Further Maths/Lessons/Polar Coordinates/Area & Tangents
Core Pure · Polar Coordinates · Year 2

Area Enclosed & Tangents

The sector-area formula for polar regions, and how to locate tangents parallel or perpendicular to the initial line.

14 min read 3 worked examples

Area enclosed by a polar curve

Theorem 5.3 — sector area formula

The area enclosed by the curve $r=f(\theta)$ between the half-lines $\theta=\alpha$ and $\theta=\beta$ (measured from the pole) is

$$A=\frac12\int_\alpha^\beta r^2\,d\theta$$

ProofDerivation from elementary sectors
1
Divide the angle range $[\alpha,\beta]$ into $n$ thin strips of width $\delta\theta=\dfrac{\beta-\alpha}n$. Each strip, bounded by the curve and two nearby radii, is approximately a circular sector of radius $r$ and angle $\delta\theta$.
2
The area of a circular sector of radius $r$ and angle $\delta\theta$ is $\dfrac12r^2\delta\theta$ (a fraction $\dfrac{\delta\theta}{2\pi}$ of the full circle's area $\pi r^2$).
3
Summing over all $n$ strips gives an approximation $A\approx\displaystyle\sum\frac12r^2\,\delta\theta$.
4
As $n\to\infty$ (so $\delta\theta\to0$), the sum becomes exact and converges to the integral $A=\dfrac12\displaystyle\int_\alpha^\beta r^2\,d\theta$.
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Example 1 · Area of a single loop
Find the area enclosed by the cardioid $r=4(1+\cos\theta)$.

The full curve is traced as $\theta$ runs from $-\pi$ to $\pi$, so

$$A=\frac12\int_{-\pi}^\pi 16(1+\cos\theta)^2\,d\theta=8\int_{-\pi}^\pi(1+2\cos\theta+\cos^2\theta)\,d\theta$$

Using $\cos^2\theta=\dfrac12(1+\cos2\theta)$:

$$A=8\int_{-\pi}^\pi\left(\frac32+2\cos\theta+\frac12\cos2\theta\right)d\theta=8\left[\frac32\theta+2\sin\theta+\frac14\sin2\theta\right]_{-\pi}^\pi$$

$$A=8\left(\frac{3\pi}2-\left(-\frac{3\pi}2\right)\right)=8\times3\pi=24\pi$$

Example 2 · Area between two curves
The curves $C_1:r=6$ and $C_2:r=6(1-\cos\theta)$ intersect where $\theta=\pm\dfrac\pi2$. Find the area of the region that lies inside both curves.

By symmetry about the initial line, consider $0\leq\theta\leq\dfrac\pi2$ and double.

For $0\leq\theta\leq\dfrac\pi2$, $C_2$ is the inner boundary (check at $\theta=0$: $C_2$ gives $r=0$, well inside $C_1$'s $r=6$), so the region inside both is bounded by $C_2$ here.

For $\dfrac\pi2\leq\theta\leq\pi$, $C_1$ becomes the inner boundary (at $\theta=\pi$, $C_2$ gives $r=12>6=C_1$).

By the symmetry of the whole figure about the initial line:

$$A=2\left[\frac12\int_0^{\pi/2}36(1-\cos\theta)^2\,d\theta+\frac12\int_{\pi/2}^\pi 36\,d\theta\right]$$

The second integral: $\dfrac12\times36\times\dfrac\pi2=9\pi$.

The first: $18\displaystyle\int_0^{\pi/2}(1-2\cos\theta+\cos^2\theta)\,d\theta=18\int_0^{\pi/2}\left(\frac32-2\cos\theta+\frac12\cos2\theta\right)d\theta$

$$=18\left[\frac32\theta-2\sin\theta+\frac14\sin2\theta\right]_0^{\pi/2}=18\left(\frac{3\pi}4-2\right)=\frac{27\pi}2-36$$

So $A=2\left(\dfrac{27\pi}2-36+9\pi\right)=2\left(\dfrac{45\pi}2-36\right)=45\pi-72$.

Tangents to polar curves

Method — tangents parallel or perpendicular to the initial line

Convert to Cartesian parametric form using $\theta$ as the parameter: $x=r\cos\theta=f(\theta)\cos\theta$, $y=r\sin\theta=f(\theta)\sin\theta$.

· The tangent is parallel to the initial line (horizontal) where $\dfrac{dy}{d\theta}=0$.
· The tangent is perpendicular to the initial line (vertical) where $\dfrac{dx}{d\theta}=0$.

