Surface Areas and Volumes — Study Notes

Overview

Surface Areas and Volumes is a critical pillar of the SOF IMO Mathematical Reasoning section, directly testing your ability to visualize three-dimensional objects and apply precise formulas. This topic appears consistently in IMO papers, often in multi-step word problems requiring you to calculate material costs, capacities, or conversions between different solid shapes.

Mastery here requires two skills: **formula fluency** (instant recall of all surface area and volume formulas for six basic solids) and **problem decomposition** (breaking complex real-world scenarios into geometric parts). IMO questions frequently combine shapes—such as a cylinder with hemispherical ends or a cone carved out of a cylinder—testing whether you can identify components and apply appropriate formulas. The Achievers Section elevates this further with optimization problems (maximum volume for given surface area) or cost calculations involving thickness and materials.

Your exam strategy should prioritize accurate formula recall first, then practice identifying which parts of a composite solid contribute to exposed surface area versus hidden interfaces. Dimensional consistency (converting all measurements to the same unit before calculation) prevents 80% of errors in this topic.

Key Concepts

• **Lateral vs. Total Surface Area**: Lateral surface area (LSA or CSA—curved surface area) covers only the sides, excluding top and bottom bases. Total surface area (TSA) includes all exposed faces. For a cylinder, CSA = 2πrh; TSA = 2πr(r + h).

• **Volume as Capacity**: Volume measures the space inside a solid, directly translating to capacity for containers. 1 cubic metre = 1000 litres, and 1 cubic centimetre = 1 millilitre—critical for word problems involving water tanks or paint coverage.

• **Composite Solids**: Real IMO problems present combinations like a tent (cone + cylinder), a capsule (cylinder + two hemispheres), or a toy (cone on hemisphere). Calculate each part separately, then sum for total volume; for surface area, exclude contact regions.

• **Dimension Relationships**: Radius (r), diameter (d = 2r), height (h), and slant height (l) are interconnected. For a cone, l² = r² + h² by Pythagoras. Always check which measurements the problem provides versus what the formula needs.

• **Unit Conversions**: IMO loves mixing units—radius in cm, height in m, asking answer in litres. Convert everything to one unit system before plugging into formulas. 1 m = 100 cm; 1 m³ = 10⁶ cm³.

• **Cost and Material Problems**: If painting costs ₹5 per m² and TSA is 120 m², total cost = 5 × 120 = ₹600. When thickness is involved (hollow cylinders, walls), treat as the difference of two volumes: outer solid minus inner cavity.

• **Sphere vs. Hemisphere**: A hemisphere is exactly half a sphere by volume, but its TSA includes the flat circular base. Hemisphere TSA = 2πr² (curved) + πr² (base) = 3πr².

• **Melting and Recasting**: When a solid is melted and recast into a different shape, volume remains constant but surface area changes. Set volume₁ = volume₂ and solve for the unknown dimension.

Formulas / Key Facts

**Cuboid**

• Volume = l × b × h
• TSA = 2(lb + bh + hl)
• LSA = 2h(l + b)

**Cube** (special cuboid where l = b = h = a)

• Volume = a³
• TSA = 6a²
• LSA = 4a²

**Cylinder**

• Volume = πr²h
• TSA = 2πr(r + h)
• CSA = 2πrh

**Cone**

• Volume = (1/3)πr²h
• TSA = πr(l + r), where l = √(r² + h²)
• CSA = πrl

**Sphere**

• Volume = (4/3)πr³
• Surface Area = 4πr²

**Hemisphere**

• Volume = (2/3)πr³
• TSA = 3πr² (curved + base)
• CSA = 2πr²

**Key Fact**: For a cone, if only radius and height are given, calculate slant height l using l = √(r² + h²) before applying surface area formulas.

Worked Examples

**Example 1: TSA of a Closed Cylinder** A cylindrical tank has radius 7 cm and height 10 cm. Find its total surface area. (Use π = 22/7)

*Solution*: TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = 748 cm².

**Example 2: Volume of Composite Solid** A toy consists of a cone of radius 3 cm and height 4 cm mounted on a hemisphere of the same radius. Find total volume.

*Solution*: Cone volume = (1/3)πr²h = (1/3)π(3²)(4) = 12π cm³. Hemisphere volume = (2/3)πr³ = (2/3)π(3³) = 18π cm³. Total volume = 12π + 18π = 30π = 30 × 3.14 ≈ 94.2 cm³.

**Example 3: Melting and Recasting** A solid sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.

*Solution*: Volume of sphere = (4/3)π(6³) = 288π cm³. Volume of cylinder = πr²h = π(4²)h = 16πh. Equate: 16πh = 288π → h = 288/16 = 18 cm.

Common Mistakes

**Using diameter instead of radius** → Always halve the diameter before substituting into formulas. If d = 14 cm, then r = 7 cm.

**Forgetting to calculate slant height for cones** → Cone TSA needs l = √(r² + h²), not just h. Skipping this step gives wrong surface area.

**Adding base areas in composite solids** → When a cone sits on a cylinder, the circular base where they join is not part of the exposed surface. Exclude it from TSA.

**Mixing units mid-calculation** → If radius is 50 cm and height is 2 m, convert both to cm (200 cm) or both to m (0.5 m) before applying formulas.

**Confusing hemisphere formulas** → Hemisphere TSA is 3πr², not 2πr². The extra πr² is the flat circular base. Don't drop it for closed hemispheres.

**Volume formula denominators** → Cone and pyramid volumes have 1/3; sphere volume has 4/3. Mixing these up ruins the answer. Memorize the exact fractions.

Quick Reference

• **Cube**: Volume = a³, TSA = 6a². If edge doubles, volume increases 8×, surface area 4×.
• **Cylinder**: CSA = 2πrh, TSA = 2πr(r + h). Open top/bottom? Subtract πr² per missing base.
• **Cone slant height**: l = √(r² + h²). Must calculate before finding TSA.
• **Sphere**: SA = 4πr³, Volume = (4/3)πr³. Hemisphere volume is exactly half, but TSA = 3πr².
• **Melting problems**: Volume before = Volume after. Solve for unknown dimension.
• **Unit conversions**: 1 m³ = 1000 litres = 10⁶ cm³. Always convert before calculating.