Class 9 › Mathematics › Chapter

Math of Space: Surface Area and Volume

Surface area and volume of cuboids, cubes, right circular cylinders, cones, pyramids, spheres and hemispheres; how the formulas arise; how changing dimensions changes area and volume; and estimating quantities with sensible assumptions.

1. Surface Area, Volume and Units

Surface area: the total area of the surface of a solid (the "covering"). It is measured in square units.
Volume: the amount of three-dimensional space a solid or a container occupies. It is measured in cubic units.

Unit cube: a cube with each edge $1$ unit has volume $1$ cubic unit (for example $1\ \text{cm}^3$ or $1\ \text{m}^3$). The volume of a solid is the number of unit cubes that fit into it.
Nets: cutting a solid open and flattening it gives its net. Surface area is the area of the net, so the faces of a solid can be added up one by one.
Unit conversions: $1\ \text{m} = 100\ \text{cm}$, so $1\ \text{m}^2 = 10{,}000\ \text{cm}^2$ and $1\ \text{m}^3 = 10{,}00{,}000\ \text{cm}^3$. Also $1000\ \text{cm}^3 = 1$ litre. For length, $1\ \text{ft} \approx 30.48\ \text{cm}$, so $1\ \text{ft}^3 \approx 28{,}317\ \text{cm}^3 \approx 28.3$ litres.
Consistent units: convert all lengths to one unit before using a formula. A rate "per $100\ \text{cm}^2$" needs the area in $\text{cm}^2$.
Value of $\pi$: use $\pi \approx \dfrac{22}{7}$ or $\pi \approx 3.14$ as the problem suggests. Leaving the answer in terms of $\pi$ (for example $100\pi\ \text{cm}^3$) keeps it exact.
Cross-section: an extremely thin slice of a solid is its cross-section. MRI scans work like this: they show the inside of the body in slices.

2. Cuboids

Cuboid: a solid with six rectangular faces, described by three quantities: length $l$, width $w$ and height $h$. It is the 3D version of a rectangle.

Total surface area (TSA) $$\text{TSA} = 2(wl + hl + hw)$$ Volume $$V = lwh = (\text{area of base}) \times \text{height}$$

A cuboid is a pile of thin rectangular sheets, like a pack of cards: volume $=$ area of one sheet $\times$ number of sheets.

Surface area: a cuboid has three pairs of equal faces: two of area $wl$, two of area $hl$, two of area $hw$. Adding all six gives the TSA.
Why $V = lwh$: the base has $l \times w$ unit squares. Stacking $h$ layers of this kind gives $lwh$ unit cubes.
Compare with a rectangle: area of a rectangle $= \text{length} \times \text{width}$; volume of a cuboid $= \text{length} \times \text{width} \times \text{height}$.
Packing boxes into a container: if the boxes tile the container exactly, the number of boxes is $\dfrac{\text{volume of container}}{\text{volume of one box}}$. Otherwise count how many fit along each edge, using the edge length divided by the box edge, and multiply the three counts.
Cuboid from its face areas: if the three face areas meeting at a corner are $lw$, $wh$, $lh$, their product is $(lwh)^2$. So $V = \sqrt{\text{product of the three face areas}}$.
Adding a slab: increasing one dimension by $1$ unit adds a slab of volume equal to the product of the other two (for example, length $+1$ adds $wh$ cubic units).
Same volume, different surface area: a cube and a cuboid of equal volume have different surface areas, and for a given volume the cube has the least surface area of all cuboids. A cube needs the least covering material, while a cuboid exposes more area to its surroundings. This matters in packaging, heat transfer and engineering.

3. Cubes

Cube of side $a$ $$\text{TSA} = 6a^2, \qquad V = a^3$$ These are the cuboid formulas with $l = w = h = a$: $2(a^2 + a^2 + a^2) = 6a^2$ and $a \cdot a \cdot a = a^3$.

