Surface Areas and Volumes
Why This Matters
Look around and almost nothing is a pure shape. A capsule of medicine is a cylinder with a rounded cap at each end. A circus tent is a cylinder with a cone for a roof. An ice-cream cone topped with a scoop is a cone plus a hemisphere. A water truck is a cylinder with two domed ends. None of these is a single solid you met in Class 9 — each is a combination of two or three of them stuck together.
So how much canvas does the tent need? How much medicine fits in the capsule? How much paint covers the toy? You already know the surface area and volume of every basic solid. This chapter is about the one new skill: how to combine those known answers correctly when the solids are joined.
And there’s a second, beautiful idea. When you melt a metal sphere and pour it into a mould to make wire, the metal doesn’t vanish — its volume stays the same. That single fact (“volume is conserved”) lets you solve a whole family of “recasting” problems without any new formula at all.
The Big Idea
When solids are joined, two different rules apply. For surface area, add up only the exposed outer surfaces — the faces hidden inside the join disappear, so you do not add the full surface areas of the pieces. For volume, just add the volumes of all the pieces — nothing is lost. And when one solid is melted and recast into another shape, the volume is conserved: volume before = volume after, even though the shape (and surface area) changes completely.
Let’s Break It Down
First, the basic formulas you’ll combine
Every problem in this chapter is built from these five solids. Here r is the radius, h the height, and l the slant height of a cone, with l = √(r² + h²). CSA means curved surface area (the rounded sides only) and TSA means total surface area (every face, including flat ends). All of these come from Class 9 — keep this table handy.
| Solid | CSA | TSA | Volume |
|---|---|---|---|
| Cylinder | 2πrh | 2πrh + 2πr² | πr²h |
| Cone | πrl | πrl + πr² | (1/3)πr²h |
| Sphere | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere | 2πr² | 2πr² + πr² = 3πr² | (2/3)πr³ |
| Cube / Cuboid | — | 6a² / 2(lb+bh+hl) | a³ / l×b×h |
Two things to notice. A hemisphere has two different surface numbers: its curved part is 2πr² and its flat circular face is πr², so its total surface is 3πr². Which one you use depends on whether that flat face is exposed or buried in a join. Also, for a sphere CSA and TSA are the same (a ball has no flat face).
A cone has base radius 3 cm and height 4 cm. What is its slant height l?
Surface area of a combination — add only the exposed parts
When you glue two solids together, the surfaces that touch at the join vanish from the outside — you can’t paint a face that’s hidden inside. So the rule is: add the surface areas of only the parts you can still see.
Take a “rocket” — a cone sitting on a cylinder. The cylinder’s top circle is hidden under the cone, and the cone has no separate base (it merges into the cylinder). What’s left outside is the curved surface of the cone plus the curved surface of the cylinder (and, if you also paint the bottom, the cylinder’s base circle).
A tent is a cylinder topped by a cone. The cylindrical part is 2.1 m high with diameter 4 m, and the slant height of the conical top is 2.8 m. Find the area of canvas needed (the floor is not covered). Take π = 22/7.
- Radius r = 4/2 = 2 m. Cylinder height h = 2.1 m, cone slant height l = 2.8 m. The canvas covers the cone’s curved surface and the cylinder’s curved surface — the floor and the hidden join are left out.
- CSA of cone = πrl = (22/7) × 2 × 2.8 = (22 × 5.6)/7 = 123.2/7 = 17.6 m².
- CSA of cylinder = 2πrh = 2 × (22/7) × 2 × 2.1 = (2 × 22 × 4.2)/7 = 184.8/7 = 26.4 m².
- Canvas area = 17.6 + 26.4 = 44 m².
Sometimes a piece is scooped out instead of added on. A hemispherical hollow cut into a cube removes the flat circle of the cube’s top face but adds the curved inside of the hemisphere — same principle, just bookkeeping which surfaces appear and disappear.
A decorative block is a cube of edge 5 cm with a hemisphere of diameter 4.2 cm fixed on top. Find the total surface area of the block. Take π = 22/7.
- The cube alone has TSA = 6a² = 6 × 5 × 5 = 150 cm². But the hemisphere sits on the top face, hiding a circle of radius r = 4.2/2 = 2.1 cm.
- Surface of block = TSA of cube − base circle of hemisphere + curved surface of hemisphere = 6a² − πr² + 2πr² = 6a² + πr².
- πr² = (22/7) × 2.1 × 2.1 = (22 × 4.41)/7 = 97.02/7 = 13.86 cm².
- Total surface area = 150 + 13.86 = 163.86 cm². (Note: this is NOT 150 + the hemisphere’s full surface — the buried circle is subtracted, then the curved part added.)
Volume of a combination — just add the volumes
Volume behaves more simply. When you join solids, no volume is lost at the join, so the total volume is just the sum of the volumes of the pieces. And if a part is hollowed out, you subtract that part’s volume.
A toy is a hemisphere with a cone on top. The cone's height is 2 cm and the common base diameter is 4 cm. Find the volume of the toy. Take π = 3.14.
- Radius r = 4/2 = 2 cm (shared by cone and hemisphere). Cone height h = 2 cm. Total volume = volume of cone + volume of hemisphere.
- Volume of cone = (1/3)πr²h = (1/3) × 3.14 × 2² × 2 = (1/3) × 3.14 × 8 = 25.12/3 ≈ 8.37 cm³.
- Volume of hemisphere = (2/3)πr³ = (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = 50.24/3 ≈ 16.75 cm³.
- Total volume = 8.37 + 16.75 = 25.12 cm³ (= 25.12 cm³ exactly, since (1/3 + 2/3)×3.14×8 = 3.14×8 here works out neatly).
A glass is a cylinder of inner diameter 5 cm and height 10 cm, but its bottom has a hemispherical raised portion that reduces capacity. Find the actual capacity. Take π = 3.14.
- Radius r = 5/2 = 2.5 cm. The apparent capacity (ignoring the dome) is the cylinder’s volume; the actual capacity is that minus the hemisphere pushed up into it.
- Apparent capacity = πr²h = 3.14 × 2.5 × 2.5 × 10 = 3.14 × 62.5 = 196.25 cm³.
- Volume of hemisphere = (2/3)πr³ = (2/3) × 3.14 × 2.5³ = (2/3) × 3.14 × 15.625 = 98.125 × 2/3 ≈ 32.71 cm³.
- Actual capacity = 196.25 − 32.71 = 163.54 cm³.
Conversion of solids — volume is conserved
Here’s the second big idea. If you take a lump of clay (or melt some metal) and reshape it into something else, the amount of stuff doesn’t change — only its shape does. So:
Volume of the original solid = Volume of the new solid.
This is the whole secret to “a sphere is melted and recast into a wire / smaller balls / a cone” problems. Set the two volumes equal and solve for the unknown dimension. The surface area, of course, changes — but the volume is your bridge.
A solid metal sphere of radius 4.2 cm is melted and recast into a solid cylinder of radius 6 cm. Find the height of the cylinder. Take π = 22/7.
- No metal is gained or lost, so volume of sphere = volume of cylinder: (4/3)πR³ = πr²h, where R = 4.2 cm (sphere) and r = 6 cm (cylinder).
- Cancel π from both sides: (4/3) × 4.2³ = 6² × h, i.e. (4/3) × 74.088 = 36h.
- (4/3) × 74.088 = 98.784, so 98.784 = 36h.
- h = 98.784 / 36 = 2.744 cm. (The shape changed completely, but the volume stayed equal — that’s the only equation we needed.)
A spherical ball of radius 3 cm is melted and recast into small spherical balls each of radius 0.5 cm. How many small balls are made?
- Total metal is conserved, so volume of big sphere = number of small balls × volume of one small ball. Let the number be n.
- (4/3)π × 3³ = n × (4/3)π × (0.5)³. The (4/3)π cancels from both sides — it always does when both shapes are spheres.
- 3³ = n × 0.5³ → 27 = n × 0.125.
- n = 27 / 0.125 = 216 small balls.
A solid is melted and recast into a different shape. Which quantity is guaranteed to stay the same — its volume or its surface area?
Common Mistakes
For a combined solid, the surface area is the sum of the total surface areas of the separate pieces.
Each piece has a tidy TSA formula, so adding TSA + TSA feels like the natural way to 'combine' them.
The faces at the join are hidden and must NOT be counted. Add only the EXPOSED surfaces — usually curved parts plus any still-visible flat faces. For a cone on a cylinder, that's πrl + 2πrh, not (πrl + πr²) + (2πrh + 2πr²).
Use a hemisphere's surface as 2πr² in every problem.
2πr² is the formula you memorised for a hemisphere, so it seems like the one true answer to plug in.
A hemisphere has a curved part (2πr²) AND a flat circular face (πr²). Use 2πr² when only the dome shows, but 3πr² (curved + flat) when the flat face is also exposed — and subtract πr² when that flat face gets buried in a join.
When melting one solid into another, set their surface areas equal.
Both objects 'look' the same size after recasting, and surface area is the quantity you spent the chapter computing, so equating it feels right.
Melting conserves VOLUME, not surface area. Always write 'volume before = volume after'. The surface area changes when the shape changes — it is not the conserved quantity.
You can add a length in cm to a length in m, or mix cm³ with m³, as long as the numbers are right.
The arithmetic still 'works' on the calculator, so a units slip is easy to miss.
Convert everything to ONE unit before computing. 1 m = 100 cm, so 1 m³ = 100³ = 1,000,000 cm³ (not 100). Diameter must be halved to radius first, too — many errors are really 'used d instead of r'.
Quick Check
A cone is mounted on a cylinder of the same radius (a rocket shape). Which gives the area of canvas/paint on the outside (excluding the base)?
A solid hemisphere rests flat-side-down on a table, so only its dome shows. Which is its exposed surface area?
A metal cube is melted and recast into a sphere. What stays the same?
To find the VOLUME of a capsule (a cylinder with a hemisphere stuck on each end), you should:
Practice Problems
Easy
Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Each cube has volume 64 = a³, so the edge a = 4 cm.
Joining two cubes end to end gives a cuboid of length 8 cm, breadth 4 cm, height 4 cm.
Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 2 × 80 = 160 cm².
(Notice it is not 2 × 96 = 192 cm², the sum of the two cubes’ surface areas — the two faces glued together vanish.)
A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find its total surface area. Take π = 22/7.
Radius r = 3.5 cm. The hemisphere’s height equals its radius = 3.5 cm, so the cone’s height h = 15.5 − 3.5 = 12 cm.
Slant height l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm.
Total surface area = CSA of cone + CSA of hemisphere = πrl + 2πr² = (22/7) × 3.5 × 12.5 + 2 × (22/7) × 3.5 × 3.5 = (22/7) × 3.5 × (12.5 + 2 × 3.5) = (22/7) × 3.5 × 19.5 = 11 × 19.5 = 214.5 cm².
Medium
A vessel is a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Take π = 22/7.
Radius r = 14/2 = 7 cm. The hemisphere’s depth equals its radius = 7 cm, so the cylinder’s height h = 13 − 7 = 6 cm.
The inside surface = curved inner surface of cylinder + curved inner surface of hemisphere = 2πrh + 2πr² = 2 × (22/7) × 7 × 6 + 2 × (22/7) × 7 × 7 = 2 × (22/7) × 7 × (6 + 7) = 2 × 22 × 13 = 572 cm².
A cylindrical container of radius 6 cm and height 15 cm is full of ice cream, which is distributed into 10 identical cones, each having a hemispherical top of the same radius. If the height of each cone is 12 cm, find the radius of the ice-cream cone. Take π = 22/7.
All the ice cream is conserved: volume of cylinder = 10 × (volume of one cone + one hemisphere).
Volume of cylinder = πR²H = π × 6² × 15 = π × 36 × 15 = 540π cm³.
One cone + hemisphere (radius r, cone height 12) = (1/3)πr²(12) + (2/3)πr³ = 4πr² + (2/3)πr³.
So 540π = 10 × [4πr² + (2/3)πr³]. Cancel π: 540 = 40r² + (20/3)r³.
Try r = 3: 40(9) + (20/3)(27) = 360 + 180 = 540 ✓.
So the radius of each ice-cream cone is 3 cm.
Challenge
A gulab jamun is shaped like a cylinder with a hemisphere at each end. Its total length is 5 cm and its diameter is 2.8 cm. It contains sugar syrup up to about 30% of its volume. Find approximately how much syrup is in 45 such gulab jamuns. Take π = 22/7.
Radius r = 2.8/2 = 1.4 cm. The two hemispheres take up 2 × 1.4 = 2.8 cm of the length, so the cylinder’s length h = 5 − 2.8 = 2.2 cm.
Volume of one gulab jamun = πr²h + (4/3)πr³ (two hemispheres = one sphere) = (22/7) × 1.4² × 2.2 + (4/3) × (22/7) × 1.4³ = (22/7) × 1.96 × 2.2 + (4/3) × (22/7) × 2.744 = 13.55 + 11.50 ≈ 25.05 cm³.
Volume of 45 gulab jamuns ≈ 45 × 25.05 = 1127.25 cm³.
Syrup is 30% of this: 0.30 × 1127.25 ≈ 338 cm³ (about 338 cm³ of syrup).
A solid cone of height 120 cm and radius 60 cm stands on a hemisphere of radius 60 cm. This solid is placed upright in a cylinder full of water (radius 60 cm, height 180 cm) so that it touches the bottom. Find the volume of water left in the cylinder. Take π = 3.14.
The water that spills/remains is found by subtracting the solid’s volume from the cylinder’s volume.
Volume of cylinder = πr²h = 3.14 × 60² × 180 = 3.14 × 3600 × 180 = 2,034,720 cm³.
Volume of solid = cone + hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3) × 3.14 × 60² × 120 + (2/3) × 3.14 × 60³ = (1/3) × 3.14 × 3600 × 120 + (2/3) × 3.14 × 216000 = 452,160 + 452,160 = 904,320 cm³.
Water left = 2,034,720 − 904,320 = 1,130,400 cm³ (= 1.1304 m³).
Summary
You should now be able to explain:
- A combination of solids is two or more basic solids (cuboid, cube, cylinder, cone, sphere, hemisphere) joined together.
- Surface area of a combination = sum of only the exposed surfaces. The faces at the join disappear, so you add curved surfaces (and any visible flat faces) — not the full TSAs.
- Volume of a combination = sum of the volumes of all pieces; subtract the volume of any part that is hollowed out.
- A hemisphere has a curved surface 2πr² and a flat face πr² (total 3πr²) — pick the right one for which faces are exposed.
- Key base formulas: cylinder πr²h, 2πrh; cone (1/3)πr²h, πrl with l = √(r² + h²); sphere (4/3)πr³, 4πr²; hemisphere (2/3)πr³, 2πr².
- Conversion of solids: melting and recasting conserves volume — set volume before = volume after to find an unknown dimension or count.
- Always convert to one unit and use the radius (not the diameter) before computing.
What’s Next
You’ve spent this chapter measuring shapes. Next, in Statistics, you’ll turn from shapes to data — large lists of numbers — and learn to summarise them with a single representative value. You’ll find the mean, median and mode of grouped data, building directly on the kind of careful, organised calculation you practised here.