Area in polar coordinates
Objective. Integrate ½r² with correct angular bounds.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
The area enclosed by a polar curve r = f(θ) between angles α and β is given by the integral A = ∫ ½r² dθ from α to β, derived by summing the areas of infinitesimally thin triangular sectors each with area ½r²dθ. Choosing the correct bounds α and β is essential, often found from where the curve crosses the pole or completes one loop. For example, the area of one loop of r = cos(2θ) uses the bounds where r=0 at the start and end of that loop.
Worked example
Find the area enclosed by one loop of the polar curve r = 3 sin(2θ), for θ between 0 and π/2.
- 1Area formula: A = ∫ ½r² dθ from 0 to π/2, with r² = 9 sin²(2θ)
- 2Use the identity sin²(2θ) = (1−cos(4θ))/2, so ½r² = (9/4)(1−cos(4θ))
- 3Integrate: ∫(9/4)(1−cos(4θ))dθ = (9/4)[θ − sin(4θ)/4] from 0 to π/2
- 4Evaluate: (9/4)[(π/2 − 0) − (0 − 0)] = 9π/8
Answer: 9π/8
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Start practisingCommon mistake. Students use the bounds for a full revolution (0 to 2π) even when only one loop or petal of the curve is required, which double-counts or includes unwanted regions. Always identify the correct start and end angles for the specific region asked about.
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