---
title: "Volume or surface area of cycloid about base"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:volume-or-surface-cycloid-base"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/volume-or-surface-cycloid-base
site: BCT Engineering
---

# Volume or surface area of cycloid about base

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2071 Chaitra
- Video: [Surface of solid of revolution of cycloid](https://www.youtube.com/watch?v=j-m-p25tvro) — smart study helper
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/volume-or-surface-cycloid-base
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the volume or surface area of solid generated by revolving the cycloid x = a(θ+sinθ), y = a(1+cosθ) about its base.

## Short answer

Revolve one arch of the cycloid about its base (x-axis). Volume V = π∫ y² dx = 5π²a³ cubic units. Surface area S = 2π∫ y ds = 64πa²/3 square units.

## Answer

Cycloid: x = a(θ + sin θ), y = a(1 + cos θ). One arch runs from θ = −π to π, with cusps at (∓πa, 0) on the x-axis, so the base is the x-axis. We have dx/dθ = a(1 + cos θ) = 2a cos²(θ/2) and dy/dθ = −a sin θ. The arc element is

ds = √(x′² + y′²) dθ = 2a cos(θ/2) dθ.

Also y = 2a cos²(θ/2).

#### Volume about the base (x-axis)
```text
V = π ∫ y² dx = π ∫_(−π)^π a²(1+cosθ)² · a(1+cosθ) dθ
  = πa³ ∫_(−π)^π (1 + cosθ)³ dθ
  = 8πa³ ∫_(−π)^π cos⁶(θ/2) dθ
```
Put θ/2 = u: ∫_(−π)^π cos⁶(θ/2) dθ = 2∫_(−π/2)^(π/2) cos⁶u du = 4∫₀^(π/2) cos⁶u du = 4 · 5π/32 = 5π/8.

**V = 8πa³ · 5π/8 = 5π²a³ cubic units.**

#### Surface area about the base
```text
S = 2π ∫ y ds = 2π ∫_(−π)^π 2a cos²(θ/2) · 2a cos(θ/2) dθ
  = 8πa² ∫_(−π)^π cos³(θ/2) dθ
```
With θ/2 = u: ∫ cos³(θ/2) dθ = 2∫_(−π/2)^(π/2) cos³u du = 4∫₀^(π/2) cos³u du = 4 · 2/3 = 8/3.

**S = 8πa² · 8/3 = 64πa²/3 square units.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The volume about the x-axis is
   A. 2π integral y ds
   B. π integral y^2 dx
   C. π integral y dx
   D. π integral x^2 dy
   Answer: B. Disc method. (Key point: Volume formula)

2. For this cycloid ds equals
   A. 2a sin(θ/2) dθ
   B. a(1+cos θ) dθ
   C. 2a cos(θ/2) dθ
   D. a cos θ dθ
   Answer: C. sqrt(x'^2 + y'^2) = 2a cos(θ/2). (Key point: Arc element)

3. The volume is
   A. 5π a^3
   B. π^2 a^3
   C. 8π^2 a^3
   D. 5π^2 a^3
   Answer: D. 8πa^3 · 5π/8. (Key point: Volume)

4. The surface area is
   A. 64πa^2/3
   B. 32πa^2/3
   C. 8πa^2/3
   D. 64πa^2
   Answer: A. 8πa^2 · 8/3. (Key point: Surface)

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