Home / A-Level Further Maths 9231 / Polar coordinates / Area in polar coordinates
Cambridge International AS & A Level · Topic 1.5
A-Level Further Maths 9231 / Topic 1.5 / AS & A Level

Sweep out an area, one thin sector at a time.

A region bounded by a polar curve and two rays from the pole is a fan of very thin sectors. Each one has area about \(\tfrac{1}{2}r^{2}\,\delta\theta\), and adding them up gives one formula, \(A = \tfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^{2}\,d\theta\), that finds the area of a cardioid, a petal of a rose or the region between two curves.

Topic 1.5All polar coordinates lessons →
The key idea

The area of the region bounded by the curve \(r = f(\theta)\) and the half-lines \(\theta = \alpha\) and \(\theta = \beta\) is \(A = \tfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^{2}\,d\theta\), with \(\alpha < \beta\). It comes from adding thin sectors of area \(\tfrac{1}{2}r^{2}\,\delta\theta\). The limits are the angles that bound the region: \(0\) to \(2\pi\) for a whole closed curve traced once, and two consecutive zeros of \(r\) for one petal or loop. Square \(r\) first, then turn \(\cos^{2}\) and \(\sin^{2}\) into \(\tfrac{1}{2}(1 \pm \cos 2\theta)\) before integrating. For the region between two curves, find the angles where they meet and integrate \(\tfrac{1}{2}(r_{1}^{2} - r_{2}^{2})\).

initial line θ = α θ = β r δθ thin sector area ≈ ½r²δθ r = 2(1 + cos θ) shaded: α ≤ θ ≤ β O
Fig. 1: the cardioid r = 2(1 + cos θ). The shaded region lies between the half-lines θ = α and θ = β and the curve. It is a fan of thin sectors; the one drawn has radius r and angle δθ, so its area is about ½r²δθ
Section 01

Thin sectors add up to an integral.

A sector of a circle with radius \(r\) and angle \(\theta\) radians has area \(\tfrac{1}{2}r^{2}\theta\). A region bounded by a polar curve is not a sector, because \(r\) changes as \(\theta\) changes. But cut it into very thin slices between the rays \(\theta\) and \(\theta + \delta\theta\): over such a small angle \(r\) hardly changes, so each slice is almost a sector of area \(\tfrac{1}{2}r^{2}\,\delta\theta\) (Fig. 1).

Adding the slices and letting \(\delta\theta \to 0\) turns the sum into an integral:

\[A = \lim_{\delta\theta \to 0} \sum \tfrac{1}{2}r^{2}\,\delta\theta = \frac{1}{2}\int_{\alpha}^{\beta} r^{2}\,d\theta.\]
Worked example A spiral
Find the area of the region bounded by the spiral \(r = 2\theta\), for \(0 \le \theta \le \pi\), and the initial line.
  1. The spiral starts at the pole (\(\theta = 0\)) and ends on the half-line \(\theta = \pi\); the region is swept as \(\theta\) goes from \(0\) to \(\pi\).
  2. \(r^{2} = 4\theta^{2}\), so \(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi} 4\theta^{2}\,d\theta = 2\Big[\dfrac{\theta^{3}}{3}\Big]_{0}^{\pi}\).
  3. \(= \dfrac{2\pi^{3}}{3}\).
Answer: \(\dfrac{2\pi^{3}}{3} \approx 20.7\)
Worked example The formula agrees with a triangle
The curve \(r = \sec\theta\) for \(0 \le \theta \le \dfrac{\pi}{4}\) is part of the straight line \(x = 1\). Use the area formula to find the area of the region bounded by this curve, the initial line and the half-line \(\theta = \dfrac{\pi}{4}\), and check your answer.
  1. \(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/4} \sec^{2}\theta\,d\theta = \dfrac{1}{2}\Big[\tan\theta\Big]_{0}^{\pi/4}\).
  2. \(= \dfrac{1}{2}(1 - 0) = \dfrac{1}{2}\).
  3. Check: \(r\cos\theta = 1\) is the line \(x = 1\), and \(\theta = \dfrac{\pi}{4}\) is \(y = x\). The region is the triangle with vertices \(O\), \((1, 0)\) and \((1, 1)\), of area \(\tfrac{1}{2} \times 1 \times 1 = \tfrac{1}{2}\).
Answer: \(\dfrac{1}{2}\), the area of the triangle \(O(0, 0)\), \((1, 0)\), \((1, 1)\)
Your turn
Use the formula to find the area of the region bounded by the circle \(r = 3\) and the half-lines \(\theta = 0\) and \(\theta = \dfrac{2\pi}{3}\).
Examiner trap

Two marks vanish with the formula half-remembered. It is \(\tfrac{1}{2}\int r^{2}\,d\theta\): not \(\int r\,d\theta\) (no square) and not \(\int r^{2}\,d\theta\) (no half). And the limits are values of \(\theta\), never values of \(r\), \(x\) or \(y\).