Example 3 · Tangents parallel to the initial line
Find the value of $\theta$, with $0<\theta<\dfrac\pi2$, at which the tangent to $r=5(1+\cos\theta)$ is parallel to the initial line, and find the corresponding value of $r$.

$y=r\sin\theta=5(1+\cos\theta)\sin\theta=5\sin\theta+5\sin\theta\cos\theta=5\sin\theta+\dfrac52\sin2\theta$.

$$\frac{dy}{d\theta}=5\cos\theta+5\cos2\theta=5\cos\theta+5(2\cos^2\theta-1)=10\cos^2\theta+5\cos\theta-5$$

Set to zero: $2\cos^2\theta+\cos\theta-1=0$, so $(2\cos\theta-1)(\cos\theta+1)=0$, giving $\cos\theta=\dfrac12$ or $\cos\theta=-1$.

Since $0<\theta<\dfrac\pi2$, we need $\cos\theta=\dfrac12$, so $\theta=\dfrac\pi3$.

Then $r=5\left(1+\dfrac12\right)=\dfrac{15}2$.

ProofThe dimple condition for r = a(p + q cosθ)
1
To justify the classification from the previous lesson, consider $r=a(p+q\cos\theta)$ with $p,q\gt0$ and examine where $\dfrac{dx}{d\theta}=0$ for $0\lt\theta\lt\pi$.
$x=r\cos\theta=a(p+q\cos\theta)\cos\theta=ap\cos\theta+aq\cos^2\theta$.
2
$\dfrac{dx}{d\theta}=-ap\sin\theta-2aq\cos\theta\sin\theta=-a\sin\theta(p+2q\cos\theta)$.
3
For $0\lt\theta\lt\pi$, $\sin\theta\neq0$, so stationary points of $x$ occur where $p+2q\cos\theta=0$, i.e. $\cos\theta=-\dfrac p{2q}$.
4
This equation has a solution with $-1\lt\cos\theta\lt1$ precisely when $-1\lt-\dfrac p{2q}\lt1$, i.e. when $p\lt2q$ (since $p,q\gt0$ already gives $-\dfrac p{2q}\lt0\lt1$). When such a $\theta$ exists strictly between $0$ and $\pi$, the boundary curve doubles back on itself in the $x$-direction there, creating the concave "dimple". When $p\geq2q$, no such interior stationary point exists and the curve remains convex.
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Exam-style questions

Question 1 6 marks

Find the area enclosed by the curve $r=3\sin2\theta$, for $0\leq\theta\leq\dfrac\pi2$ (one petal).

$$A=\frac12\int_0^{\pi/2}9\sin^22\theta\,d\theta=\frac92\int_0^{\pi/2}\frac{1-\cos4\theta}2\,d\theta$$ (M1, A1, M1)

$$=\frac94\left[\theta-\frac14\sin4\theta\right]_0^{\pi/2}=\frac94\times\frac\pi2=\frac{9\pi}8$$ (A1, A1, A1)

Question 2 7 marks

The curve $C$ has polar equation $r=2+\cos\theta$. Find the value of $\theta$, with $0<\theta<\pi$, at which the tangent to $C$ is perpendicular to the initial line, giving your answer to 3 significant figures if it is not exact.

$x=r\cos\theta=(2+\cos\theta)\cos\theta=2\cos\theta+\cos^2\theta$. (M1, A1)

$$\frac{dx}{d\theta}=-2\sin\theta-2\cos\theta\sin\theta=-2\sin\theta(1+\cos\theta)$$ (M1, A1)

Set to zero: $\sin\theta=0$ or $\cos\theta=-1$. For $0<\theta<\pi$, $\sin\theta\neq0$, and $\cos\theta=-1$ gives $\theta=\pi$, which is outside the open interval. (M1)

So there is no solution with $0<\theta<\pi$ — the tangent is never perpendicular to the initial line on this open interval (the curve is convex throughout, consistent with $p=2\geq2q=2$ in the dimple condition). (A1, A1)

Question 3 5 marks

Find the area of the region enclosed by one petal of the curve $r=6\cos3\theta$.

One petal runs from $\theta=-\dfrac\pi6$ to $\theta=\dfrac\pi6$. (B1)

$$A=\frac12\int_{-\pi/6}^{\pi/6}36\cos^23\theta\,d\theta=18\int_{-\pi/6}^{\pi/6}\frac{1+\cos6\theta}2\,d\theta$$ (M1, A1)

$$=9\left[\theta+\frac16\sin6\theta\right]_{-\pi/6}^{\pi/6}=9\left(\frac\pi6+\frac\pi6\right)=3\pi$$ (A1, A1)

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