Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Choose a once-traced polar region and accumulate sectors with the one-half squared-radius factor.

College Board Topic 9.8 · CED p. 173 (PDF p. 178) · CED effective Fall 2020; project course outline 2026–27.

A circular sector of radius r and angle Δθ has area $r^{2} Δ θ / 2$ . For a polar boundary $r = f (θ)$ , adding thin sectors gives $A = \frac{1}{2} {\int_{α}}^{β} (f (θ))^{2} d θ$ . This is a radial area, not a rectangle integral. Squaring the radius gives square units and makes the factor 1/2 essential.

Choose angles that sweep the intended region exactly once. Zeros of r often mark petals or loops; symmetry can reduce the work, but explain what region you multiply. For $r = 2 \sin (3 θ)$ , the interval $0 \leq θ \leq π / 3$ traces one petal because r is nonnegative between consecutive zeros. Integrating from 0 to 2π traces every petal twice; blindly using that interval overcounts.

Negative radius does not make negative area, but it sends points to the opposite ray. Make a trace table or sketch to justify bounds. A circle $r = 2 \cos θ$ is traced once on [ $- π / 2, π / 2$ ]; extending to 2π retraces it. Name the actual region before evaluating.

For an area swept over a specified sector with nonnegative r, the formula works directly. For a closed curve with loops, break the region into separately traced pieces so overlapping sweeps do not masquerade as one area.

Polar area accumulates sectors, so its differential is one half r² dθ.

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For one petal of $r = 2 \sin (3 θ)$ , choose $0 \leq θ \leq π / 3$ . Why: Consecutive radius zeros bound one nonnegative petal.
Set $A = \frac{1}{2} {\int_{0}}^{π / 3} 4 \sin^{2} (3 θ) d θ$ . Why: Each thin sector has area r²dθ/2.
Using $\sin^{2} u = (1 - \cos (2 u)) / 2$ , obtain $A = π / 3$ . Why: The cosine term integrates to zero over this full petal.

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Question 1

What integral gives the full disk inside

r = 3

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Question 1: prompt and worked reasoning

What integral gives the full disk inside

r = 3

\frac{1}{2} {\int_{0}}^{2 π} 9 d θ = 9 π

A full turn sweeps the disk once.

Question 2: prompt and worked reasoning

Why check the tracing interval before computing polar area?

A longer interval may trace a loop or petal repeatedly.

The integral counts the sweeps specified by the angle interval.

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Retrieval 1

What integral gives the full disk inside

r = 3

Worked answer and reason

\frac{1}{2} {\int_{0}}^{2 π} 9 d θ = 9 π

A full turn sweeps the disk once.

Retrieval 2

Why check the tracing interval before computing polar area?

Worked answer and reason

A longer interval may trace a loop or petal repeatedly.

The integral counts the sweeps specified by the angle interval.

Catch the tempting mistake

Any full turn gives the area inside a polar curve once.: Some curves retrace or overlap during a full turn. Bounds must correspond to a single sweep of the named region.

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Unit 9 problem 23 · No calculator · analytic
Unit 9 problem 24 · No calculator · multiple
Unit 9 problem 25 · No calculator · analytic
Unit 9 problem 26 · No calculator · analytic

For the teacher

Ask learners to highlight the region and mark consecutive radius zeros before integration. Make symmetry multipliers explicit.

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Glossary and deeper reading · 1 refreshers

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Integral
Integration can mean accumulating a function over an interval or finding an antiderivative. A definite integral gives a number, while an indefinite integral represents a family of antiderivatives. For example, integrating 2x indefinitely gives x²+C, and integrating it from 0 to 2 gives 4. The Fundamental Theorem of Calculus connects these two operations under its hypotheses.
Read more: 6.7 · Integral →

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