Cube numbers: $1, 8, 27, 64, \dots$ are the volumes of cubes with sides $1, 2, 3, 4, \dots$, that is, the number of unit cubes inside each. Cube roots reverse this: $\sqrt[3]{64} = 4$.
Surface area and volume in terms of each other: from $V = a^3$ we get $a = V^{1/3}$ and $\text{TSA} = 6V^{2/3}$. From $S = 6a^2$ we get $a = \sqrt{S/6}$ and $V = \left(\dfrac{S}{6}\right)^{3/2}$.
Growing a cube: the number of unit cubes to be added to a cube of side $a$ to make a cube of side $a + 1$ is $(a+1)^3 - a^3 = 3a^2 + 3a + 1$.
Cutting a cube into smaller cubes: cut a cube of side $a$ into small cubes of side $\dfrac{a}{n}$. There are $n^3$ small cubes, the total volume is unchanged, but the total surface area becomes $n$ times the original ($n^3 \times 6\left(\frac{a}{n}\right)^2 = 6a^2 n$). The surface area grows a lot while the volume stays the same. This has major consequences in the biological world.
Painted cube cut into unit cubes (side $n \ge 2$): $8$ corner cubes have $3$ faces painted; $12(n-2)$ edge cubes have $2$ faces painted; $6(n-2)^2$ face cubes have $1$ face painted; $(n-2)^3$ inner cubes have no paint.
Joining cubes: when solids are joined, the faces that touch are hidden and are not counted in the surface area. The volume is simply the sum.

4. Right Circular Cylinder

Right circular cylinder: a cylinder whose cross-section parallel to the base is a circle ("circular") and whose axis is perpendicular to the base ("right"). A cylinder with a slanted axis is an oblique cylinder; we study only right ones.

Radius $r$, height $h$ $$\text{CSA} = 2\pi rh \qquad V = \pi r^2 h$$ $$\text{TSA (closed at both ends)} = 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$$ $$\text{Surface area (open at one end)} = 2\pi rh + \pi r^2 = \pi r(2h + r)$$

The curved surface opens out into a rectangle: its length is the circumference of the base and its width is the height.

Curved surface area: cut the curved surface along the height and flatten it. It becomes a rectangle of length $2\pi r$ and width $h$, so $\text{CSA} = \text{circumference} \times \text{height}$.
Closed or open: each flat circular end adds $\pi r^2$. Closed at both ends (biscuit tin): add $2\pi r^2$. Open at one end (tumbler): add $\pi r^2$. A pipe open at both ends has only the curved surface.
Volume: $V = \text{cross-sectional area} \times \text{height}$. Think of the cylinder as a stack of thin circular discs (like khakhras), each of area $\pi r^2$.
Cross-sections: a slice parallel to the base is a circle; a slice parallel to the axis is a rectangle.
How CSA and volume depend on $r$ and $h$: $\text{CSA} \propto rh$ and $V \propto r^2 h$. If the radius is doubled and the height halved, the CSA is unchanged but the volume doubles.
Constant volume: if the radius is multiplied by a factor $p$ and the volume must stay the same, the height must be multiplied by $\dfrac{1}{p^2}$. For example, a radius up by $10\%$ ($p = 1.1$) needs the height multiplied by $\dfrac{1}{1.21}$.
Largest cylinder in a cube of edge $a$: radius $\dfrac{a}{2}$ and height $a$, so $V = \pi\left(\dfrac{a}{2}\right)^2 a = \dfrac{\pi a^3}{4}$.
Change in volume when the radius grows by $1$: $\pi(r+1)^2 h - \pi r^2 h = \pi h(2r + 1) = 2\pi rh + \pi h$.
Material for a rolled-up sheet: a sheet used to make the curved part (a kaleidoscope, a pipe) needs area equal to the CSA.

5. Cones

Right circular cone: the solid formed when a right triangle is rotated through $360^\circ$ about one of its legs. That leg is the axis (the height $h$), the other leg sweeps out the base (the radius $r$) and the hypotenuse sweeps out the curved surface (the slant height $l$).

Radius $r$, height $h$, slant height $l$ $$l^2 = h^2 + r^2 \quad (\text{Baudhāyana–Pythagoras theorem})$$ $$\text{CSA} = \pi rl, \qquad \text{TSA} = \pi rl + \pi r^2 = \pi r(l + r), \qquad V = \dfrac{1}{3}\pi r^2 h$$

The curved surface of a cone unrolls into a sector of a circle of radius $l$.

Why CSA $= \pi r l$

Method 1 (thin triangles): unroll the surface and cut it into many very thin sectors, each almost an isosceles triangle of height $l$ and base $b_i$. The area is $\tfrac{1}{2}b_1 l + \tfrac{1}{2}b_2 l + \cdots = \tfrac{1}{2}(b_1 + b_2 + \cdots)\,l$. The bases add up to the circumference $2\pi r$, so $\text{CSA} = \tfrac{1}{2}(2\pi r)\,l = \pi r l$.

Method 2 (sector): the sector has radius $l$ and angle $\theta$, so its area is $\pi l^2 \times \dfrac{\theta}{360^\circ}$. Its arc length equals the base circumference: $\dfrac{\theta}{360^\circ} \times 2\pi l = 2\pi r$, so $\dfrac{\theta}{360^\circ} = \dfrac{r}{l}$. Hence $\text{CSA} = \pi l^2 \times \dfrac{r}{l} = \pi r l$.

Volume vs cylinder: a cone has exactly $\dfrac{1}{3}$ of the volume of the cylinder with the same base and the same height. $V = \dfrac{1}{3} \times \text{area of base} \times \text{height}$.
Experiments: fill a paper cone with salt and pour it into a cylinder of the same radius and height; it takes three cones to fill the cylinder. Or mould clay into a cylinder and reshape it into identical cones of the same radius and height; you get exactly three.
Why the $\frac{1}{3}$: explaining it fully needs calculus (Grade 12). Archimedes (c. 225 BCE) first found the cone's volume formula.
Finding a missing length: any one of $r$, $h$, $l$ follows from the other two using $l^2 = h^2 + r^2$. When a right triangle is rotated about one leg, that leg is $h$, the other leg is $r$ and the hypotenuse is $l$.
Sector angle of the unrolled cone: $\theta = 360^\circ \times \dfrac{r}{l}$, and the radius of the sector is $l$.
Sheet and cloth problems: the material for the curved part (a cap, a tent) equals the CSA. For a strip of cloth of width $b$, length $= \dfrac{\text{CSA}}{b}$, plus any extra needed for stitching or wastage.
Cone filled to half its depth: the water forms a smaller cone with half the height and also half the radius, so its volume is $\left(\tfrac{1}{2}\right)^3 = \dfrac{1}{8}$ of the whole cone. In general, filling to a fraction $k$ of the depth gives $k^3$ of the volume.

6. Pyramidal Shapes

Pyramid: a solid whose base is a triangle, a square or any closed flat shape, with every point of the base joined to a single point, the apex, that lies outside the plane of the base.

Volume of a pyramidal shape $$V = \dfrac{1}{3} \times (\text{area of base}) \times \text{height}$$

Same formula for every base: the formula holds whether the base is triangular, square or any other flat shape.
A cone is a pyramid with a circular base, which is why its volume formula is the same.
Surface area: add up the areas of all the faces (the base and the triangular or polygonal side faces). No separate formula is needed.
Height: the height is the perpendicular distance from the apex to the plane of the base.
In the world: the Egyptian pyramids (square base), and pyramid-like mountain peaks such as Shivling Peak in Uttarakhand and the Matterhorn in the Alps (shaped by glacier erosion over millennia).
Square pyramid inside a cylinder: if its square base is inscribed in the cylinder's base circle (radius $r$) and its apex is at the top, the base area is $2r^2$, so $V_{\text{pyramid}} = \dfrac{1}{3}(2r^2)h$. Compared with the cylinder's $\pi r^2 h$, the ratio is $\dfrac{2}{3\pi}$.

