Volume of combination of solids

An ice-cream cone topped with a hemispherical scoop, a cylindrical tank capped by two conical ends, a wooden cylinder with a conical cavity drilled out — everyday objects are rarely a single textbook solid. They are combinations of basic solids: cuboids, cylinders, cones, spheres and hemispheres, joined or hollowed out of one another. Finding the volume of such a composite figure is simpler than finding its surface area, because volume never double-counts anything at a shared joint. If a part is stuck on, its volume is added; if a part is carved out, its volume is subtracted. This section turns that add-or-subtract idea into a reliable method, works through a build-up example and a scooped-out example in full, and flags the mistakes that most often cost marks in this chapter of CBSE Class 10 Mathematics.

The core insight

The volume of a composite solid is the sum of the volumes of its added parts minus the volumes of any parts removed, since volume — unlike surface area — has no hidden joint to worry about.

The core idea

A composite solid is built from two or more basic solids meeting at a common face. Because volume measures enclosed space rather than exposed surface, there is no correction needed at the joint: every cubic centimetre of material is counted exactly once, whether it sits at the surface or buried inside where two pieces meet.

When solids are joined (added). If a cone sits on a cylinder, or a hemisphere caps a cone, the total volume is just the sum of the individual volumes:

V_{\text{combination}} = V_1 + V_2 + \cdots

where: $V_1, V_2, \ldots$ are the volumes of each basic solid making up the figure, computed independently using its own formula.

When material is removed (subtracted). If a conical or hemispherical cavity is scooped, drilled, or carved out of a larger solid, the volume of the cavity must be taken away from the volume of the original solid:

V_{\text{remaining}} = V_{\text{original}} - V_{\text{removed}}

where: $V_{\text{original}}$ is the volume of the solid before any material is removed, and $V_{\text{removed}}$ is the volume of the cavity, computed as if it were a solid of the same shape and dimensions.

The five formulae you need. Every composite figure in this chapter is built from just these basic solids (radius $r$ , height $h$ , and for a cuboid, length $l$ and breadth $b$ ):

V_{\text{cuboid}} = lbh, \quad V_{\text{cylinder}} = \pi r^2 h, \quad V_{\text{cone}} = \frac{1}{3}\pi r^2 h, \quad V_{\text{sphere}} = \frac{4}{3}\pi r^3, \quad V_{\text{hemisphere}} = \frac{2}{3}\pi r^3

where: $\pi \approx \tfrac{22}{7}$ or $3.14$ , as instructed by the problem. Notice a cone is exactly one-third of the cylinder sharing its base and height, and a hemisphere is exactly half of the full sphere of the same radius — both facts are worth memorising, since they let you sanity-check an answer at a glance.

Reading the figure correctly matters more than the arithmetic. In most composite figures, one dimension — usually the radius — is shared between the joined solids, while the heights differ and belong to different parts. Before substituting any numbers, identify each basic solid in the figure, decide whether it is added or removed, and note which radius/height value belongs to which part.

Derivation

Derivation AVolume of a composite solid built from a cone, cylinder and hemisphere of common radius

The goal is to show, step by step and without skipping the algebra, exactly how the three separate volume formulae for a cone, a cylinder and a hemisphere sharing one radius combine into a single total volume by straightforward addition — the reasoning that justifies simply writing $V = V_{\text{cone}} + V_{\text{cylinder}} + V_{\text{hemisphere}}$ instead of deriving some new combined formula from scratch.

Step 1. Set up the shared quantity. Let the common radius of the base be $r$ , the cylinder's height be $h_1$ , and the cone's height be $h_2$ . Since all three solids are attached along circular cross-sections of the same radius $r$ , that one value of $r$ appears in every formula — this is exactly why problems in this chapter usually give you a single radius rather than three separate ones.

Step 2. Write the three volumes separately, using each solid's own formula: $V_{\text{cylinder}} = \pi r^2 h_1$ , $V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h_2$ , and $V_{\text{hemisphere}} = \tfrac{2}{3}\pi r^3$ . Each formula is applied exactly as it would be for that solid standing alone — attaching it to the others does not change its own volume, because no material is lost or gained by the act of joining.

Step 3. Justify the addition. Volume measures the three-dimensional space enclosed by the solid's boundary. When the cone, cylinder and hemisphere are joined face-to-face, the interior points of the composite figure are exactly the union of the interior points of the three pieces, with no overlap (the joining faces have zero thickness and so add zero extra volume). Therefore the total enclosed space is simply the sum of the three separate enclosed spaces.

Step 4. Combine into the working formula: $V_{\text{total}} = \pi r^2 h_1 + \tfrac{1}{3}\pi r^2 h_2 + \tfrac{2}{3}\pi r^3$ , or, factoring $\pi r^2$ from the first two terms, $V_{\text{total}} = \pi r^2\left(h_1 + \tfrac{1}{3}h_2\right) + \tfrac{2}{3}\pi r^3$ . Factoring is only a computational shortcut, not a new rule — and the same add-or-subtract logic extends unchanged to any combination of cuboids, cylinders, cones, spheres and hemispheres.

