Math of Space: Surface Area and Volume

Chapter at a glance

Chapter 14 gives the surface area and volume formulas for cuboids, cubes, right circular cylinders, cones, pyramids, spheres and hemispheres, explains why they work (stacks of cards and khakhras, unrolled cones, Archimedes’ cylinder), and finishes with guesstimate problems.

Formula sheet

Volume is measured in cubic units (cm³, m³). 1000 cm³ = 1 litre; 1 ft³ ≈ 28.3 litres.

14.1 Cuboids and cubes

A cuboid is a 3D version of a rectangle: six rectangular faces. Its volume is the number of unit cubes that fit inside. Think of it as a stack of thin l × w cards, so V = base area × height = lwh.

Cube numbers 1, 8, 27, 64 are the volumes of cubes of side 1, 2, 3, 4, and cube roots go the other way (∛64 = 4).

Cutting a 4 cm cube into 1 cm cubes keeps the volume (64 cm³) but raises the surface area from 96 cm² to 64 × 6 = 384 cm², a ratio of 1 : 4. This surface-to-volume effect has big consequences in biology.

14.2 Right circular cylinder

Right: the axis is perpendicular to the base. Circular: the cross-section is a circle. (In an oblique cylinder the axis is tilted; it is not studied here.) A cross-section is a thin slice: parallel to the base you get a circle, parallel to the axis a rectangle. MRI scans work in a similar way, in slices.

• Curved surface: cut and unroll it into a rectangle of width 2πr and height h, so CSA = 2πrh.
• Volume: a stack of thin circular khakhras gives cross-section × height = πr²h.
• Example: a kaleidoscope 25 cm long with r = 3.5 cm needs 2 × (22/7) × 3.5 × 25 = 550 cm² of chart paper.

14.3 Cones

Rotate a right triangle about one of its legs and you get a right circular cone. By Baudhāyana–Pythagoras, l² = h² + r².

• CSA = πrl. Unroll the cone into a sector of radius l, cut it into thin triangles of height l, and the bases add up to the circumference 2πr: area = ½ × 2πr × l. (The sector-area formula gives the same result.)
• Volume = ⅓πr²h, exactly one-third of the cylinder on the same base and height. Check it with salt: three cones of salt fill the cylinder. Or mould clay: one cylinder makes exactly three cones.
• The formula was first found by Archimedes (c. 225 BCE).
• Example: the 5-12-13 triangle rotated about the 12 cm side gives r = 5 and h = 12, so V = ⅓π × 25 × 12 = 100π cm³.

14.4 Pyramidal shapes

A base of any shape (triangle, square, …) is joined to a single apex in another plane. Examples: the Egyptian pyramids, Rubik’s triangular pyramid puzzle, the Shivling peak (Uttarakhand) and the Matterhorn. V = ⅓ × base area × height, the same as a cone, which is a pyramid with a circular base. For surface area, add up the faces.

14.5 Spheres and hemispheres

A sphere is the set of all points in space at distance r from a centre O.

String activity: wind string over half a ball, then the rest. The same string exactly fills 4 circles of the ball’s radius, so the surface area is 4πr².

Hemisphere: CSA = 2πr² and TSA = 2πr² + πr² = 3πr². A bowl of radius 3.5 cm holds ⅔ × (22/7) × 3.5³ ≈ 89.8 cm³.

14.6 Guesstimates

When exact data are missing, make an intuitive guess, model the situation, state your assumptions, approximate, calculate, and compare with your guess. The reasoning matters more than the exact answer.

• Golgappas: the pani is π × 10² × 50 = 5000π cm³. One student assumes r = 1.5 cm and 60% filling, giving about 1850 golgappas; another assumes r = 1.75 cm and 80% filling, giving about 870. Different assumptions, different estimates.
• Tennis balls in a classroom (30 × 30 × 15 ft, ball r = 3 cm): volume ÷ ball volume ≈ 33 lakh is only an upper bound, because spheres leave gaps. A simple grid packing of 150 × 150 × 75 gives about 16.9 lakh. That is why cubes, not spheres, are used as units of volume.

Practice questions with answers

1. A cube has volume 64 cm³. TSA? Side 4 cm, TSA 96 cm²
2. How many 20 cm cubes fit in a 2 m cubical box? 10 × 10 × 10 = 1000
3. Three faces at a corner of a cuboid have areas 6, 15 and 10 cm². Volume? (lw)(wh)(hl) = (lwh)² = 900, so V = 30 cm³
4. A 5 cm painted cube is cut into 1 cm cubes. How many have 3, 2, 1 and 0 painted faces? 8, 36, 54, 27
5. A conical cup is filled with water to half its depth. What fraction is full? The water cone has half the radius and height: (½)³ = 1/8
6. A sphere’s radius rises by x% and its volume by 72.8%. Find x. 1.728 = 1.2³, so x = 20
7. Show a 10% increase in radius raises a sphere’s volume by about 33.1%. 1.1³ = 1.331
8. A glass (r = 4 cm) has water at 16 cm; the crow needs 20 cm. How many 1 cm marbles? Needed volume π × 16 × 4 = 64π; one marble is ⁴⁄₃π; 48 marbles
9. A string 1 m longer than the equator is held evenly around the Earth. How high is it? 2π(R + h) − 2πR = 1, so h = 1/(2π) ≈ 16 cm, the same for the Moon, Jupiter or a volleyball!

Exam tips

• Check whether the question wants curved or total surface area, and whether the cylinder is open or closed.
• For cones, find the slant height first using l = √(h² + r²) when only h and r are given.
• Keep units consistent; convert m to cm before multiplying. Volume scales as length³ and area as length².
• Use the π value the question asks for (22/7 or 3.14).