7. Spheres

Sphere: the collection of all points in 3D space that are at a fixed distance $r$ (the radius) from a fixed point $O$ (the centre). Rotating a circle about a diameter gives a (hollow) sphere. It is the 3D equivalent of a circle.

Sphere of radius $r$ $$\text{Surface area} = 4\pi r^2, \qquad V = \dfrac{4}{3}\pi r^3$$ $$V = \dfrac{1}{3} \times (\text{surface area}) \times r$$

Volumes of a cone, a sphere and a cylinder with the same radius $r$ (and height $2r$ for the cone and cylinder) are in the ratio $1 : 2 : 3$.

Archimedes' surface area: the surface area of a sphere equals the curved surface area of the cylinder of radius $r$ and height $h = 2r$ that fits tightly over it: $2\pi r \cdot 2r = 4\pi r^2$.
Map-making: imagine a cylinder wrapped around the Earth touching it at the equator. Projecting the Earth's surface onto the cylinder and unwrapping it gives a rectangular world map. It is convenient, but shapes are greatly distorted near the poles (high latitudes) and hardly at all near the equator.
Connecting volume to the cone: divide the sphere's surface into tiny regions and join each to the centre. This gives many thin cones of height $r$. The sphere's volume is the sum of their volumes: $\dfrac{r}{3} \times (\text{sum of base areas}) = \dfrac{r}{3} \times 4\pi r^2 = \dfrac{4}{3}\pi r^3$.
String activity: wind string tightly over the whole surface of a ball, then use it to fill four circles drawn with the ball's radius. The string covers exactly four circles, which shows surface area $= 4 \times \pi r^2$.
Smallest enclosing cylinder: its radius is $r$ and height $2r$, so its volume is $2\pi r^3$. The sphere's volume $\dfrac{4}{3}\pi r^3$ is $\dfrac{2}{3}$ of it. A cone of radius $r$ and height $2r$ has volume $\dfrac{2}{3}\pi r^3$. Hence cone : sphere : cylinder $= 1 : 2 : 3$.
Same material: for solid spheres of the same metal, weight is proportional to volume, so the ratio of weights equals the ratio of $r^3$. A weight ratio of $8 : 1$ means a radius ratio of $2 : 1$.
Packing spheres: spheres leave gaps, so dividing volumes only gives an upper bound. Count how many spheres fit along each edge using the diameter.
String around the equator: since $C = 2\pi r$, increasing the circumference by $1$ metre increases the radius by $\dfrac{1}{2\pi}$ metre ($\approx 16$ cm), whatever the size of the sphere. The gap is the same for the Earth, the Moon, Jupiter or a volleyball.

8. Hemispheres

Hemisphere: half of a sphere, formed when a sphere is cut by a plane through its centre. It has one curved surface and one flat circular base. Examples: bowls, dome-shaped covers, planetariums, igloos.

Hemisphere of radius $r$ $$\text{CSA} = 2\pi r^2, \qquad \text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2, \qquad V = \dfrac{2}{3}\pi r^3$$

Why: the curved surface is half of $4\pi r^2$, and the volume is half of $\dfrac{4}{3}\pi r^3$. The flat base is a circle of area $\pi r^2$.
Open bowl or dome: if the base is open (a bowl, a painted dome), only the curved surface $2\pi r^2$ counts.
Radius from circumference: if the circumference of the base is given, use $2\pi r = \text{circumference}$ to find $r$ first.
Capacity: the water a hemispherical bowl holds is $\dfrac{2}{3}\pi r^3$.
Cost of painting: cost $=$ area $\times$ rate. Make sure the area unit matches the unit in the rate.

