Surface Areas and Volumes — Cylinder, Cone, Sphere, Frustum

The Big Picture

Surface area and volume problems boil down to one skill: picking the right formula for the right shape. Once you know which shape you’re dealing with (and whether the question asks for surface area, curved surface area, or volume), it’s straight substitution and arithmetic.

For CBSE Classes 9 and 10, this chapter carries 8-10 marks in board exams. Class 9 covers basic solids (cylinder, cone, sphere). Class 10 adds combinations of solids and the frustum of a cone.

graph TD
    A[3D Shape Problem] --> B{Which shape?}
    B -->|Cylinder| C[CSA = 2πrh]
    B -->|Cone| D[CSA = πrl]
    B -->|Sphere| E[SA = 4πr²]
    B -->|Hemisphere| F[CSA = 2πr²]
    B -->|Frustum| G[Use frustum formulas]
    A --> H{What quantity?}
    H -->|Volume| I[Pick volume formula]
    H -->|Surface area| J{CSA or TSA?}
    J -->|Curved only| K[Exclude bases]
    J -->|Total| L[CSA + base areas]
    A --> M{Combination solid?}
    M -->|Yes| N[Split into parts]
    N --> O[Add volumes / exposed areas]

Key Terms & Definitions

Curved Surface Area (CSA) — The area of just the curved part of a solid, excluding flat bases. For a cylinder, think of the “label” wrapped around a can.

Total Surface Area (TSA) — CSA plus the area of all flat faces. A cylinder’s TSA = CSA + two circular bases.

Volume — The space enclosed inside the solid. Measured in cubic units (cm $^3$ , m $^3$ ).

Slant Height ( $l$ ) — For a cone, the distance from the apex to any point on the base circle along the surface. Related by $l = \sqrt{h^2 + r^2}$ .

Frustum — The portion of a cone remaining when the top is cut off by a plane parallel to the base. It has two circular faces of different radii.

All Formulas at a Glance

Quantity	Formula
CSA	$2\pi rh$
TSA	$2\pi r(r + h)$
Volume	$\pi r^2 h$

Quantity	Formula
Slant height	$l = \sqrt{h^2 + r^2}$
CSA	$\pi r l$
TSA	$\pi r(r + l)$
Volume	$\frac{1}{3}\pi r^2 h$

Quantity	Formula
Surface Area	$4\pi r^2$
Volume	$\frac{4}{3}\pi r^3$

Quantity	Formula
CSA	$2\pi r^2$
TSA	$3\pi r^2$
Volume	$\frac{2}{3}\pi r^3$

Quantity	Formula
Slant height	$l = \sqrt{h^2 + (r_1 - r_2)^2}$
CSA	$\pi(r_1 + r_2)l$
TSA	$\pi(r_1 + r_2)l + \pi r_1^2 + \pi r_2^2$
Volume	$\frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1 r_2)$

A cone’s volume is exactly $\frac{1}{3}$ of the cylinder with the same base and height. A hemisphere’s volume is $\frac{2}{3}$ of its enclosing cylinder. These ratios help you cross-check formulas in the exam.

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE Class 9)

Find the volume and TSA of a cylinder with radius 7 cm and height 10 cm.

Volume $= \pi r^2 h = \frac{22}{7} \times 49 \times 10 = \mathbf{1540 \text{ cm}^3}$

TSA $= 2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times 17 = \mathbf{748 \text{ cm}^2}$

Example 2 (Medium — CBSE Class 9)

A cone has slant height 13 cm and base diameter 10 cm. Find CSA and volume.

Radius $r = 5$ cm, slant height $l = 13$ cm.

Height: $h = \sqrt{l^2 - r^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm

CSA $= \pi r l = \pi \times 5 \times 13 = 65\pi \approx \mathbf{204.3 \text{ cm}^2}$

Volume $= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 25 \times 12 = 100\pi \approx \mathbf{314.2 \text{ cm}^3}$

Example 3 (Medium — CBSE Class 10)

A solid is made by placing a hemisphere on top of a cylinder. Both have radius 7 cm. Total height is 17 cm. Find the TSA.

