Volume of solids

A water tank, a brick, a steel rod melted down and poured into a mould — every one of these is a question about volume: how much three-dimensional space a solid takes up. In Class 8 you learn to measure that space for exactly three shapes — the cube, the cuboid, and the right circular cylinder — and to use volume in real situations like digging a pit, filling a tank, or melting one metal object and recasting it into another shape. The same three formulas that describe surface area come back here in a different role. Instead of adding up the areas of outer faces, volume asks how many unit cubes would pack the inside solid. This article builds the three volume formulas from a single idea — base area stacked through a height — works through a melt-and-recast problem step by step, and flags the mix-ups that cost the most marks in this chapter.

The core insight

The volume of a cube, cuboid, or right circular cylinder is always the area of its base multiplied by its height, and this volume stays constant when the same solid is melted and recast into a different shape.

The core idea

Volume measures the amount of three-dimensional space enclosed by a solid, in cubic units such as $\text{cm}^3$ or $\text{m}^3$ . Unlike surface area, which adds up flat outer faces in square units, volume is about how much the solid fills. For a cuboid with length $l$ , breadth $b$ , and height $h$ , the volume is:

V_{cuboid} = l \times b \times h

where: $l$ , $b$ , $h$ are the three edge lengths of the box, all measured in the same unit, so that $V$ comes out in that unit cubed. A cube is just a special cuboid where all three edges are equal, $l = b = h = a$ :

V_{cube} = a^3

where: $a$ is the edge length of the cube. For a right circular cylinder of radius $r$ and height $h$ , the volume is the area of the circular base multiplied by the height:

V_{cylinder} = \pi r^2 h

where: $r$ is the base radius, $h$ is the perpendicular height between the two circular ends, and $\pi \approx \dfrac{22}{7}$ or $3.14$ . Look closely and all three formulas say the same thing: base area times height. For the cuboid, $l \times b$ is the area of the rectangular base, and $h$ stacks that base upward. For the cylinder, $\pi r^2$ is the area of the circular base, stacked through $h$ the same way. Seeing every volume formula as "base area $\times$ height" makes them far easier to recall correctly, and stops the common error of confusing a volume formula (which multiplies lengths to get cubic units) with a surface-area formula (which adds up face areas to get square units). A useful scaling check: in the cylinder formula, $r$ is squared but $h$ is not. So doubling the radius multiplies the volume by $2^2 = 4$ , while doubling the height only doubles the volume. The same asymmetry holds for a cuboid — doubling one edge doubles the volume, but doubling two edges at once (say $l$ and $b$ ) multiplies it by 4. Estimating this way, before calculating exactly, catches many arithmetic slips.

Figure

A cuboid shown as five thin horizontal layers stacked to a height h above a rectangular base of area length times breadth, illustrating that volume is base area multiplied by height

Derivation AVolume is base area stacked through height

Show that the volume of a cuboid — and, as special or extended cases, a cube and a right circular cylinder — can be obtained by imagining the solid as a stack of many thin identical layers, each shaped like the base, piled up to the full height, so that the total volume equals the base area multiplied by the height.

Step 1. Start with a rectangular base of length $l$ and breadth $b$ . Its area, from the ordinary formula for a rectangle, is $A_{base} = l \times b$ . This is the flat "footprint" of the cuboid — the shape you would see if you looked straight down on it.

Step 2. Imagine slicing the solid cuboid into many thin horizontal layers, each of thickness $\delta$ , stacked one on top of another up to the total height $h$ . Every layer has exactly the same footprint as the base, so every layer has the same area, $A_{base} = l \times b$ , and a tiny volume of $A_{base} \times \delta$ .

Step 3. The number of such layers needed to reach height $h$ is $h / \delta$ . Adding up the volumes of all the layers gives the total volume:

V = \left(\frac{h}{\delta}\right) \times \left(A_{base} \times \delta\right) = A_{base} \times h

The thickness $\delta$ cancels out, which is why the result does not depend on how thin you imagine slicing the layers — the volume is simply base area times height.

