Class 9 Mathematics Chapter 14: Math of Space: Surface Area and Volume

Volume measures the space inside the solid and is given in cubic units such as $\text{cm}^3$. Surface area measures the skin of the solid and is given in square units such as $\text{cm}^2$. Filling questions want volume; covering questions want surface area.

Volume $= 12 \times 8 \times 5 = 480\ \text{cm}^3$. Surface area $= 2(12 \times 8 + 8 \times 5 + 5 \times 12) = 2(96 + 40 + 60) = 392\ \text{cm}^2$.

Volume $= 7^3 = 343\ \text{cm}^3$ and surface area $= 6 \times 7^2 = 294\ \text{cm}^2$. A cube is a cuboid with $l = w = h = a$, so $lwh$ becomes $a^3$ and $2(lw + wh + hl)$ becomes $2(3a^2) = 6a^2$.

Cut the curved surface and unroll it. It flattens into a rectangle whose height is the height of the cylinder, $h$, and whose width is the distance all the way round the circular base, which is the circumference $2\pi r$. The area of a rectangle is its two sides multiplied, so the curved surface area is $2\pi r \times h$. Nothing is squared because only two lengths are being multiplied, unlike the circular ends, whose area $\pi r^2$ does involve a square.

The pipe has only the curved surface, so its area is $2\pi r h$. The tin has the curved surface and two circular ends, so its total surface area is $2\pi r h + 2\pi r^2 = 2\pi r(r + h)$. The difference is $2\pi r^2$, the area of the two ends.

CSA $= 2 \times \frac{22}{7} \times 7 \times 10 = 440\ \text{cm}^2$. TSA $= 2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times 17 = 748\ \text{cm}^2$. Volume $= \frac{22}{7} \times 49 \times 10 = 1540\ \text{cm}^3$.

CSA $= 2 \times \frac{22}{7} \times 14 \times 5 = 440\ \text{cm}^2$ and volume $= \frac{22}{7} \times 196 \times 5 = 3080\ \text{cm}^3$.

The first cylinder has CSA $440\ \text{cm}^2$ and volume $1540\ \text{cm}^3$. So the two have exactly the same curved surface area while the second holds twice as much.

The curved surface area is $2\pi r h$, in which $r$ appears once. Doubling $r$ multiplies it by 2 and halving $h$ divides it by 2, so the two changes cancel exactly.

The volume is $\pi r^2 h$, in which $r$ appears squared. Doubling $r$ multiplies the volume by $2^2 = 4$, and halving $h$ then divides by 2, leaving a factor of 2. So the volume doubles. The difference is simply the power of $r$ in each formula.

Volume $= \frac{22}{7} \times 49 \times 20 = 3080\ \text{cm}^3$. Since $1000\ \text{cm}^3 = 1$ litre, it holds $3.08$ litres.

The diameter must be halved to give the radius, $r = 14$ cm, because every formula for a cylinder is written in terms of $r$ and not the diameter. Using 28 in place of 14 would make the curved surface area twice as large and the volume four times as large.

A solid takes up space and also has a skin, and those are two different measurements answering two different questions. Volume asks how much space is inside, so it suits filling, holding and capacity, and it is measured in cubic units. Surface area asks how much skin there is, so it suits painting, wrapping and covering, and it is measured in square units. The units themselves are a check: a volume ending in $\text{cm}^2$ means the wrong formula was used.

For a cuboid the volume is the base area multiplied by the height, $V = lwh$. For the box that is $12 \times 8 \times 5 = 480\ \text{cm}^3$.

Its surface area needs all six faces, and they come in three matching pairs — front with back, left with right, top with bottom — so $\text{SA} = 2(lw + wh + hl)$. For the box that is $2(96 + 40 + 60) = 2 \times 196 = 392\ \text{cm}^2$.

A cube is just a cuboid whose three edges are equal, so putting $l = w = h = a$ into both formulas gives $V = a^3$ and $\text{SA} = 6a^2$. For the cube of edge 7 cm that is $343\ \text{cm}^3$ and $6 \times 49 = 294\ \text{cm}^2$.

Notice that the cube has the smaller volume of the two and the smaller surface area, but there is no rule forcing those to move together, which the cylinder work makes clearer still.

First, what the words mean. A cylinder is CIRCULAR when its cross-section is a circle, and RIGHT when its axis stands perpendicular to that base. A cylinder whose axis leans is oblique, and only the right circular cylinder is treated here.

For the curved surface, cut it down its height and unroll it. It flattens into a rectangle. Its height is the height of the cylinder, $h$. Its width is the distance all the way round the circular base, which is the circumference $2\pi r$. The area of a rectangle is the two sides multiplied, so

$\text{CSA} = 2\pi r h$.

This is also why the formula contains no square: only two lengths are being multiplied. The circular ends are different — each is a circle of area $\pi r^2$, and that does involve a square.

A closed cylinder has the curved surface and both ends, so

$\text{TSA} = 2\pi r h + 2\pi r^2 = 2\pi r(r + h)$.

Which to use is decided by the object. A pipe open at both ends has no circular faces at all, so its surface is the curved part alone and the CSA is the answer. A closed tin has two ends, so the TSA applies. The difference between the two is exactly $2\pi r^2$, and choosing the wrong one is a mistake about the object rather than about the arithmetic.

Taking $\pi = \frac{22}{7}$ throughout.

First cylinder: CSA $= 2 \times \frac{22}{7} \times 7 \times 10 = 440\ \text{cm}^2$, and volume $= \frac{22}{7} \times 7^2 \times 10 = \frac{22}{7} \times 490 = 1540\ \text{cm}^3$.

Second cylinder: CSA $= 2 \times \frac{22}{7} \times 14 \times 5 = 440\ \text{cm}^2$, and volume $= \frac{22}{7} \times 14^2 \times 5 = \frac{22}{7} \times 980 = 3080\ \text{cm}^3$.

So the two have exactly the same curved surface area while the second holds twice as much. Practically, a tin twice as wide and half as tall needs the same metal for its side and carries double the contents.

The reason is the power of $r$ in each formula. In $2\pi r h$ the radius appears once, so doubling $r$ and halving $h$ cancel exactly. In $\pi r^2 h$ it appears squared, so doubling $r$ multiplies the volume by four, and halving $h$ brings that back to two.

The comparison therefore shows that surface area and volume are not two ways of saying the same thing: one can be held fixed while the other changes.

For the cylinder, the volume is $\pi r^2 h = \frac{22}{7} \times 49 \times 20$. Here $\frac{22}{7} \times 49 = 154$, and $154 \times 20 = 3080$, so the volume is $3080\ \text{cm}^3$.

For the cuboid, the volume is $12 \times 8 \times 5 = 480\ \text{cm}^3$.

Capacity is quoted in litres, and $1000\ \text{cm}^3 = 1$ litre, so divide each by 1000. The cylinder holds $3.08$ litres and the cuboid holds $0.48$ litres.

The cylinder holds more — about six and a half times as much. It is worth noticing that the comparison had to be made in the same unit before it meant anything: $3080$ against $480$ is the same comparison as $3.08$ against $0.48$, but $3.08$ against $480$ would not be. Mixing $\text{cm}^3$ with litres is the commonest way to get a capacity question wrong by a factor of a thousand.

The base radius $r$ is measured from the centre of the circular base to its rim. The height $h$ is measured straight up from the centre of the base to the apex. The slant height $l$ is measured along the sloping surface from the rim to the apex.

Since $l^2 = r^2 + h^2$, we have $h^2 = 25^2 - 7^2 = 625 - 49 = 576$, so $h = 24$ cm.

$l^2 = 21^2 + 28^2 = 441 + 784 = 1225$, so $l = 35$ cm.

Because the height is perpendicular to the base, the radius, height and slant height form a right triangle in which the slant height is the hypotenuse. The hypotenuse is the longest side of a right triangle, so $l$ exceeds both $r$ and $h$. A slant height smaller than the height is therefore a sign that two lengths have been confused.