Cylinder height $= 17 - 7 = 10$ cm (subtract hemisphere radius).

TSA = CSA of cylinder + CSA of hemisphere + one base of cylinder (the bottom only — the top is internal).

\text{TSA} = 2\pi(7)(10) + 2\pi(7)^2 + \pi(7)^2 = 140\pi + 98\pi + 49\pi = 287\pi \approx \mathbf{901.7 \text{ cm}^2}

Example 4 (Hard — CBSE Class 10)

A bucket shaped like a frustum has top and bottom diameters of 40 cm and 20 cm, and depth 12 cm. Find the capacity and cost of tin sheet at Rs 10 per cm $^2$ .

$r_1 = 20$ cm, $r_2 = 10$ cm, $h = 12$ cm.

Volume $= \frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1 r_2) = \frac{1}{3}\pi(12)(400 + 100 + 200) = 2800\pi \approx \mathbf{8796.5 \text{ cm}^3}$

Slant height: $l = \sqrt{12^2 + (20-10)^2} = \sqrt{244} \approx 15.62$ cm

Tin sheet area (open at top) = CSA + bottom base $= \pi(20+10)(15.62) + \pi(10)^2 \approx 568.6\pi \approx 1786$ cm $^2$

Cost $= 1786 \times 10 = \mathbf{Rs\ 17,860}$

Exam-Specific Tips

CBSE Class 9: Focus on direct formula questions — CSA, TSA, volume of individual solids. The most common question involves melting and recasting (equate volumes). Also practise finding the remaining height/radius when two quantities are given.

CBSE Class 10: Combination of solids is the main topic. Shapes like cylinder + hemisphere, cone + hemisphere (toy), cylinder + cone (tent), and frustum (bucket) appear every year. Always sketch the solid and mark which surfaces are exposed.

Common Mistakes to Avoid

Mistake 1 — Using diameter instead of radius. When a problem says “diameter 14 cm”, plug in $r = 7$ , not 14. This is the single most common error.

Mistake 2 — Confusing slant height with vertical height. For cones and frustums, $l \neq h$ . Always check which one the problem gives and use $l = \sqrt{h^2 + r^2}$ if needed.

Mistake 3 — Counting hidden surfaces in combination solids. When a hemisphere sits on a cylinder, the flat face of the hemisphere and the top face of the cylinder are internal — don’t include them in TSA.

Mistake 4 — Volume stays the same when reshaping, surface area does not. When a solid is melted and recast, equate volumes (not surface areas).

Mistake 5 — Forgetting units. Surface area is in cm $^2$ , volume in cm $^3$ . In cost problems, check whether the rate is per cm $^2$ or per m $^2$ .

Practice Questions

Q1. Find the volume of a sphere of radius 21 cm.

$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 9261 = 38808$ cm $^3$ .

Q2. A cone has height 24 cm and radius 7 cm. Find slant height and CSA.

$l = \sqrt{576 + 49} = 25$ cm. CSA $= \pi \times 7 \times 25 = 550$ cm $^2$ .

Q3. A metallic sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the cylinder’s height.

Equate volumes: $\frac{4}{3}\pi(6)^3 = \pi(4)^2 h$ . So $288 = 16h$ , giving $h = 18$ cm.

Q4. Find the TSA of a solid hemisphere of radius 10 cm.

TSA $= 3\pi r^2 = 3\pi(100) = 300\pi \approx 942.5$ cm $^2$ .

Q5. A toy is a cone mounted on a hemisphere. Both have radius 3.5 cm. Total height is 15.5 cm. Find the TSA.

Cone height $= 12$ cm. $l = \sqrt{144 + 12.25} = 12.5$ cm. TSA $= \pi(3.5)(12.5) + 2\pi(3.5)^2 = 43.75\pi + 24.5\pi = 68.25\pi \approx 214.5$ cm $^2$ .

Q6. A frustum has top radius 5 cm, bottom radius 10 cm, height 6 cm. Find the volume.

$V = \frac{1}{3}\pi(6)(100 + 25 + 50) = 350\pi \approx 1099.6$ cm $^3$ .