Step 4. Substitute $A_{base} = l \times b$ to recover the cuboid formula, $V_{cuboid} = l \times b \times h$ . Setting $l = b = h = a$ collapses this to the cube, $V_{cube} = a^3$ . The identical stacking argument applies to a right circular cylinder: its base is a circle of area $\pi r^2$ instead of a rectangle, so stacking that circular base through height $h$ gives $V_{cylinder} = \pi r^2 h$ . All three solids share one underlying idea — a fixed base shape, repeated (stacked) through a height.

Worked example

A solid metal cube of edge 12 cm is melted and recast into a right circular cylinder of radius 6 cm. Find the height of the cylinder, correct to two decimal places. (Take $\pi = \dfrac{22}{7}$ .)

Step 1. When a solid is melted and recast into a different shape, no metal is lost or gained, so the volume is conserved: $V_{cube} = V_{cylinder}$ .

Step 2. Find the volume of the cube using $V_{cube} = a^3$ with $a = 12$ cm:

V_{cube} = 12^3 = 1728\ \text{cm}^3

Step 3. Set this equal to the cylinder's volume formula, $\pi r^2 h$ , with $r = 6$ cm:

\pi r^2 h = 1728

\frac{22}{7} \times 6^2 \times h = 1728

\frac{22}{7} \times 36 \times h = 1728 \implies \frac{792}{7} \times h = 1728

Step 4. Solve for $h$ by multiplying both sides by $\dfrac{7}{792}$ :

h = 1728 \times \frac{7}{792} = \frac{12096}{792} \approx 15.27\ \text{cm}

The height of the cylinder is approximately $15.27$ cm.

Competitive stretch

Suppose instead the same cube of volume $1728\ \text{cm}^3$ is recast into two identical smaller cylinders of the same radius, $r = 6$ cm, each sharing the melted volume equally. Each cylinder then holds $1728 / 2 = 864\ \text{cm}^3$ , so $\dfrac{22}{7} \times 36 \times h = 864$ , giving $h = 864 \times \dfrac{7}{792} \approx 7.64\ \text{cm}$ — exactly half the single-cylinder height, since the base area is unchanged and only the volume (and therefore the height) has been halved. This confirms the "base area $\times$ height" picture: with $A_{base}$ fixed, $h$ scales directly with $V$ .

Common traps

Mixing up surface area and volume formulas. Writing $2(lb+bh+hl)$ (surface area) when the question asks for volume ( $lbh$ ), or the reverse. Check the units asked for: cubic units mean volume, square units mean surface area.
Forgetting to cube the edge for a cube's volume. Some students compute $a \times 3$ instead of $a \times a \times a = a^3$ .
Mixing units within one calculation. Substituting a length in metres and another in centimetres without converting first gives a wildly wrong volume; convert every dimension to the same unit before multiplying.
Equating surface areas instead of volumes in melt-and-recast problems. The quantity conserved when a solid is reshaped is volume, not surface area — surface area typically changes because the outer shape is different.

Ask the tutor: If a melt-and-recast question or a mixed-unit problem is confusing you, ask the AI tutor to walk through it with you step by step, or to generate fresh practice problems on cube, cuboid, and cylinder volume.

Why does volume use cubic units while surface area uses square units

Volume measures how much three-dimensional space a solid fills, so it is built from three lengths multiplied together (length times breadth times height), which is why the unit is cubed, such as $\text{cm}^3$ . Surface area instead adds up the areas of flat outer faces, and each face's area is just two lengths multiplied together, giving square units like $\text{cm}^2$ . Always check which one a question is asking for before choosing a formula.

In a melt-and-recast problem, why is volume conserved but not surface area

Melting and recasting a solid changes its shape but not the amount of material in it — the same metal is simply poured into a new mould. Since volume measures the amount of space that material occupies, it stays the same before and after. Surface area, however, depends on the outer shape, and a cube and a cylinder of equal volume can have quite different surface areas, since surface area is about the boundary, not the amount of material inside it.

How do I know whether to use the cube, cuboid, or cylinder formula in a word problem

Look at the shape described. A box, room, pit, or tank with straight rectangular edges is a cuboid (or a cube if all edges are equal); anything round like a pipe, drum, pillar, or well is a right circular cylinder. Reading the problem for words like "circular cross-section" or "radius" signals a cylinder, while "length, breadth, height" signals a cuboid or cube. Sketching the solid quickly before choosing a formula avoids picking the wrong one.

Volume of solids

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads