---
title: "Double integral r sinθ over cardioid"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:double-integral-r-sin-over"
source: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-r-sin-over
site: BCT Engineering
---

# Double integral r sinθ over cardioid

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2072 Magh, 5 marks
- Video: [Double Integral of f( r, theta) around  cardioid / Problems on Multiple Integral](https://www.youtube.com/watch?v=4hh8bfsVL4s) — JNTU Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-r-sin-over
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate: ∬ r sin θ dr dθ over the area of the cardioid r = a(1+cos θ) above the initial line.

## Short answer

Integrate r sinθ dr dθ with r from 0 to a(1+cosθ), θ from 0 to π. The r-integral yields a²(1+cosθ)²/2; substitute t=1+cosθ to get 4a²/3.

## Answer

**Region.** The part of the cardioid r = a(1 + cos θ) above the initial line (θ = 0) is traced for θ from 0 to π. For each θ, r runs from 0 to a(1 + cos θ).

**Reading.** The integrand is taken exactly as printed, r sin θ, with the integration element dr dθ.

```text
I = ∫₀^π ∫₀^{a(1+cosθ)} r sinθ dr dθ
```

#### r-integration
```text
∫₀^{a(1+cosθ)} r dr = a²(1 + cos θ)²/2
```

#### θ-integration
Put t = 1 + cos θ, dt = −sin θ dθ. When θ = 0, t = 2; when θ = π, t = 0.

```text
I = (a²/2) ∫₀^π (1 + cos θ)² sin θ dθ
  = (a²/2) ∫₀² t² dt
  = (a²/2)(8/3)
  = 4a²/3
```

**Answer: 4a²/3.**

If the area element r dr dθ were also included (that is, the integral of y = r sin θ over the area), the value would be 4a³/3. With the integral as printed, the answer is 4a²/3.

## Check yourself

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

1. θ from:
   A. 0 to π
   B. 0 to π/2
   C. −π to π
   D. 0 to 2π
   Answer: A. Above initial line. (Key point: Limits)

2. t = 1+cosθ turns it into:
   A. ∫₀² t² dt
   B. ∫₀¹ t² dt
   C. ∫₀² t dt
   D. ∫₀^π t² dt
   Answer: A. With limits 2 to 0 flipped. (Key point: Substitution)

3. As printed (no extra r):
   A. 2a²/3
   B. 8a²/3
   C. 4a²/3
   D. 4a³/3
   Answer: C. (a²/2)(8/3). (Key point: Result)

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