Q7. How many lead shots of diameter 4.2 cm from a block of $66 \times 42 \times 21$ cm?

Block volume $= 58212$ cm $^3$ . Shot volume $= \frac{4}{3}\pi(2.1)^3 \approx 38.808$ cm $^3$ . Count $= 58212/38.808 = 1500$ .

Q8. A cylindrical pipe (diameter 5 mm) delivers water at 10 m/min. How long to fill a cone of radius 40 cm and depth 24 cm?

Pipe radius $= 0.25$ cm. Flow $= \pi(0.0625)(1000) = 62.5\pi$ cm $^3$ /min. Cone volume $= \frac{1}{3}\pi(1600)(24) = 12800\pi$ cm $^3$ . Time $= 12800/62.5 \approx 205$ min.

Conversion Between Solids — Melting and Recasting

One of the most important CBSE problem types: a solid of one shape is melted and recast into another shape. The key principle is volume is conserved — surface area is NOT.

Worked Example — Metal Sphere to Wire

A solid sphere of radius 6 cm is melted and drawn into a wire of radius 0.2 cm. Find the length of the wire.

V_{sphere} = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi \times 216 = 288\pi \text{ cm}^3

Wire is a cylinder with radius 0.2 cm and length $l$ :

V_{wire} = \pi (0.2)^2 \times l = 0.04\pi l

288\pi = 0.04\pi l \implies l = \frac{288}{0.04} = 7200 \text{ cm} = 72 \text{ m}

Worked Example — Water Flowing into a Tank

A cylindrical pipe of diameter 7 cm delivers water at a rate of 12 m/min. How long will it take to fill a cuboidal tank of dimensions $22 \text{ m} \times 14 \text{ m} \times 10.5 \text{ m}$ ?

Pipe radius $= 3.5$ cm $= 0.035$ m. Water flows at 12 m/min.

Volume per minute $= \pi r^2 \times \text{speed} = \frac{22}{7} \times (0.035)^2 \times 12 = \frac{22}{7} \times 0.001225 \times 12 = 0.0462$ m³/min.

$V = 22 \times 14 \times 10.5 = 3234$ m³.

$t = \frac{3234}{0.0462} = 70000$ min $= \frac{70000}{60} \approx 1167$ hours. This highlights how small a pipe’s capacity is compared to a large tank.

Cone volume $= \frac{1}{3} \times$ Cylinder volume (same base, same height)
Hemisphere volume $= \frac{2}{3} \times$ Cylinder volume (same base, height = radius)
Sphere volume $= \frac{4}{3}\pi r^3 = 2 \times$ hemisphere volume
When reshaping: $V_{\text{old}} = V_{\text{new}}$ (equate volumes to find unknowns)

CBSE Class 10 board exams 2023 and 2024 both had a combination solid problem worth 5 marks. The standard approach: (1) sketch the solid, (2) identify which surfaces are external, (3) for volume: add all component volumes, (4) for TSA: add only exposed surfaces. Never count the internal joining surfaces.

FAQs

How do I decide between CSA and TSA?

If the object is open (bucket without lid) or a surface is hidden (hemisphere on cylinder), those absent/hidden faces are excluded. Use TSA only when all surfaces are exposed.

Why is a cone’s volume one-third of a cylinder?

This can be proved with calculus. Intuitively, a cone tapers from the full base to a point, filling only one-third of the cylinder space. The ratio comes from integrating circular cross-sections that shrink linearly.

How do I handle combination problems?

Sketch the solid. For volume, add all component volumes. For TSA, add only the exposed surface areas — skip surfaces where two shapes meet internally.

What is the relation between litres and cm $^3$ ?

$1$ litre $= 1000$ cm $^3$ . Also, $1$ m $^3 = 1000$ litres. This conversion appears in water tank and container problems.

How do we find the dimensions of a frustum from the original cone?

If a cone of height $H$ and radius $R$ is cut at height $h$ from the base, the top radius of the frustum is $r = R \cdot \frac{H-h}{H}$ by similar triangles. The frustum height is $h$ .