AS & A Level · Practice questions · Area in polar coordinates

Square, halve, integrate.

Six original Cambridge-style questions on area in polar coordinates: whole curves, petals and loops, regions cut off by a half-line, regions between two curves and a general result for rose curves.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch, choose the angles, then \(\tfrac{1}{2}\int r^{2}\,d\theta\).

Sketch the region and mark the half-lines that bound it.
Limits: \(0\) to \(2\pi\) for a whole curve traced once; consecutive zeros of \(r\) for a petal; solve \(r_{1} = r_{2}\) where two curves meet.
Square \(r\) and replace \(\cos^{2}\) and \(\sin^{2}\) with \(\tfrac{1}{2}(1 \pm \cos 2\theta)\).
Use symmetry to halve the work, then double.
Two curves: \(\tfrac{1}{2}\int (r_{1}^{2} - r_{2}^{2})\,d\theta\), outer minus inner, and give exact answers in \(\pi\) and surds.

Recall

[2 marks]

The curve \(r = 6\sin\theta\), for \(0 \le \theta \le \pi\), is a circle. Use integration to find the area it encloses.

\(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi} 36\sin^{2}\theta\,d\theta = 9\displaystyle\int_{0}^{\pi} (1 - \cos 2\theta)\,d\theta = 9\Big[\theta - \tfrac{1}{2}\sin 2\theta\Big]_{0}^{\pi}\). ✓
\(= 9\pi\) (a circle of radius 3, as expected). ✓

Calculation

[3 marks]

Find the exact area enclosed by the curve \(r = 2 + \cos\theta\), for \(0 \le \theta < 2\pi\).

\(r > 0\) for every \(\theta\), so the curve is traced once: \(A = \dfrac{1}{2}\displaystyle\int_{0}^{2\pi} (4 + 4\cos\theta + \cos^{2}\theta)\,d\theta\). ✓
\(= \dfrac{1}{2}\displaystyle\int_{0}^{2\pi} \big(\tfrac{9}{2} + 4\cos\theta + \tfrac{1}{2}\cos 2\theta\big)\,d\theta = \dfrac{1}{2}\Big[\tfrac{9}{2}\theta + 4\sin\theta + \tfrac{1}{4}\sin 2\theta\Big]_{0}^{2\pi}\). ✓
\(= \dfrac{1}{2} \times 9\pi = \dfrac{9\pi}{2}\). ✓

Calculation

[4 marks]

The curve \(r = 3\sin 2\theta\), where \(r \ge 0\), has a loop for \(0 \le \theta \le \dfrac{\pi}{2}\).

(a) Find the exact area of this loop.

(b) The full curve \(r = 3\sin 2\theta\) (allowing \(r\) to be negative) has four congruent petals. Write down the total area they enclose.

(a) \(r = 0\) at \(\theta = 0\) and \(\theta = \dfrac{\pi}{2}\), with \(r > 0\) in between. ✓
\(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/2} 9\sin^{2} 2\theta\,d\theta = \dfrac{9}{4}\displaystyle\int_{0}^{\pi/2} (1 - \cos 4\theta)\,d\theta = \dfrac{9}{4}\Big[\theta - \tfrac{1}{4}\sin 4\theta\Big]_{0}^{\pi/2}\). ✓
\(= \dfrac{9}{4} \times \dfrac{\pi}{2} = \dfrac{9\pi}{8}\). ✓
(b) \(4 \times \dfrac{9\pi}{8} = \dfrac{9\pi}{2}\). ✓

Analysis

[4 marks]

The circle \(C\) has polar equation \(r = 4\cos\theta\), for \(-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\).

(a) Find the exact area of the region bounded by \(C\) and the half-line \(\theta = \dfrac{\pi}{4}\), for \(\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{2}\).

(b) \(C\) has centre \((2, 0)\) and radius 2. Confirm your answer using the geometry of the circle.