CSA $= \pi r l = \frac{22}{7} \times 7 \times 25 = 550\ \text{cm}^2$. The base has area $\pi r^2 = \frac{22}{7} \times 49 = 154\ \text{cm}^2$, so TSA $= 550 + 154 = 704\ \text{cm}^2$.

The formula $\pi r l$ needs the SLANT height, and only the height was given, so find $l$ first: $l^2 = 7^2 + 24^2 = 49 + 576 = 625$, giving $l = 25$ cm. Then CSA $= \frac{22}{7} \times 7 \times 25 = 550\ \text{cm}^2$. Using 24 in place of 25 would have been the commonest error in this chapter.

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232\ \text{cm}^3$.

A cone and a cylinder standing on the same circular base with the same height stand in the ratio 1 to 3 by volume, so $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$ while $V_{\text{cylinder}} = \pi r^2 h$.

It can be checked directly: fill the cone with water and empty it into the cylinder. Three coneloads fill the cylinder exactly, and no measurement of either solid is needed beyond having them share a base and a height.

For the first, the base area is $6^2 = 36\ \text{m}^2$, so $V = \frac{1}{3} \times 36 \times 10 = 120\ \text{m}^3$.

For the second, the base area is given as $18\ \text{cm}^2$, so $V = \frac{1}{3} \times 18 \times 7 = 42\ \text{cm}^3$. Nothing about the triangle's sides was needed — only its area.

Because the base of a pyramidal shape may be a triangle, a square or any closed shape at all, so no single side length would describe every case, while the base area always does. This chapter gives no surface-area formula for a general pyramid — only the volume.

Take $\pi = \frac{22}{7}$ throughout.

Slant height. The radius, height and slant height form a right triangle with $l$ as the hypotenuse, so $l^2 = r^2 + h^2 = 49 + 576 = 625$ and $l = 25$ cm.

Curved surface area. $\text{CSA} = \pi r l$, which uses the SLANT height, so $\frac{22}{7} \times 7 \times 25 = 550\ \text{cm}^2$. The height may not be used here, and doing so is the commonest error in the chapter.

Total surface area. A closed cone adds its circular base, of area $\pi r^2 = \frac{22}{7} \times 49 = 154\ \text{cm}^2$, so $\text{TSA} = 550 + 154 = 704\ \text{cm}^2$, which is also $\pi r(l + r) = \frac{22}{7} \times 7 \times 32$.

Volume. $V = \frac{1}{3}\pi r^2 h$, which uses the HEIGHT, so $\frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232\ \text{cm}^3$.

So the curved surface uses $l$, the total surface uses $l$ and the base, and the volume uses $h$. Reading which length a formula wants, before substituting anything, is what keeps the four answers apart.

A cone standing on the same circular base as a cylinder, with the same height, has exactly one third of its volume. That is a fact anyone can check by filling the cone with water and emptying it into the cylinder three times. It gives $V = \frac{1}{3}\pi r^2 h$, and since $\pi r^2$ is just the area of the circular base, the formula can be written

$V = \frac{1}{3} \times \text{base area} \times \text{height}$.

A pyramidal shape has a flat base with every point joined to a single apex, and its volume is given by precisely the same expression. It has to be written in terms of base AREA, because the base may be a triangle, a square or any closed shape, and no single side length would cover every case.

For the cone: $V = \frac{1}{3} \times 154 \times 24 = 1232\ \text{cm}^3$, which agrees with $\frac{1}{3} \times \frac{22}{7} \times 49 \times 24$ as it must, since $\frac{22}{7} \times 49 = 154$.

For the pyramid: the base area is $36\ \text{m}^2$, so $V = \frac{1}{3} \times 36 \times 10 = 120\ \text{m}^3$.

The two are the same rule because a cone IS a pyramidal shape whose base happens to be a circle. What differs is the surface area: the cone has the formulas $\pi r l$ and $\pi r(l + r)$, while this chapter gives no surface-area formula for a general pyramid at all.

A cone is described by its base radius $r$, its height $h$ measured straight up from the centre of the base to the apex, and its slant height $l$ measured along the sloping surface from the rim to the apex.

Because the height stands perpendicular to the base, those three lengths form a right triangle in which $r$ and $h$ are the shorter sides and $l$ is the hypotenuse, so $l^2 = r^2 + h^2$. It follows that the slant height is always the longest of the three, which is a useful check: a slant height smaller than the height means two lengths have been confused.

The mixing-up matters because the formulas take different lengths. The curved surface area $\pi r l$ takes the slant height, while the volume $\frac{1}{3}\pi r^2 h$ takes the height. Substituting $h$ into $\pi r l$ gives a wrong answer that looks perfectly reasonable, with no clue in the units, which is why it is so often missed.

For $r = 21$ and $h = 28$: $l^2 = 441 + 784 = 1225$, so $l = 35$. The curved surface area is then $\frac{22}{7} \times 21 \times 35 = 2310$, while the volume is $\frac{1}{3} \times \frac{22}{7} \times 441 \times 28 = 12936$. Using 28 in the surface-area formula would have given 1848, and nothing about that number would have looked wrong.

A pyramidal shape has a flat base, and every point of that base is joined to a single point, the apex, which lies outside the plane of the base. The base can be a triangle, a square, or any other closed shape at all at all.

They occur widely. The famous pyramids of Egypt have square bases. A triangular-based one appears in the puzzle made by the Hungarian architect Erno Rubik. And some mountains have been worn into pyramidal peaks by glacial erosion over many thousands of years — Shivling Peak in Uttarakhand, and the Matterhorn in the Alps.

The volume is

$V = \frac{1}{3} \times \text{base area} \times \text{height}$,

the same one-third rule as the cone. It has to be written with the base AREA rather than a side length, precisely because the base can be any shape: a formula in terms of a side would describe only square-based pyramids.

What the chapter does NOT give is a surface-area formula for a general pyramid. The cone, whose base happens to be a circle, has both $\pi r l$ and $\pi r(l + r)$, but for a pyramid on an arbitrary base only the volume is provided, so a question asking for a general pyramid's surface area is outside what this chapter supports.

A sphere is the set of all points in three-dimensional space at a given distance $r$ from a given point $O$, the centre. It is the same definition as a circle, which is the set of points in a PLANE at a given distance from a given point — the only change is the space it lives in.

He fitted a cylinder tightly over the sphere. Such a cylinder has the same radius $r$, and its height must be $2r$ because the sphere touches both of its ends. He showed that the sphere's surface area equals the CURVED surface area of that cylinder.

That can be checked directly: the cylinder's curved surface is $2\pi r h$, and substituting $h = 2r$ gives $2\pi r(2r) = 4\pi r^2$, which is exactly the sphere's surface area.

Surface area $= 4 \times \frac{22}{7} \times 49 = 616\ \text{cm}^2$. Volume $= \frac{4}{3} \times \frac{22}{7} \times 9261 = 38808\ \text{cm}^3$.

The radius is SQUARED in the surface area $4\pi r^2$ and CUBED in the volume $\frac{4}{3}\pi r^3$. Doubling the radius therefore multiplies the surface area by $2^2 = 4$ and the volume by $2^3 = 8$.

A hemisphere is half a sphere, formed when a plane through the centre cuts the sphere into two equal halves. It has two surfaces: one curved surface, and one flat circular base created by the cut.

The curved surface is half of the sphere's, so $\text{CSA} = \frac{1}{2} \times 4\pi r^2 = 2\pi r^2$. The total adds the flat circular base, of area $\pi r^2$, giving $\text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2$.

For $r = 7$: CSA $= 2 \times \frac{22}{7} \times 49 = 308\ \text{cm}^2$ and TSA $= 3 \times \frac{22}{7} \times 49 = 462\ \text{cm}^2$.

The varnished hemisphere needs its TOTAL surface area, $3\pi r^2$, because varnishing all over includes the flat circular base. The open bowl needs only its CURVED surface area, $2\pi r^2$, because an open bowl has no flat face to glaze.

They differ by $\pi r^2$, the area of that base, and the difference is a fact about the two OBJECTS rather than about the arithmetic.

$V = \frac{2}{3} \times \frac{22}{7} \times 9261 = 19404\ \text{cm}^3$. The sphere of the same radius has volume $38808\ \text{cm}^3$, and $19404$ is exactly half of it, as it must be.

The box holds $20 \times 20 \times 10 = 4000\ \text{cm}^3$. One laddu occupies $\frac{4}{3} \times \frac{22}{7} \times 1.5^3 = \frac{99}{7} \approx 14.14\ \text{cm}^3$. Dividing gives about 282.

That figure is too high because it divides volume by volume, which assumes the laddus fill the box completely. Spheres cannot do that — there are always gaps between them — so the true number is appreciably smaller. A good guesstimate states that assumption rather than hiding it.

Cuboidal: a brick or a cupboard. Cylindrical: a water pipe or a gas cylinder. Conical: a party hat. Spherical: a football. Hemispherical: a serving bowl or a dome.

A sphere is every point of space at a given distance $r$ from a centre $O$. Its surface area is $A = 4\pi r^2$ and its volume is $V = \frac{4}{3}\pi r^3$. The radius is SQUARED in the first, as an area must be, and CUBED in the second, as a volume must be.

Archimedes reached the surface area by comparison rather than by calculation. Fit a cylinder tightly over the sphere. Its radius is the same $r$, and its height must be $2r$, because the sphere touches both of its flat ends. He showed that the sphere's surface area is exactly the CURVED surface area of that cylinder — not its total, which would include the two ends.

The claim is easy to verify from the cylinder formula. Its curved surface is $2\pi r h$, and putting $h = 2r$ gives $2\pi r \times 2r = 4\pi r^2$. The two expressions agree, so the sphere's skin has the same area as the label wrapping its cylinder.

For $r = 7$: $A = 4 \times \frac{22}{7} \times 49 = 616\ \text{cm}^2$. For $r = 21$: $V = \frac{4}{3} \times \frac{22}{7} \times 9261 = 38808\ \text{cm}^3$, which is $38.808$ litres.

Because the powers of $r$ differ, doubling the radius multiplies the surface area by 4 and the volume by 8 — the two do not grow together.

A hemisphere is half a sphere. Cutting a sphere with a plane through its centre gives two equal halves, and the flat circular face that the cut creates is the hemisphere's base. So a hemisphere has exactly two surfaces: one curved and one flat.

Its curved surface is half the sphere's, so $\text{CSA} = \frac{1}{2} \times 4\pi r^2 = 2\pi r^2$.

The total surface adds the base, whose area is $\pi r^2$, giving $\text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2$. That extra $\pi r^2$ is the term most often dropped.

The volume is half the sphere's: $V = \frac{2}{3}\pi r^3$. For $r = 21$ that is $19404\ \text{cm}^3$, exactly half of the sphere's $38808\ \text{cm}^3$.

Which surface area applies depends on the object. A solid wooden hemisphere varnished ALL OVER needs the total, $3\pi r^2$, since the base is varnished too — $462\ \text{cm}^2$ when $r = 7$. An OPEN bowl glazed only inside needs the curved surface alone, $2\pi r^2$, because it has no flat face — $308\ \text{cm}^2$ for the same radius. The difference of $154\ \text{cm}^2$ is precisely that circular base.