Section 02

Squares of cosines, cardioids and petals.

Most polar curves in this course contain \(\cos\theta\), \(\sin\theta\) or \(\cos n\theta\), so \(r^{2}\) contains \(\cos^{2}\) or \(\sin^{2}\). You cannot integrate those directly; use the double angle identities from Pure Mathematics 3 first.

\[\cos^{2}\theta = \tfrac{1}{2}(1 + \cos 2\theta), \qquad \sin^{2}\theta = \tfrac{1}{2}(1 - \cos 2\theta),\] \[\cos^{2} n\theta = \tfrac{1}{2}(1 + \cos 2n\theta), \qquad \sin^{2} n\theta = \tfrac{1}{2}(1 - \cos 2n\theta).\]
Worked example The whole cardioid
Show that the area enclosed by the cardioid \(r = a(1 + \cos\theta)\), where \(a > 0\), is \(\dfrac{3\pi a^{2}}{2}\).
  1. The curve is symmetric about the initial line, so \(A = 2 \times \dfrac{1}{2}\displaystyle\int_{0}^{\pi} a^{2}(1 + \cos\theta)^{2}\,d\theta = a^{2}\displaystyle\int_{0}^{\pi} (1 + 2\cos\theta + \cos^{2}\theta)\,d\theta\).
  2. Replace \(\cos^{2}\theta\): the integrand is \(1 + 2\cos\theta + \tfrac{1}{2} + \tfrac{1}{2}\cos 2\theta = \tfrac{3}{2} + 2\cos\theta + \tfrac{1}{2}\cos 2\theta\).
  3. \(A = a^{2}\Big[\tfrac{3}{2}\theta + 2\sin\theta + \tfrac{1}{4}\sin 2\theta\Big]_{0}^{\pi} = a^{2}\Big(\dfrac{3\pi}{2} + 0 + 0\Big)\).
  4. So \(A = \dfrac{3\pi a^{2}}{2}\). For the curve in Fig. 1, \(a = 2\), the area is \(6\pi\).
Answer: \(A = \dfrac{3\pi a^{2}}{2}\)
Worked example One petal of a rose
Find the area of one petal of the rose \(r = 2\sin 3\theta\).
  1. \(r = 0\) when \(\sin 3\theta = 0\), so at \(\theta = 0\) and \(\theta = \dfrac{\pi}{3}\), and \(r > 0\) in between: that is one petal.
  2. \(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/3} 4\sin^{2} 3\theta\,d\theta = \displaystyle\int_{0}^{\pi/3} (1 - \cos 6\theta)\,d\theta\).
  3. \(= \Big[\theta - \tfrac{1}{6}\sin 6\theta\Big]_{0}^{\pi/3} = \dfrac{\pi}{3} - \tfrac{1}{6}\sin 2\pi = \dfrac{\pi}{3}\).
Answer: \(\dfrac{\pi}{3}\) (so the three petals together enclose \(\pi\))
Your turn
The curve \(r = 4\cos\theta\), for \(-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\), is a circle. Find the area it encloses by integration, and check it.
Examiner trap

Do not integrate over more angle than the curve needs. The rose \(r = \sin 3\theta\) is traced completely for \(0 \le \theta \le \pi\); integrating \(\tfrac{1}{2}\sin^{2} 3\theta\) from \(0\) to \(2\pi\) gives \(\dfrac{\pi}{2}\), twice the true total \(\dfrac{\pi}{4}\), because every petal is counted twice. Also, \(\cos^{2}\theta = \tfrac{1}{2}(1 + \cos 2\theta)\), not \(\tfrac{1}{2}(1 + \cos\theta)\).

