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ICSE • Class 8 • Mathematics • Ch 23
Estimated Time: 45 Mins
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Surface Area and Volume of Solids

In ICSE Class 8 Mathematics, "Surface Area and Volume of Solids" provides an authoritative, mathematically rigorous master study guide investigating three-dimensional spatial mensuration for cubes, cuboids, and right circular cylinders. This comprehensive chapter explores Spatial Fundamentals (Difference between perimeter [1D: $\text{cm}$], area [2D: $\text{cm}^2$], and volume [3D: $\text{cm}^3$]; Capacity vs Volume: internal volume of a container; Metric volume-capacity conversions: $1\text{ m}^3 = 1000\text{ liters}$, $1\text{ liter} = 1000\text{ cm}^3$, $1\text{ mL} = 1\text{ cm}^3$), Mensuration of Cuboid (Dimensions $l, b, h$: 1. Total Surface Area [TSA]: $2(lb + bh + hl)$, 2. Lateral Surface Area [LSA / Area of 4 Walls]: $2h(l + b)$, 3. Volume: $l \times b \times h$, 4. Space Diagonal [Length of longest rod]: $\sqrt{l^2 + b^2 + h^2}$), Mensuration of Cube (Edge length $a$: 1. TSA: $6a^2$, 2. LSA: $4a^2$, 3. Volume: $a^3$, 4. Space Diagonal: $a\sqrt{3}$), Mensuration of Right Circular Cylinder (Radius $r$, height $h$: 1. Curved Surface Area [CSA]: $2\pi rh$, 2. Total Surface Area [TSA]: $2\pi r(h + r)$, 3. Volume: $\pi r^2 h$), Hollow Cylinders / Pipes (External radius $R$, internal radius $r$, thickness $R - r$: 1. Volume of material: $\pi(R^2 - r^2)h$, 2. Total Surface Area: $2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2)$), and Practical Applied Problems (Excavation of earth to dig wells, melting and recasting solids where volume is conserved, cost of plastering, painting, and painting walls and ceilings) aligned with the 2026–27 CISCE ICSE curriculum.

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How Did Archimedes Use the Master Cylinder Law Inscribed on His Tombstone to Revolutionize 3D Geometry?

When the greatest mathematician of antiquity, Archimedes of Syracuse, was asked what he considered his greatest discovery, he did not choose his military catapults or his famous buoyancy principle that made him run naked through the streets shouting "Eureka!". Instead, he chose a geometric relationship between a sphere and a cylinder: that a sphere inscribed inside a cylinder occupies exactly two-thirds of its volume and two-thirds of its surface area! He was so proud of this discovery that he requested a cylinder and sphere be carved onto his gravestone! In three-dimensional space, formulas are not arbitrary: if you take a cylinder of radius $r$ and height $h$, and slice its curved wall vertically, it unrolls into a PERFECT RECTANGLE whose length is the base circumference $2\pi r$ and breadth is the height $h$! That is why Curved Surface Area is simply: $2\pi r \times h = \mathbf{2\pi rh}$! What is the longest fishing rod you can pack diagonally inside a room? What happens to total volume when metallic blocks are melted down? Let's master surface area and volume of solids.

Why This Chapter Matters

3D mensuration governs architecture (HVAC airflow capacity, building room volumes), packaging industrial design (beverage cans, shipping cartons), civil hydraulic engineering (well digging, reservoir storage dams), and mining geology. Mastering cylinder and cuboid mensuration is a core requirement for ICSE Class 8, 9, and 10 board examinations.

Before You Begin (Prerequisites)

  • Circle circumference and area ($2\pi r, \pi r^2$) from Chapters 21 and 22.
  • Units of measurement and conversions.
  • Pythagoras theorem in 3D.

What You Will Learn (Core Objectives)

  • Convert between metric units of volume and capacity ($1\text{ m}^3 = 1000\text{ L}, 1\text{ L} = 1000\text{ cm}^3$).
  • Calculate TSA, LSA, volume, and space diagonal of a cuboid.
  • Calculate TSA, LSA, volume, and diagonal of a cube.
  • Calculate CSA, TSA, and volume of solid right circular cylinders.
  • Calculate material volume and TSA of hollow cylindrical pipes.
  • Solve conservation-of-volume problems (melting, recasting, well excavation).

