one symbol throughout: the cone contributes $V_{\text{cone}} = \tfrac13\pi r^2h$ and the hemisphere contributes $V_{\text{hemi}} = \tfrac23\pi r^3$ . Because the toy is formed by joining these two pieces at their flat circular faces, and joining always adds volume, the total is simply their sum with no correction term: $V_{\text{toy}} = \tfrac13\pi r^2h + \tfrac23\pi r^3$ .

Step 2. Identify every surface that survives on the outside of the combined solid. The hemisphere contributes its curved dome, $2\pi r^2$ ; the cone contributes its slanted surface, $\pi r l$ , where $l=\sqrt{r^2+h^2}$ by Pythagoras applied to the cone's radius and height. The hemisphere's flat circular face and the cone's flat circular base are the same circle, pressed together at the join — both vanish from the outer surface simultaneously, which is why no separate "join area" term is subtracted; it was simply never added in the first place.

Step 3. Assemble the total surface area from only the two curved pieces identified in Step 2: $\text{TSA}_{\text{toy}} = \pi r l + 2\pi r^2$ , with $l = \sqrt{r^2+h^2}$ substituted from Step 2's Pythagorean relation. Notice this formula has no flat-circle term at all, unlike the cone-on-cylinder case, because here both flat faces disappear rather than one of two surviving.

Step 4. Sanity-check the logic by comparing to the cone-on-cylinder case from the core idea section: there, the cylinder's own bottom base stays exposed, so one $\pi r^2$ term survives; here, both flat faces (cone's base and hemisphere's flat face) coincide and cancel, so zero flat-area terms survive. The general lesson is that the number of surviving flat circles equals the number of flat faces in the whole composite minus twice the number of internal joins.

Worked example

A toy is shaped as a cone mounted on a hemisphere of the same radius. The base radius is $r = 3.5\text{ cm}$ , and the cone's height (measured above the hemisphere) is $h = 12\text{ cm}$ . Find the toy's volume and total surface area, taking $\pi = \dfrac{22}{7}$ .

Step 1. Find the cone's slant height using the Pythagorean relation from the derivation above: $l = \sqrt{r^2+h^2} = \sqrt{3.5^2+12^2} = \sqrt{12.25+144} = \sqrt{156.25} = 12.5\text{ cm}$ .

Step 2. Apply the volume formula from Step 1 of the derivation: $V_{\text{cone}} = \tfrac13\pi r^2h = \tfrac13\times\tfrac{22}{7}\times3.5^2\times12 = 154\text{ cm}^3$ , and $V_{\text{hemi}} = \tfrac23\pi r^3 = \tfrac23\times\tfrac{22}{7}\times3.5^3 \approx 89.83\text{ cm}^3$ . Adding, since the pieces are joined, $V_{\text{toy}} = 154 + 89.83 = 243.83\text{ cm}^3$ .

Step 3. Apply the TSA formula from Step 3 of the derivation: $\text{CSA}_{\text{cone}} = \pi rl = \tfrac{22}{7}\times3.5\times12.5 = 137.5\text{ cm}^2$ and $\text{CSA}_{\text{hemi}} = 2\pi r^2 = 2\times\tfrac{22}{7}\times3.5^2 = 77\text{ cm}^2$ . No flat-circle term is added, per Step 4 of the derivation, so $\text{TSA}_{\text{toy}} = 137.5 + 77 = 214.5\text{ cm}^2$ .

The toy's volume is $243.83\text{ cm}^3$ and its total surface area is $214.5\text{ cm}^2$ .

Competitive stretch

If the same toy is instead placed inside a snugly fitting cylindrical box of radius $3.5\text{ cm}$ and height equal to the toy's total height ( $12+3.5=15.5\text{ cm}$ , since the hemisphere adds its radius as height), the empty space (air gap) left inside the box is the cylinder's volume minus the toy's volume: $V_{\text{cyl}} = \pi r^2h_{\text{box}} = \tfrac{22}{7}\times3.5^2\times15.5 \approx 596.75\text{ cm}^3$ , so the air gap is $596.75 - 243.83 \approx 352.92\text{ cm}^3$ — a classic hollowed-out-solid extension of the same two rules.

Common traps

Adding the two pieces' surface areas as if nothing were joined, instead of dropping the flat circle(s) at the join.
Forgetting that a hollowed-out cavity adds its own newly exposed curved or flat surface to the TSA, rather than simply subtracting the removed piece's surface area.
Confusing a "total height" given in the problem with the height of just one part — it is often the sum of a cylinder's height and a cone's height, or a cone's height plus a hemisphere's radius.
Using the wrong radius when the two joined solids do not actually share the same radius, which is stated explicitly whenever it happens.

Ask the tutor: Sketch your own combination-of-solids problem and ask the AI tutor to walk through which surfaces survive and which volumes add or subtract.

Why don't we just add the surface areas of the two solids like we add their volumes

Because a surface area formula like $2\pi rh$ already counts every bit of that solid's outer boundary, including the flat circular face at the join. When two solids are glued together, that particular circle stops being an outer boundary — it becomes buried between the pieces, touching nothing but the other solid. Volume behaves differently because volume is about the amount of material or space enclosed, and gluing two chunks of material together never removes any material, so the volumes still add cleanly even though one surface disappears.

How do I know whether a problem wants me to add or subtract volumes

Read for whether the object is built up or hollowed out. Words like "mounted on," "surmounted by," "attached to," or "a toy made of a cone and a hemisphere" describe joining, so volumes add. Words like "a cavity is drilled," "scooped out," "carved from," or "the greatest sphere is cut from a cube" describe removal, so you subtract the smaller volume from the larger one. The picture or description almost always makes this unambiguous — check it before writing any formula.

Do I need frustum formulas for this ICSE topic

No. The frustum of a cone (a cone with its tip sliced off) is a separate mensuration topic outside the ICSE Class 10 combination-of-solids syllabus. Every question here uses only whole cylinders, whole cones, and spheres or hemispheres, joined or hollowed according to the two rules in this article. If a problem mentions a frustum, it belongs to a different unit and different formulas.