Stage 1 · Learn

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The area bounded by \(r = f(\theta)\) and the half-lines \(\theta = \alpha\), \(\theta = \beta\) (with \(\alpha < \beta\)) is

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Which identity lets you integrate \(\cos^{2}\theta\)?

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Find the area of the region bounded by the circle \(r = 2\) and the half-lines \(\theta = 0\) and \(\theta = \dfrac{\pi}{2}\).

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Which limits give the area of one petal of the rose \(r = \cos 2\theta\)?

Section 03

Between a curve and a line, and between two curves.

Exam regions are often only part of a curve. Draw a sketch, mark the half-lines that bound the region, and read the limits from it. When the region lies between two curves, first find the angles where the curves meet; those are the limits.

Worked example A cardioid and a half-line
Find the exact area of the region bounded by the curve \(r = 2(1 + \cos\theta)\), the initial line and the half-line \(\theta = \dfrac{\pi}{2}\).
  1. The region is swept from \(\theta = 0\) to \(\theta = \dfrac{\pi}{2}\): \(A = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/2} 4(1 + \cos\theta)^{2}\,d\theta = 2\displaystyle\int_{0}^{\pi/2} \big(\tfrac{3}{2} + 2\cos\theta + \tfrac{1}{2}\cos 2\theta\big)\,d\theta\).
  2. \(= 2\Big[\tfrac{3}{2}\theta + 2\sin\theta + \tfrac{1}{4}\sin 2\theta\Big]_{0}^{\pi/2} = 2\Big(\dfrac{3\pi}{4} + 2 + 0\Big)\).
  3. \(= \dfrac{3\pi}{2} + 4\). (The \(\sin\theta\) term no longer vanishes, so the answer is not simply a quarter of \(6\pi\).)
Answer: \(\dfrac{3\pi}{2} + 4 \approx 8.71\)
Worked example Inside a circle, outside a cardioid
Find the area of the region inside the circle \(r = 3\cos\theta\) and outside the cardioid \(r = 1 + \cos\theta\).
  1. Meet: \(3\cos\theta = 1 + \cos\theta\), so \(\cos\theta = \tfrac{1}{2}\) and \(\theta = \pm\dfrac{\pi}{3}\). Between these the circle is outside the cardioid (at \(\theta = 0\): \(3 > 2\)).
  2. By symmetry about the initial line, \(A = 2 \times \dfrac{1}{2}\displaystyle\int_{0}^{\pi/3} \big(9\cos^{2}\theta - (1 + \cos\theta)^{2}\big)\,d\theta = \displaystyle\int_{0}^{\pi/3} (8\cos^{2}\theta - 2\cos\theta - 1)\,d\theta\).
  3. \(8\cos^{2}\theta = 4 + 4\cos 2\theta\), so the integrand is \(3 + 4\cos 2\theta - 2\cos\theta\).
  4. \(A = \Big[3\theta + 2\sin 2\theta - 2\sin\theta\Big]_{0}^{\pi/3} = \pi + 2 \cdot \dfrac{\sqrt{3}}{2} - 2 \cdot \dfrac{\sqrt{3}}{2} = \pi\).
Answer: \(\pi\)
Your turn
Find the exact area of the region inside the circle \(r = 2\sin\theta\) and outside the circle \(r = 1\).
Examiner trap

Subtract the squares, not the radii: the area between two curves is \(\tfrac{1}{2}\int (r_{1}^{2} - r_{2}^{2})\,d\theta\), not \(\tfrac{1}{2}\int (r_{1} - r_{2})^{2}\,d\theta\). The thin slice between the curves is a sector minus a sector, so its area is \(\tfrac{1}{2}r_{1}^{2}\,\delta\theta - \tfrac{1}{2}r_{2}^{2}\,\delta\theta\).

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Find the area enclosed by the curve \(r = 3(1 + \cos\theta)\).

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Find the area of one petal of the curve \(r = 4\cos 3\theta\).

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Find the area of the region bounded by the curve \(r = 2\theta\), for \(0 \le \theta \le \dfrac{\pi}{2}\), and the half-line \(\theta = \dfrac{\pi}{2}\).

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Find the exact area of the region inside the circle \(r = 4\sin\theta\) and outside the circle \(r = 2\).

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Areas swept out, one sector at a time.

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Six original Cambridge-style questions
A circle through the pole, a limaçon, the petals of a four-leaved rose, a segment checked by geometry, a cardioid against a circle, and a proof for every rose.