Chapter Roadmap & Progression

1 1. Cuboid Mensuration: TSA, LSA & S...
2 2. Cube Mensuration: Edge Length $a...
3 3. Right Circular Cylinder Mensurat...
4 4. Hollow Cylindrical Pipes & Conse...

Complete Concept Guide (100% Curriculum Coverage)

1. Cuboid Mensuration: TSA, LSA & Space Diagonal

Understand
A. Formulas for a Cuboid ($l \times b \times h$):
  • Volume: Space occupied: $$\mathbf{V = l \times b \times h}$$
  • Total Surface Area (TSA): Sum of all 6 rectangular faces: $$\mathbf{\text{TSA} = 2(lb + bh + hl)}$$
  • Lateral Surface Area (LSA / Area of 4 Walls): $$\mathbf{\text{LSA} = 2h(l + b) = (\text{Perimeter of Base}) \times h}$$
  • Space Diagonal (Longest Rod inside Room): $$\mathbf{D = \sqrt{l^2 + b^2 + h^2}}$$

2. Cube Mensuration: Edge Length $a$

Cube Formulas

When all dimensions are identical ($l = b = h = a$):

$$\mathbf{V = a^3}$$ $$\mathbf{\text{TSA} = 6a^2}$$ $$\mathbf{\text{LSA} = 4a^2}$$ $$\mathbf{\text{Space Diagonal} = a\sqrt{3}}$$

3. Right Circular Cylinder Mensuration

Cylinder
A. Solid Cylinder (Radius $r$, Height $h$):
  • Volume: Base Area $\times$ Height: $$\mathbf{V = \pi r^2 h}$$
  • Curved Surface Area (CSA): Base Circumference $\times$ Height: $$\mathbf{\text{CSA} = 2\pi rh}$$
  • Total Surface Area (TSA): CSA $+ 2 \times \text{Base Area}$: $$\mathbf{\text{TSA} = 2\pi rh + 2\pi r^2 = 2\pi r(h + r)}$$
B. Metric Volume-Capacity Invariants:
$$\mathbf{1\text{ m}^3 = 1000\text{ Liters} = 1\text{ Kiloliter}}$$ $$\mathbf{1\text{ Liter} = 1000\text{ cm}^3 \quad \Longleftrightarrow \quad 1\text{ mL} = 1\text{ cm}^3}$$

4. Hollow Cylindrical Pipes & Conservation of Volume

Advanced Solids
A. Hollow Cylinder / Metallic Pipe:

External radius $R$, Internal radius $r$, Height $h$:

  • $$\mathbf{\text{Volume of Metal} = \pi (R^2 - r^2) h}$$
  • $$\mathbf{\text{TSA} = 2\pi Rh + 2\pi rh + 2\pi (R^2 - r^2)}$$
B. Conservation of Volume in Melting / Recasting:
$$\mathbf{\text{Volume of Original Solid(s)} = \text{Total Volume of Recast Solid(s)}}$$

Key Formulas, Identities & Theorems

Cylinder Volume & TSA
$$V = \pi r^2 h, \quad \text{TSA} = 2\pi r(h + r)$$
Radius r and height h of circular cylinder.
Metric Capacity Invariant
$$1\text{ m}^3 = 1000\text{ L}, \quad 1\text{ L} = 1000\text{ cm}^3$$
Standard conversion between volumetric space and fluid capacity.

Mensuration: Cuboid & Unrolled Cylinder Anatomy

Surface Area & Volume: Solid Mensuration Laws CUBOID (l × b × h) Length l h b • Volume = l × b × h • TSA = 2(lb + bh + hl) • 4 Walls (LSA) = 2h(l + b) • Longest Rod = √(l2 + b2 + h2) CYLINDER (UNROLLED RECTANGLE) h CSA = 2πrh Circumference = 2πr • Volume = πr2h • CSA = 2πrh • TSA = 2πr(h + r) • 1 m3 = 1,000 Liters • 1 Liter = 1,000 cm3 • Hollow Pipe Metal Vol = π(R2 - r2)h CUBOID V = lbh • CYLINDER V = πr^2 h • TSA = 2πr(h+r) • 1 m^3 = 1000 L • MELTING CONSERVES VOLUME

