Polar coordinates – further maths revision notes to download as PDF

Example 1

Convert \(r=4\cos\theta\) to Cartesian form.

\(r^2=4r\cos\theta\Rightarrow x^2+y^2=4x\Rightarrow(x-2)^2+y^2=4\), a circle of radius 2 centred at \((2,0)\). Similarly the line \(x=3\) is \(r=3\sec\theta\), and \(y=\sqrt3x\) is \(\theta=\frac\pi3\).

Example 2: cardioid

Find the area of \(r=a(1+\cos\theta)\).

\(A=\tfrac12a^2\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)d\theta=\tfrac12a^2(2\pi+0+\pi)=\tfrac32\pi a^2\), using \(\cos^2\theta=\tfrac12(1+\cos2\theta)\).

Example 3: one petal

Find the area of one petal of \(r=\sin3\theta\).

One petal is \(0\le\theta\le\frac\pi3\). \(A=\tfrac12\int_0^{\pi/3}\sin^23\theta\,d\theta=\tfrac12\cdot\tfrac\pi6=\tfrac\pi{12}\).

Example 4: limaçon

Describe and find the area of \(r=2+\cos\theta\).

The curve never reaches the pole: \(r\) varies from 1 (at \(\theta=\pi\)) to 3 (at \(\theta=0\)), and it is symmetric about the initial line. \(A=\tfrac12\int_0^{2\pi}(2+\cos\theta)^2d\theta=\tfrac{9\pi}2\).

Example 5

Find the greatest height of the cardioid \(r=a(1+\cos\theta)\) above the initial line.

\(y=a(1+\cos\theta)\sin\theta\) and \(\frac{dy}{d\theta}=a(\cos\theta+\cos2\theta)=a(2\cos\theta-1)(\cos\theta+1)=0\). So \(\theta=\frac\pi3\), giving \(y=\frac{3\sqrt3}4a\).

Worked example 6: a straight line in polar form

Find the polar equation of the line \(y=2\) and the area of the region bounded by the line and the half-lines \(\theta=\tfrac\pi4\) and \(\theta=\tfrac\pi2\).

Since \(y=r\sin\theta\), the line is \(r\sin\theta=2\), so \(r=2\operatorname{cosec}\theta\). The area is \(\tfrac12\int_{\pi/4}^{\pi/2}4\operatorname{cosec}^2\theta\,d\theta=2\left[-\cot\theta\right]_{\pi/4}^{\pi/2}=2(0+1)=2\). As a check, the region is the triangle with vertices at the pole, \((2,2)\) and \((0,2)\), which has area \(\tfrac12\times2\times2=2\). This shows that the polar area formula works for any region bounded by two half-lines from the pole, even when the boundary is straight.

Worked example 7: inside a cardioid and outside a circle

Find the area of the region inside \(r=2(1+\cos\theta)\) and outside \(r=3\).

The curves meet when \(2+2\cos\theta=3\), so \(\cos\theta=\tfrac12\) and \(\theta=\pm\tfrac\pi3\). Between these angles the cardioid is outside the circle. Using symmetry, the area is \(2\times\tfrac12\int_0^{\pi/3}\left(4(1+\cos\theta)^2-9\right)d\theta=\int_0^{\pi/3}(-3+8\cos\theta+2\cos2\theta)\,d\theta\). This equals \(\left[-3\theta+8\sin\theta+\sin2\theta\right]_0^{\pi/3}=-\pi+4\sqrt3+\tfrac{\sqrt3}2=\tfrac92\sqrt3-\pi\), which is about 4.65 square units. Notice that we squared each radius before subtracting: the expression \((2+2\cos\theta-3)^2\) would give a wrong answer.

Worked example 8: a tangent perpendicular to the initial line

Find the points on \(r=2+\cos\theta\) where the tangent is perpendicular to the initial line.

We need \(\dfrac{d}{d\theta}(r\cos\theta)=0\). Here \(x=2\cos\theta+\cos^2\theta\), so \(\dfrac{dx}{d\theta}=-2\sin\theta-2\sin\theta\cos\theta=-2\sin\theta(1+\cos\theta)\). This is zero when \(\sin\theta=0\) or \(\cos\theta=-1\), that is at \(\theta=0\) and \(\theta=\pi\). The points are \((3,0)\) and \((1,\pi)\), the two ends of the curve on the initial line. Since \(2\ge2\times1\), this limaçon is convex and has no dimple, so there are no other vertical tangents.

Worked example 9: sketching a rose curve

Sketch the curve \(r=3\cos2\theta\) for \(r\ge0\) and find the total area that it encloses.

The curve exists only where \(\cos2\theta\ge0\), that is for \(-\tfrac\pi4\le\theta\le\tfrac\pi4\) and \(\tfrac{3\pi}4\le\theta\le\tfrac{5\pi}4\). In each of these intervals \(r\) rises from 0 to a maximum of 3 and falls back to 0, so the curve consists of two petals, one pointing along the initial line and one pointing in the opposite direction. A table of values at intervals of \(\tfrac\pi{12}\) helps to draw each petal smoothly. The area of one petal is \(\tfrac12\int_{-\pi/4}^{\pi/4}9\cos^22\theta\,d\theta=\tfrac94\int_{-\pi/4}^{\pi/4}(1+\cos4\theta)\,d\theta=\tfrac94\times\tfrac\pi2=\tfrac{9\pi}8\). The total area is therefore \(\tfrac{9\pi}4\). If negative values of \(r\) were allowed, two further petals would appear along the half-lines \(\theta=\pm\tfrac\pi2\), and the total area would double.