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

# Volume of cycloid about tangent at vertex

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2070 Chaitra
- Video: [Find the volume of the solid revolving the cycloid x=a(θ+sinθ),y=a(1+cosθ) ;about cycloid it's base.](https://www.youtube.com/watch?v=g2P8lOFx4XU) — Doctor of Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/volume-cycloid-about-tangent-at-vertex
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the volume of the solid formed by the revolution of the cycloid x = a(θ+sinθ), y = a(1−cosθ) about the tangent at the vertex.

## Short answer

Revolve the region between one arch and the line joining the cusps (y=2a) about the x-axis (tangent at vertex). V = π∫(4a²-y²)dx from θ=-π to π gives 7π²a³ cubic units.

## Answer

**Reading of the question.** The curve x = a(θ + sin θ), y = a(1 − cos θ) has its cusps at y = 2a (at θ = ±π) and its vertex at the origin, where the tangent is the x-axis (y = 0). The usual textbook problem (answer 7π²a³) asks for the solid formed when the area between one arch and the line joining its cusps is revolved about the tangent at the vertex. We solve that version.

**Setup.** One arch runs for θ from −π to π, and dx = a(1 + cos θ) dθ. The area enclosed between the arch and the cusp line y = 2a is revolved about y = 0. The strip at x has outer radius 2a and inner radius y, so

```text
V = π ∫ (4a² − y²) dx
  = π ∫_(−π)^π [4a² − a²(1 − cosθ)²] · a(1 + cosθ) dθ
```
Split the integral:

- ∫ 4a² · a(1 + cos θ) dθ over (−π, π) = 4a³ · 2π = 8πa³.
- ∫ a²(1 − cos θ)² a(1 + cos θ) dθ = a³ ∫ (1 − cos θ) sin²θ dθ = a³ [∫ sin²θ dθ − ∫ cos θ sin²θ dθ] = a³ (π − 0) = πa³.

```text
V = π (8πa³ − πa³)
```
**V = 7π²a³ cubic units.**

(If instead the region between the arch and the tangent y = 0, bounded by x = ±πa, is revolved about y = 0, the volume is π∫y² dx = π²a³. The standard result 7π²a³ is for the region between the arch and the cusp line.)

## Check yourself

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

1. The standard problem revolves
   A. the arch curve only
   B. the region between one arch and the cusp line
   C. the area under the arch
   D. the cusp line only
   Answer: B. Answer 7π^2 a^3 refers to this region. (Key point: Region)

2. The strip has outer radius 2a and inner radius
   A. 2a - y
   B. 0
   C. y
   D. a
   Answer: C. Area between y and 2a. (Key point: Radii)

3. The volume is
   A. π integral y^2 dx
   B. 2π integral y dx
   C. π integral (2a - y) dx
   D. π integral (4a^2 - y^2) dx
   Answer: D. Washer difference. (Key point: Formula)

4. V equals
   A. 7π^2 a^3
   B. π^2 a^3
   C. 8π^2 a^3
   D. 5π^2 a^3
   Answer: A. π(8πa^3 - πa^3). (Key point: Result)

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