Chapter Summary & 10 Key Takeaways

Takeaway 1
Volume measures three-dimensional space occupied; surface area measures the total 2D boundary wrapper.
Takeaway 2
Capacity is internal volume: 1 m^3 = 1000 Liters, and 1 Liter = 1000 cm^3.
Takeaway 3
Cuboid: Volume = lbh, TSA = 2(lb + bh + hl), LSA = 2h(l + b).
Takeaway 4
Space diagonal of cuboid (longest rod inside room): D = sqrt(l^2 + b^2 + h^2).
Takeaway 5
Cube: Volume = a^3, TSA = 6a^2, LSA = 4a^2, Diagonal = a*sqrt(3).
Takeaway 6
Cylinder: Volume = pi * r^2 * h, CSA = 2*pi*r*h, TSA = 2*pi*r(h + r).
Takeaway 7
The curved surface of a cylinder unrolls into a flat rectangle of length 2*pi*r and breadth h.
Takeaway 8
Hollow cylinder metal volume: V = pi * (R^2 - r^2) * h.
Takeaway 9
When solid shapes are melted and recast, total volume remains strictly conserved.
Takeaway 10
Area of four walls and ceiling (painting a room): Area = 2h(l + b) + lb.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A godown measures $40\text{ m} \times 25\text{ m} \times 15\text{ m}$. Find the maximum number of wooden crates each measuring $1.5\text{ m} \times 1.25\text{ m} \times 0.5\text{ m}$ that can be stored in the godown.
Reveal Answer & Explanation
Answer: Step 1: Calculate the volume of the godown:
$$V_1 = 40 \times 25 \times 15 = \mathbf{15,000\text{ m}^3}$$
Step 2: Calculate the volume of one wooden crate:
$$V_2 = 1.5 \times 1.25 \times 0.5 = \frac{3}{2} \times \frac{5}{4} \times \frac{1}{2} = \frac{15}{16} = \mathbf{0.9375\text{ m}^3}$$
Step 3: Number of crates $= \frac{\text{Volume of Godown}}{\text{Volume of One Crate}}$:
$$N = \frac{15,000}{\frac{15}{16}} = 15,000 \times \frac{16}{15} = 1000 \times 16 = \mathbf{16,000\text{ crates}}$$.
$N = (40 \times 25 \times 15) / (1.5 \times 1.25 \times 0.5) = 15000 / (15/16) = 16,000\text{ crates}$.
2
Find the curved surface area and total surface area of a right circular cylinder of diameter $14\text{ cm}$ and height $20\text{ cm}$. (Take $\pi = \frac{22}{7}$).
Reveal Answer & Explanation
Answer: Step 1: Radius $r = \frac{14}{2} = 7\text{ cm}$, height $h = 20\text{ cm}$.
Step 2: Curved Surface Area (CSA):
$$\text{CSA} = 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 20 = 2 \times 22 \times 20 = \mathbf{880\text{ cm}^2}$$
Step 3: Total Surface Area (TSA):
$$\text{TSA} = 2\pi r(h + r) = 2 \times \frac{22}{7} \times 7 \times (20 + 7) = 44 \times 27 = \mathbf{1,188\text{ cm}^2}$$.
$\\text{CSA} = 2 \times (22/7) \times 7 \times 20 = 880\text{ cm}^2$. $\\text{TSA} = 44 \times 27 = 1188\text{ cm}^2$.
3
What is the length of the longest pole that can be put in a room of dimensions $10\text{ m} \times 10\text{ m} \times 5\text{ m}$?
Reveal Answer & Explanation
Answer:

The longest pole is along the space diagonal of the cuboid:

$$D = \sqrt{l^2 + b^2 + h^2}$$


$$D = \sqrt{10^2 + 10^2 + 5^2} = \sqrt{100 + 100 + 25} = \sqrt{225} = \mathbf{15\text{ meters}}$$

.
The longest pole has length $15\text{ meters}$.


