---
title: "Surface area of cardioid by revolution"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:surface-area-cardioid-revolution"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/surface-area-cardioid-revolution
site: BCT Engineering
---

# Surface area of cardioid by revolution

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2067 Asoj
- Video: [Find the area of the surface of revolution formed by revolving the curve r=a(1+cosθ) about the init](https://www.youtube.com/watch?v=NZ4IziY_lww) — 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/surface-area-cardioid-revolution
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the surface area of solid generated by the revolution of cardioid r = a(1+cosθ).

## Short answer

Use S=2π∫₀^π r sinθ √(r²+(dr/dθ)²) dθ. With r=a(1+cosθ), ds=2a cos(θ/2)dθ, integration yields 32πa²/5 square units.

## Answer

**Reading:** the surface generated when the cardioid r = a(1 + cos θ) is revolved about the initial line (θ = 0). The upper half 0 ≤ θ ≤ π generates the whole surface.

For revolution about the initial line, S = 2π ∫ y ds = 2π ∫ r sin θ · √(r² + (dr/dθ)²) dθ.

With r = a(1 + cos θ), dr/dθ = −a sin θ:

r² + (dr/dθ)² = a²[(1 + cos θ)² + sin²θ] = 2a²(1 + cos θ) = 4a² cos²(θ/2)

so ds = 2a cos(θ/2) dθ. Also r = 2a cos²(θ/2) and sin θ = 2 sin(θ/2) cos(θ/2).

```text
S = 2π ∫₀^π 2a cos²(θ/2) · 2 sin(θ/2) cos(θ/2) · 2a cos(θ/2) dθ
  = 16π a² ∫₀^π cos⁴(θ/2) sin(θ/2) dθ
```
Put u = cos(θ/2), du = −½ sin(θ/2) dθ. Limits: θ = 0 → u = 1; θ = π → u = 0.

```text
S = 16πa² · 2 ∫₀¹ u⁴ du
  = 32πa² · (1/5)
```
**S = 32πa²/5 square units.**

## Check yourself

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

1. Which integral gives the surface area when a polar curve is revolved about the initial line?
   A. 2π ∫ r cosθ · sqrt(r^2 + (dr/dθ)^2) dθ
   B. 2π ∫ r sinθ · sqrt(r^2 + (dr/dθ)^2) dθ
   C. π ∫ r^2 sinθ dθ
   D. 2π ∫ r sinθ dθ
   Answer: B. The surface area is 2π ∫ y ds with y = r sinθ and ds = sqrt(r^2 + (dr/dθ)^2) dθ. (Key point: Surface of revolution formula)

2. For r = a(1+cosθ), what is r^2 + (dr/dθ)^2 and hence ds?
   A. 4a^2 cos^2(θ/2), so ds = 2a cos(θ/2) dθ
   B. a^2(1+cosθ)^2, so ds = a(1+cosθ) dθ
   C. 2a^2 sin^2(θ/2), so ds = a sqrt2 sin(θ/2) dθ
   D. a^2, so ds = a dθ
   Answer: A. (1+cosθ)^2 + sin^2θ = 2(1+cosθ) = 4cos^2(θ/2), giving ds = 2a cos(θ/2) dθ. (Key point: Arc element of the cardioid)

3. Why is θ integrated only from 0 to π?
   A. The cardioid only exists for 0 ≤ θ ≤ π
   B. The integrand is zero for θ > π
   C. Because the surface is only half a solid
   D. The upper half of the cardioid, when revolved about the initial line, generates the whole surface
   Answer: D. Revolving the upper half 0 ≤ θ ≤ π about the initial line sweeps out the entire surface; going to 2π would count it twice. (Key point: Limits of integration)

4. With u = cos(θ/2) the integral reduces to 32πa^2 ∫₀¹ u^4 du. What is the surface area?
   A. 16πa^2/5
   B. 8πa^2/3
   C. 32πa^2/5
   D. 32πa^2/3
   Answer: C. ∫₀¹ u^4 du = 1/5, so S = 32πa^2/5 square units. (Key point: Substitution and final result)

## Related in Mathematics I

- [Volume of cardioid r=a(1−cosθ) about initial line](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Avolume-cardioid-1-minus-cos) — Mathematics I, chapter 2, asked 6×
- [Volume of cardioid r=a(1+cosθ) about initial line](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Avolume-cardioid-1-plus-cos-initial-line) — Mathematics I, chapter 2, asked 6×
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- [Chord of curvature through pole of cardioid](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Acurvature-chord-through-pole-cardioid) — Mathematics I, chapter 1

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