(a) \(A = \dfrac{1}{2}\displaystyle\int_{\pi/4}^{\pi/2} 16\cos^{2}\theta\,d\theta = 4\displaystyle\int_{\pi/4}^{\pi/2} (1 + \cos 2\theta)\,d\theta = 4\Big[\theta + \tfrac{1}{2}\sin 2\theta\Big]_{\pi/4}^{\pi/2}\). ✓
\(= 4\Big[\Big(\dfrac{\pi}{2} + 0\Big) - \Big(\dfrac{\pi}{4} + \dfrac{1}{2}\Big)\Big] = 4\Big(\dfrac{\pi}{4} - \dfrac{1}{2}\Big) = \pi - 2\). ✓
(b) The half-line \(\theta = \dfrac{\pi}{4}\) meets \(C\) where \(r = 4\cos\dfrac{\pi}{4} = 2\sqrt{2}\), at the point \((2, 2)\). The region is the segment cut off by the chord from \(O\) to \((2, 2)\). ✓
With centre \(K(2, 0)\), the radii \(KO\) and \(K(2, 2)\) are perpendicular, so the segment is a quarter circle minus a right triangle: \(\dfrac{1}{4}\pi(2)^{2} - \dfrac{1}{2}(2)(2) = \pi - 2\), as found. ✓

Analysis

[5 marks]

The cardioid \(C_{1}\) has equation \(r = 2(1 + \cos\theta)\) and the circle \(C_{2}\) has equation \(r = 2\).

(a) Find the polar coordinates of the points where \(C_{1}\) and \(C_{2}\) meet.

(b) Find the exact area of the region inside \(C_{1}\) and outside \(C_{2}\).

(a) \(2(1 + \cos\theta) = 2\) gives \(\cos\theta = 0\): the points \(\Big(2, \dfrac{\pi}{2}\Big)\) and \(\Big(2, -\dfrac{\pi}{2}\Big)\). ✓
(b) For \(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\) the cardioid is outside the circle. By symmetry \(A = 2 \times \dfrac{1}{2}\displaystyle\int_{0}^{\pi/2} \big(4(1 + \cos\theta)^{2} - 4\big)\,d\theta = 4\displaystyle\int_{0}^{\pi/2} (2\cos\theta + \cos^{2}\theta)\,d\theta\). ✓
\(= 4\Big[2\sin\theta + \tfrac{1}{2}\theta + \tfrac{1}{4}\sin 2\theta\Big]_{0}^{\pi/2} = 4\Big(2 + \dfrac{\pi}{4}\Big) = 8 + \pi\). ✓
(c) The whole cardioid encloses \(\dfrac{3\pi(2)^{2}}{2} = 6\pi\). ✓
Inside both \(= 6\pi - (8 + \pi) = 5\pi - 8\). ✓

Proof

[5 marks]

Let \(a > 0\) and let \(n\) be a positive integer.

(a) Show that one petal of the rose \(r = a\sin n\theta\) has area \(\dfrac{\pi a^{2}}{4n}\).

(b) When \(n\) is odd the rose has \(n\) petals; when \(n\) is even it has \(2n\) petals. Deduce the total area enclosed in each case.

(a) \(r = 0\) when \(\sin n\theta = 0\); the first petal runs from \(\theta = 0\) to \(\theta = \dfrac{\pi}{n}\), with \(r > 0\) in between. ✓
\(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/n} a^{2}\sin^{2} n\theta\,d\theta = \dfrac{a^{2}}{4}\displaystyle\int_{0}^{\pi/n} (1 - \cos 2n\theta)\,d\theta = \dfrac{a^{2}}{4}\Big[\theta - \dfrac{\sin 2n\theta}{2n}\Big]_{0}^{\pi/n}\). ✓
\(\sin 2\pi = 0\), so \(A = \dfrac{a^{2}}{4} \times \dfrac{\pi}{n} = \dfrac{\pi a^{2}}{4n}\). ✓
(b) \(n\) odd: \(n \times \dfrac{\pi a^{2}}{4n} = \dfrac{\pi a^{2}}{4}\). ✓
\(n\) even: \(2n \times \dfrac{\pi a^{2}}{4n} = \dfrac{\pi a^{2}}{2}\). Either way the total does not depend on \(n\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.