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

A thin sector of a circle of radius $r$ and small angle $\delta\theta$ (in radians) has area $\frac12r^2\delta\theta$. Adding up thin sectors from $\theta=\alpha$ to $\theta=\beta$ and letting $\delta\theta\to0$ gives the area enclosed by a polar curve and the two half-lines $\theta=\alpha$ and $\theta=\beta$: $A=\frac12\int_{\alpha}^{\beta}r^2\,d\theta.$ The angle must be in radians. The integrand is $r^2$, so the sign of $r$ does not matter. Example: for $r=2\theta$ with $0\le\theta\le\pi$, $A=\frac12\int_0^{\pi}4\theta^2\,d\theta=\left[\frac{2\theta^3}{3}\right]_0^{\pi}=\frac{2\pi^3}{3}$.

The limits $\alpha$ and $\beta$ must cover the region once only. Find where $r=0$ (the curve passes through the pole) to see where a loop starts and ends.

• Loop of $r=a\sin2\theta$: from $\theta=0$ to $\theta=\frac{\pi}{2}$ ($r=0$ at both ends).
• Cardioid $r=a(1+\cos\theta)$: one full turn, $0$ to $2\pi$ (or $-\pi$ to $\pi$).
• Circle $r=a\cos\theta$: $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Going from $0$ to $2\pi$ would trace the circle twice and double the area. If the curve is symmetrical you may integrate over half the range and double the result.

Expand $r^2$ first. For trigonometric curves you almost always need the double-angle identities: $\cos^2\theta=\frac12(1+\cos2\theta),\qquad\sin^2\theta=\frac12(1-\cos2\theta).$ Worked example, $r=3+2\cos\theta$ for $0\le\theta\le2\pi$: $A=\frac12\int_0^{2\pi}(9+12\cos\theta+4\cos^2\theta)\,d\theta=\frac12\int_0^{2\pi}(11+12\cos\theta+2\cos2\theta)\,d\theta=\frac12(22\pi)=11\pi.$ For $r=4\sin2\theta$ the identity becomes $\sin^22\theta=\frac12(1-\cos4\theta)$, so the integral involves $\sin4\theta$.

To find the area between two polar curves $r_1$ (outer) and $r_2$ (inner), subtract the areas of the sectors: $A=\frac12\int_{\alpha}^{\beta}\left(r_1^2-r_2^2\right)d\theta.$ Find $\alpha$ and $\beta$ by equating $r_1=r_2$ to get the points of intersection, then check which curve is further from the pole between them. Example: $r=3\cos\theta$ and $r=1+\cos\theta$ meet where $\cos\theta=\frac12$, so $\theta=\pm\frac{\pi}{3}$. The region inside the circle but outside the cardioid has area $2\times\frac12\int_0^{\pi/3}\left(9\cos^2\theta-(1+\cos\theta)^2\right)d\theta=\pi$.

Give exact answers in terms of $\pi$ and surds unless a decimal is asked for. An area must be positive: if you get a negative value, the limits are the wrong way round or $r_1$ and $r_2$ are swapped. Check against geometry where possible: the circle $r=a$ has area $\pi a^2$, and the circle $r=3\cos\theta$ has diameter 3, so area $\frac{9\pi}{4}$. A quick sketch shows which region you are finding, and the values of $\theta$ for which the curve passes through the pole.