9. Scaling and Ratios

If every length is multiplied by $k$ $$\text{lengths} \times k, \qquad \text{areas} \times k^2, \qquad \text{volumes} \times k^3$$

Percentage change: an increase of $x\%$ means multiplying by $1 + \dfrac{x}{100}$. A sphere whose radius increases by $10\%$ has surface area multiplied by $1.1^2 = 1.21$ ($21\%$ more) and volume multiplied by $1.1^3 = 1.331$ ($33.1\%$ more).
Working backwards: if a sphere's volume rises by $72.8\%$, the factor is $1.728 = 1.2^3$, so the radius rose by $20\%$.
Surface-to-volume ratio: for similar shapes, $\dfrac{\text{surface area}}{\text{volume}}$ is proportional to $\dfrac{1}{k}$. Small objects have a lot of surface per unit of volume. This is why cells are small, why crushed solids dissolve or react faster, and why small animals lose heat quickly.
Different scale factors: for a cylinder or cone with radius multiplied by $p$ and height by $q$, the volume is multiplied by $p^2 q$; for a cuboid with edges multiplied by $p, q, s$, the volume is multiplied by $pqs$.
Ratios of similar solids: spheres with radii in the ratio $a : b$ have surface areas in the ratio $a^2 : b^2$ and volumes in the ratio $a^3 : b^3$. For example, if one diameter is $\dfrac{1}{4}$ of another, its surface area is $\dfrac{1}{16}$ of the other's.
Comparing two cylinders: with radii $r_1, r_2$ and heights $h_1, h_2$, the ratio of CSAs is $\dfrac{r_1 h_1}{r_2 h_2}$ and of volumes is $\dfrac{r_1^2 h_1}{r_2^2 h_2}$.

10. Problem-Solving Toolkit

A routine for solid-geometry problems

1. Identify the solid(s) and which quantity is wanted (curved area, total area, volume, or a length).

2. List the given dimensions in one unit and find any missing one (for a cone use $l^2 = h^2 + r^2$; radius from diameter or circumference).

3. Choose the right formula, checking whether the solid is open or closed.

4. Compute, keeping $\pi$ symbolic until the end if possible, and attach the correct units (square units for area, cubic units for volume).

5. Sense-check: is the size reasonable? Does a ratio or scaling argument agree?

Situation	Key idea
Melting and recasting a solid	The volume is conserved, the surface area is not. Number of pieces $= \dfrac{\text{volume of the big solid}}{\text{volume of one piece}}$ (round down to complete pieces).
Objects dropped into water	The water displaced equals the volume of the submerged objects. The rise in level $\times$ area of the container's base $=$ volume added. Number of objects $=$ added volume $\div$ volume of one object.
Filling a container	Capacity is the volume. Convert to litres using $1000\ \text{cm}^3 = 1\ \text{L}$. For many identical containers, multiply the volume of one by the number.
Fitting one shape inside another	Find the relation between the dimensions (for example, the largest cylinder in a cube has diameter equal to the edge; a sphere in a snug cylinder has height $2r$).
Rotating a right triangle	About a leg gives a cone: that leg is the height, the other leg the radius, the hypotenuse the slant height. Use the Pythagorean triple or theorem to check which is which.
Covering material (tent, cap, paper)	Use the curved surface area, not the total. For cloth of fixed width, length $=$ area $\div$ width, plus allowance for stitching or wastage.
Cutting a solid into pieces	Total volume is unchanged; each cut creates new surface, so total surface area increases.
Comparing two options	Compare volumes for capacity (for example one big chocolate cube against many small ones: compare $a^3$ with $n \cdot b^3$), and areas for covering or exposed surface.

11. Estimation (Guesstimates)

Guesstimate: a problem in which you make a reasonable estimate when exact data are not available, using assumptions, approximations, logic and everyday knowledge to model the situation.

How to tackle a guesstimate

1. Guess first: make a quick intuitive guess (hundreds? thousands? more?).

2. Model: identify the quantities and how they relate, usually a ratio such as $\dfrac{\text{total amount}}{\text{amount in one item}}$.

