Areas between Two Polar Curves (College Board AP® Calculus BC): Revision Note

If one of the polar areas is part of a circle, e.g. the circle $r=a$, then it can be quicker to use the formula for the area of a circle,$\pi a^2$.

A sketch of the polar curves $r=1+\cos \theta$ and $r=3\cos \theta$ where $0\le \theta \le \frac{\pi }{2}$ is shown below.

The shaded region is labeled $R_1$ and an unshaded region is labeled $R_2$.

(a) Find the area of $R_1$, leaving your answer as a single integral.

(b) Find the area of $R_2$.

Answer:

(a)

When $\theta =0$, $r=3\cos \theta$ is $3$ and $r=1+\cos \theta$ is $2$, so $r=3\cos \theta$ is the larger curve

Find the angle at which the two curves intersect by setting $1+\cos \theta$ equal to $3\cos \theta$ and solving

$\begin{array}{rcl}1+\cos \theta & =& 3\cos \theta \\ 1& =& 2\cos \theta \\ \frac{1}{2}& =& \cos \theta \\ \theta & =& \frac{\pi }{3}\end{array}$

Draw the ray $\theta =\frac{\pi }{3}$ on the diagram to see how to form the area $R_1$

It is the difference between the two polar areas shown below

Use the formula $A={\int }_{{\theta }_1}^{{\theta }_2}\frac{1}{2}{r}^{2}d\theta$ to work out the two individual areas, then subtract the smaller area from the larger area

${\int }_0^{\frac{\pi }{3}}\frac{1}{2}{\left(3\cos \theta \right)}^{2}d\theta -{\int }_0^{\frac{\pi }{3}}\frac{1}{2}{\left(1+\cos \theta \right)}^{2}d\theta$

The limits of the integrals are the same, so the integrals can be joined to form one single integral (as requested by the question)

${\int }_0^{\frac{\pi }{3}}\frac{1}{2}\left[{\left(3\cos \theta \right)}^2-{\left(1+\cos \theta \right)}^2\right]d\theta$

(b)

$R_2$ is the sum of the two polar areas shown below

$r=1+\cos \theta$ gives the part of the area from $\theta =0$ to the point of intersection

$r=3\cos \theta$ gives the part of the area from the point of intersection to $\theta =\frac{\pi }{2}$ (where $r=0$), which is the end of the range given in the question

Use the formula $A={\int }_{{\theta }_1}^{{\theta }_2}\frac{1}{2}{r}^{2}d\theta$ to work out the two individual areas, then add them together

${\int }_0^{\frac{\pi }{3}}\frac{1}{2}{\left(1+\cos \theta \right)}^{2}d\theta +{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2}{\left(3\cos \theta \right)}^{2}d\theta$

Evaluate the integrals above on your calculator

$1.759676...+0.203818...$

$1.963$ (to 3 decimal places)