Recipe

Find the area enclosed by a polar curve

Use when: a region is bounded by a single polar curve r=f(theta) (like one loop of a rose, or the whole interior of a cardioid)

Goal: the exact area swept out, using the sector-area formula integrated over the correct sweep of theta

Taught in Area Enclosed by a Polar Curve · a later section extends this recipe with more cases

1. 1

   Determine the two theta-values where the region's boundary begins and ends, typically where r=0 (for one loop or petal) or a full period (for an entire curve).

   Example

   For one petal of r=3\sin(2\theta): r=0 when \sin(2\theta)=0, i.e. 2\theta=0,\pi, so \theta=0 and \theta=\dfrac{\pi}{2} trace out one petal.

   Why: an incorrect sweep interval either misses part of the region or double-counts another petal entirely.

2. 2

   Apply A=\dfrac12\displaystyle\int_\alpha^\beta [r(\theta)]^2\,d\theta, using the limits from Step 1.

   Example

   A = \frac12\int_0^{\pi/2}\big[3\sin(2\theta)\big]^2\,d\theta = \frac92\int_0^{\pi/2}\sin^2(2\theta)\,d\theta

   Why: the factor of \tfrac12 and the squared r both come directly from the geometric sector-area formula, applied to an infinitesimally thin wedge.

3. 3

   Rewrite \sin^2 or \cos^2 using the power-reduction identity so the integral has an elementary antiderivative.

   Example

   \sin^2(2\theta)=\dfrac12-\dfrac12\cos(4\theta), so A = \frac92\int_0^{\pi/2}\left[\frac12-\frac12\cos(4\theta)\right]d\theta

   Why: r^2 almost always contains a squared trig function, and the sector-area formula's squared r makes this identity necessary far more often than in an ordinary rectangular-coordinates area problem.

4. 4

   Antidifferentiate and evaluate at the sweep limits from Step 1.

   Example

   \frac92\left[\frac\theta2-\frac{\sin(4\theta)}{8}\right]_0^{\pi/2} = \frac92\left[\left(\frac\pi4-0\right)-0\right] = \frac{9\pi}{8}
   • the region is one petal or loop out of several identical ones — multiply this result by the total number of identical petals if the whole curve's area is wanted, instead of just one petal's.
   • the region is the curve's entire interior in one sweep (like a full cardioid, theta from 0 to 2\pi) — the integral above already gives the whole area, with no multiplication needed.

   Why: reporting only one petal's area when the whole rose was asked for (or vice versa) is a common way to be off by an exact integer factor.

5. 5

   State the final numeric or exact-form area.

   Example

   \boxed{A = \frac{9\pi}{8}}
   • Not yet: the area shared by, or between, TWO overlapping polar curves is needed instead of just one curve's interior — covered in Area Between Polar Curves and Finding Intersections

   Why: this number is the actual answer to how much area this curve encloses, the quantity the whole setup was built to find.