3. Assume: estimate each unknown size or fraction sensibly and state the assumption.

4. Calculate with convenient approximations (for example $1\ \text{ft} \approx 30\ \text{cm}$).

5. Look back: compare with your first guess and ask whether the answer is reasonable.

Answers depend on assumptions: two people can get different answers, for instance by assuming different sizes or different fillings for a golgappa. What matters most is the reasoning.
Dividing volumes gives an upper bound for spheres: the volume of a room divided by the volume of one ball counts as if there were no gaps. Balls cannot fill space without gaps, so you can say for sure that you cannot fit more than this number.
Better estimate by arrangement: count the balls along each edge as $\dfrac{\text{edge of room}}{\text{diameter of a ball}}$ and multiply the three counts. Other arrangements (such as staggered layers) pack more tightly, with higher packing efficiency.
Why cubes measure volume: cubes fill space with no gaps, while spheres do not, so spheres are not used as basic units of volume.
Useful tools: volume of a vessel $=$ base area $\times$ height; items served $=$ total volume $\div$ volume per item; the fraction of an item that is actually filled matters; convert units carefully.
Typical guesstimate themes: scoops of ice cream from a container, bricks in a wall, milk for a school, a crowd of people in one place, ink in a pen, the volume of one's house, and the thickness of a thin layer spread over a sphere (thickness $\approx \dfrac{\text{volume of layer}}{\text{surface area}}$).

12. Formulas and Quick Revision

Solid	Surface area	Volume
Cuboid ($l, w, h$)	$2(wl + hw + lh)$	$lwh$
Cube (side $a$)	$6a^2$	$a^3$
Right circular cylinder	CSA $= 2\pi rh$ Closed at one end: $2\pi rh + \pi r^2 = \pi r(2h + r)$ Closed at both ends: $2\pi rh + 2\pi r^2 = 2\pi r(h + r)$	$\pi r^2 h$
Right circular cone	CSA $= \pi rl$ TSA $= \pi r(l + r)$ with $l^2 = h^2 + r^2$	$\dfrac{1}{3}\pi r^2 h$
Pyramid	Sum of the areas of all faces	$\dfrac{1}{3} \times \text{base area} \times \text{height}$
Sphere	$4\pi r^2$	$\dfrac{4}{3}\pi r^3$
Hemisphere	CSA $= 2\pi r^2$ TSA $= 3\pi r^2$	$\dfrac{2}{3}\pi r^3$

Link

Cone, pyramid, sphere

The same $\dfrac{1}{3}$ appears: pyramid and cone $\dfrac{1}{3}\times$ base $\times$ height; sphere $\dfrac{1}{3}\times$ surface area $\times$ radius.

Link

Cylinder family

A cone and a sphere inside a cylinder of radius $r$ and height $2r$ have volumes in the ratio $1 : 2 : 3$ with the cylinder.

Scaling

$k$, $k^2$, $k^3$

Lengths scale by $k$, areas by $k^2$ and volumes by $k^3$. Surface-to-volume ratio falls as size rises.

Quick facts:
• Surface area is measured in square units; volume in cubic units; $1000\ \text{cm}^3 = 1$ litre.
• A cube has the least surface area among cuboids of equal volume.
• Cone volume $=\dfrac{1}{3}$ cylinder volume (same base and height).
• Sphere surface area $=$ curved surface of the tight cylinder, $4\pi r^2$.
• Spheres leave gaps when packed, so dividing volumes gives only an upper bound.

Common mistakes:
• Mixing units (cm with m) or forgetting square versus cubic units.
• Using the vertical height $h$ in $\pi r l$ (the CSA of a cone needs the slant height $l$).
• Forgetting the base (or the open end) when deciding between CSA, TSA and "closed at one end".
• Using the diameter in place of the radius.
• Forgetting the factor $\dfrac{1}{3}$ in cone and pyramid volumes.
• Doubling the radius and thinking the volume doubles (it multiplies by $4$ for a cylinder or cone, and by $8$ for a sphere).
• Assuming the surface area is conserved when a solid is melted and recast (only volume is conserved).
• Treating half-depth water in a cone as half the volume (it is $\dfrac{1}{8}$).
• Taking a volume division as the exact count when spheres are packed.