Worked example

A solid is in the shape of a cylinder with a cone on one end and a hemisphere on the other end of the same radius. The radius of the common base is $3.5$ cm, the cylinder's height is $10$ cm, and the cone's height is $6$ cm. Find the total volume of the solid. (Use $\pi = \tfrac{22}{7}$ .)

Step 1. Identify the parts and their shared dimension. Radius $r = 3.5$ cm $= \tfrac{7}{2}$ cm for the cylinder, cone and hemisphere alike. Cylinder height $h_{\text{cyl}} = 10$ cm. Cone height $h_{\text{cone}} = 6$ cm. All three parts are added — none is a cavity.

Step 2. Compute the cylinder's volume: $V_{\text{cyl}} = \pi r^2 h = \tfrac{22}{7} \times \left(\tfrac{7}{2}\right)^2 \times 10 = \tfrac{22}{7} \times \tfrac{49}{4} \times 10 = 385\ \text{cm}^3$ .

Step 3. Compute the cone's volume: $V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times \tfrac{22}{7} \times \tfrac{49}{4} \times 6 = \tfrac{1}{3} \times 38.5 \times 6 = 77\ \text{cm}^3$ .

Step 4. Compute the hemisphere's volume: $V_{\text{hemi}} = \tfrac{2}{3}\pi r^3 = \tfrac{2}{3} \times \tfrac{22}{7} \times \left(\tfrac{7}{2}\right)^3 = \tfrac{2}{3} \times \tfrac{22}{7} \times \tfrac{343}{8} \approx 89.83\ \text{cm}^3$ .

Step 5. Add the three volumes, since all parts are joined, not removed: $V_{\text{total}} = 385 + 77 + 89.83 \approx 551.83\ \text{cm}^3$ . The total volume of the solid is approximately $551.83\ \text{cm}^3$ .

Competitive stretch

A cylindrical block of wood has radius $6$ cm and height $15$ cm. A conical cavity of the same radius and height $9$ cm is drilled out from one flat end. Find the volume of wood left in the block. (Use $\pi = 3.14$ .)

Step 1. This time one part is removed, not added, so subtraction applies. The original solid is the full cylinder: $r = 6$ cm, $h = 15$ cm. The removed cavity is a cone of the same radius: $r = 6$ cm, $h = 9$ cm.

Step 2. Cylinder volume: $V_{\text{cyl}} = \pi r^2 h = 3.14 \times 36 \times 15 = 1695.6\ \text{cm}^3$ . Cone (removed) volume: $V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times 3.14 \times 36 \times 9 = 339.12\ \text{cm}^3$ .

Step 3. Subtract: $V_{\text{remaining}} = 1695.6 - 339.12 = 1356.48\ \text{cm}^3$ . So $1356.48\ \text{cm}^3$ of wood remains after the conical cavity is drilled out.

Common traps

Adding when material has actually been removed. Words like scooped out, drilled, carved, or cavity are a signal to subtract, not add — read the problem for these cues before writing the formula.
Mixing up which height belongs to which part. The radius is often shared between joined solids, but the heights are almost always different — re-check which height goes with which formula before substituting.
Forgetting the one-third in the cone formula, especially when a cone sits alongside a cylinder of the same radius and height, where it is tempting to treat their volumes as equal (a cone is exactly one-third of the matching cylinder, never the same).
Carrying over surface-area habits. Unlike surface area, volume needs no adjustment for hidden or shared faces at a joint — every unit of material counts exactly once.

Ask the tutor: If you are unsure whether a composite-solid problem wants you to add or subtract volumes, describe the shape to the AI tutor and it will help you visualise the figure and set up the calculation step by step.

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

Look at how the solids are physically combined. If one shape is stuck onto another so that both occupy real material — like a cone fixed on top of a cylinder — their volumes add, because all of it is solid matter. If instead a shape has been scooped out, drilled, or carved from inside another solid, leaving an empty cavity, that cavity's volume must be subtracted from the original solid's volume. Words like mounted, surmounted, or attached usually signal addition; words like hollowed, drilled, or removed signal subtraction.

Why doesn't volume need the same joint correction that surface area does?

Surface area needs a correction because the circular face where two solids meet is hidden inside the combined figure, so you must subtract that hidden face from the total exposed surface. Volume works differently: it measures the space enclosed by the whole boundary, not the boundary itself. Since the joining face has essentially zero thickness, it contributes zero extra volume either way — every bit of material inside the combined solid is counted exactly once, whether it sits near the surface or deep at the joint. That is why volumes simply add (or subtract, for cavities) with no correction term needed.

What if a composite solid has more than three basic parts, like a cuboid with a cylindrical hole and a hemisphere on top?

The same add-or-subtract rule extends to any number of parts. Identify every basic solid making up the figure one at a time, decide whether each is added (built onto the shape) or removed (a cavity carved from it), compute each volume with its own formula and its own correct radius/height, and then combine them with plus and minus signs accordingly — for example, $V_{\text{total}} = V_{\text{cuboid}} + V_{\text{hemisphere}} - V_{\text{cylindrical hole}}$ . There is no limit on how many pieces a composite solid can have; the method never changes, only the number of terms in the final sum.