$D = \sqrt{10^2 + 10^2 + 5^2} = \sqrt{225} = 15\text{ meters}$.
4
A solid metallic cube of edge $12\text{ cm}$ is melted and recast into smaller cubes each of edge $3\text{ cm}$. How many such small cubes are formed?
Reveal Answer & Explanation
Answer: Step 1: Volume of original large cube $= a_1^3 = 12^3 = 1,728\text{ cm}^3$.
Step 2: Volume of one small cube $= a_2^3 = 3^3 = 27\text{ cm}^3$.
Step 3: Number of cubes formed:
$$N = \frac{a_1^3}{a_2^3} = \left( \frac{12}{3} \right)^3 = 4^3 = \mathbf{64\text{ cubes}}$$.
$N = (12/3)^3 = 4^3 = 64\text{ small cubes}$.
5
A well of diameter $14\text{ meters}$ is dug $15\text{ meters}$ deep. Find the volume of earth taken out and the cost of plastering its inner curved surface at $\text{Rs } 30\text{ per m}^2$.
Reveal Answer & Explanation
Answer: Step 1: Radius $r = \frac{14}{2} = 7\text{ m}$, depth $h = 15\text{ m}$.
Step 2: Volume of earth excavated (Volume of cylinder):
$$V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 15 = 22 \times 7 \times 15 = 154 \times 15 = \mathbf{2,310\text{ m}^3}$$
Step 3: Inner curved surface area (CSA):
$$\text{CSA} = 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 15 = 44 \times 15 = \mathbf{660\text{ m}^2}$$
Step 4: Cost of plastering:
$$\text{Cost} = 660 \times 30 = \mathbf{\text{Rs } 19,800}$$.
Volume $= (22/7) \times 49 \times 15 = 2310\text{ m}^3$. CSA $= 44 \times 15 = 660\text{ m}^2$. Cost $= 660 \times 30 = \text{Rs } 19,800$.
6
A cylindrical water tank of radius $1.4\text{ m}$ and height $2.1\text{ m}$ is full of water. How many liters of water does it hold?
Reveal Answer & Explanation
Answer: Step 1: Calculate the volume in cubic meters ($V = \pi r^2 h$):
$$V = \frac{22}{7} \times (1.4)^2 \times 2.1 = \frac{22}{7} \times 1.96 \times 2.1 = 22 \times 1.96 \times 0.3 = \mathbf{12.936\text{ m}^3}$$
Step 2: Convert cubic meters to liters ($1\text{ m}^3 = 1000\text{ Liters}$):
$$\text{Capacity} = 12.936 \times 1000 = \mathbf{12,936\text{ Liters}}$$.
$V = (22/7) \times 1.96 \times 2.1 = 12.936\text{ m}^3$. In liters: $12.936 \times 1000 = 12,936\text{ Liters}$.
7
The total surface area of a cube is $486\text{ cm}^2$. Find its volume.
Reveal Answer & Explanation
Answer: Step 1: Total Surface Area $= 6a^2 = 486\text{ cm}^2$:
$$a^2 = \frac{486}{6} = 81$$
$$a = \sqrt{81} = 9\text{ cm}$$
Step 2: Calculate Volume ($V = a^3$):
$$V = 9^3 = 9 \times 9 \times 9 = \mathbf{729\text{ cm}^3}$$.
$6a^2 = 486 \implies a^2 = 81 \implies a = 9\text{ cm}$. Volume $= 9^3 = 729\text{ cm}^3$.
8
A metallic pipe is $77\text{ cm}$ long. The inner diameter of a cross section is $4\text{ cm}$, the outer diameter being $4.4\text{ cm}$. Find the volume of the metal.
Reveal Answer & Explanation
Answer: Step 1: Internal radius $r = \frac{4}{2} = 2\text{ cm}$; External radius $R = \frac{4.4}{2} = 2.2\text{ cm}$. Length $h = 77\text{ cm}$.
Step 2: Volume of metal in hollow cylinder:
$$V = \pi (R^2 - r^2) h = \frac{22}{7} \times \left( 2.2^2 - 2^2 \right) \times 77$$
$$= 22 \times (4.84 - 4) \times 11 = 22 \times 0.84 \times 11$$
$$= 242 \times 0.84 = \mathbf{203.28\text{ cm}^3}$$.
$V = \frac{22}{7} \times (4.84 - 4) \times 77 = 22 \times 0.84 \times 11 = 203.28\text{ cm}^3$.
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