A guesstimate is a reasonable estimate of a quantity when exact data are not available. It is not a careless guess: you choose assumptions openly, approximate the real object by a solid whose volume you can compute, and arrive at a figure you can defend. What makes it defensible is that the assumptions are STATED, so a reader can judge them. Take a box $20 \times 20 \times 10$ cm to be filled with laddus of radius 1.5 cm. The box holds $20 \times 20 \times 10 = 4000\ \text{cm}^3$. One laddu, treated as a sphere, occupies $\frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3.375 = \frac{99}{7}$, which is about $14.14\ \text{cm}^3$. Dividing, $4000 \div 14.14 \approx 282$. But 282 is the wrong answer to the real question, and seeing why is the whole point of the method. Dividing volume by volume assumes the laddus fill the box with nothing left over, and spheres cannot do that — however they are arranged, gaps remain between them. So the true number is appreciably smaller than 282, perhaps around two thirds of it. The honest answer is therefore not a different number arrived at by better arithmetic, but the same arithmetic reported WITH its assumption: "about 282 if the gaps are ignored, so rather fewer in practice." A guesstimate that hides its assumption is a guess. One that names it is a usable estimate.

Take $\pi = \frac{22}{7}$ and note $21^3 = 9261$ and $21^2 = 441$.

Volumes. The sphere holds $\frac{4}{3} \times \frac{22}{7} \times 9261 = 38808\ \text{cm}^3$. The hemisphere holds $\frac{2}{3} \times \frac{22}{7} \times 9261 = 19404\ \text{cm}^3$, exactly half, which is what cutting a solid in half must give.

Total surface areas. The sphere has $4 \times \frac{22}{7} \times 441 = 5544\ \text{cm}^2$. The hemisphere has $3 \times \frac{22}{7} \times 441 = 4158\ \text{cm}^2$, which is three quarters of the sphere's and NOT half.

That asymmetry is the interesting part. Halving the solid halves the volume exactly, because volume is simply shared between the two pieces. But it does not halve the surface, because the cut CREATES a new flat face of area $\pi r^2 = 1386\ \text{cm}^2$ on each half, which did not exist before. Each half therefore carries half the original skin plus a whole new circular base, giving $2\pi r^2 + \pi r^2 = 3\pi r^2$.

So cutting a solid in half halves what is inside and more than halves what is outside — another case of surface area and volume behaving independently.

Volume measures the space inside the solid and is given in cubic units such as $\text{cm}^3$. Surface area measures the skin of the solid and is given in square units such as $\text{cm}^2$. Filling questions want volume; covering questions want surface area.

Volume $= 12 \times 8 \times 5 = 480\ \text{cm}^3$. Surface area $= 2(12 \times 8 + 8 \times 5 + 5 \times 12) = 2(96 + 40 + 60) = 392\ \text{cm}^2$.

Volume $= 7^3 = 343\ \text{cm}^3$ and surface area $= 6 \times 7^2 = 294\ \text{cm}^2$. A cube is a cuboid with $l = w = h = a$, so $lwh$ becomes $a^3$ and $2(lw + wh + hl)$ becomes $2(3a^2) = 6a^2$.

The base radius $r$ is measured from the centre of the circular base to its rim. The height $h$ is measured straight up from the centre of the base to the apex. The slant height $l$ is measured along the sloping surface from the rim to the apex.

Since $l^2 = r^2 + h^2$, we have $h^2 = 25^2 - 7^2 = 625 - 49 = 576$, so $h = 24$ cm.

$l^2 = 21^2 + 28^2 = 441 + 784 = 1225$, so $l = 35$ cm.

A sphere is the set of all points in three-dimensional space at a given distance $r$ from a given point $O$, the centre. It is the same definition as a circle, which is the set of points in a PLANE at a given distance from a given point — the only change is the space it lives in.

He fitted a cylinder tightly over the sphere. Such a cylinder has the same radius $r$, and its height must be $2r$ because the sphere touches both of its ends. He showed that the sphere's surface area equals the CURVED surface area of that cylinder.

That can be checked directly: the cylinder's curved surface is $2\pi r h$, and substituting $h = 2r$ gives $2\pi r(2r) = 4\pi r^2$, which is exactly the sphere's surface area.

Surface area $= 4 \times \frac{22}{7} \times 49 = 616\ \text{cm}^2$. Volume $= \frac{4}{3} \times \frac{22}{7} \times 9261 = 38808\ \text{cm}^3$.