Area enclosed by a polar curveEdexcel A-Level Further Maths: Revision notes

A thin sector of angle $\delta\theta$ and radius $r$ has area about $\frac12r^2\,\delta\theta$. Adding these gives the area enclosed by a polar curve between the half-lines $\theta=\alpha$ and $\theta=\beta$: $A=\frac12\int_\alpha^\beta r^2\,d\theta.$ Angles must be in radians, and $\beta>\alpha$. Square $r$ before integrating. Example: $r=2\theta$ for $0\le\theta\le\frac{\pi}{2}$: $A=\frac12\int_0^{\pi/2}4\theta^2\,d\theta=\left[\frac23\theta^3\right]_0^{\pi/2}=\frac{\pi^3}{12}$.

Terms like $\cos^2\theta$ and $\sin^2\theta$ need the double-angle identities: $\cos^2\theta=\frac12(1+\cos2\theta),\qquad\sin^2\theta=\frac12(1-\cos2\theta).$ Example: $r=3+2\cos\theta$. $r^2=9+12\cos\theta+4\cos^2\theta=11+12\cos\theta+2\cos2\theta$. Over a full turn $A=\frac12\left[11\theta+12\sin\theta+\sin2\theta\right]_0^{2\pi}=11\pi$. Remember to expand the bracket before replacing $\cos^2\theta$, including the cross term.

• The curve may start and end at the pole: solve $r=0$ for the limits, e.g. for $r=2(1+\cos\theta)$, $r=0$ at $\theta=\pi$.
• A whole closed curve such as $r=a(1+\cos\theta)$ uses $0$ to $2\pi$ (or double the area from $0$ to $\pi$, by symmetry in the initial line).
• A quadrant or sector needs the stated half-lines. Check the curve does not pass through the pole inside the interval.
• Use symmetry to halve the integration and then double the result.

For a region between two curves, both measured from the pole, subtract the two sector areas: $A=\frac12\int_\alpha^\beta\left(r_1^2-r_2^2\right)d\theta,\quad r_1\ge r_2\text{ on }[\alpha,\beta].$ Find the limits by solving $r_1=r_2$. Example: $r_1=4\cos\theta$ and $r_2=2$ meet where $\cos\theta=\frac12$, $\theta=\pm\frac{\pi}{3}$. The region inside $r_1$ and outside $r_2$ has area $\int_0^{\pi/3}(16\cos^2\theta-4)\,d\theta=\frac{4\pi}{3}+2\sqrt3$ (doubling by symmetry). Since $r_1$ is a circle of radius $2$, the area inside both is $4\pi-\left(\frac{4\pi}{3}+2\sqrt3\right)=\frac{8\pi}{3}-2\sqrt3$.

Convert to Cartesian components along the curve: $x=r\cos\theta$ and $y=r\sin\theta$ as functions of $\theta$.

• Tangent parallel to the initial line (horizontal): $\frac{dy}{d\theta}=0$, with $y=r\sin\theta$.
• Tangent perpendicular to the initial line (vertical): $\frac{dx}{d\theta}=0$, with $x=r\cos\theta$. Example: $r=2(1+\cos\theta)$, $y=2\sin\theta+\sin2\theta$. $\frac{dy}{d\theta}=2\cos\theta+2\cos2\theta=2(2\cos\theta-1)(\cos\theta+1)$, so $\cos\theta=\frac12$, $\theta=\frac{\pi}{3}$, $r=3$ and the greatest $y$ is $\frac{3\sqrt3}{2}$. Check that $\theta$ lies in the stated range and reject points at the pole where the tangent is the line $\theta=$ constant.