Areas from a Single Polar Curve (College Board AP® Calculus BC): Revision Note

Look out for any symmetry, as it may be possible to express larger polar areas as multiples of smaller polar areas.

In the non-calculator sections of the exam, you are often asked to leave polar areas as definite integrals. In the calculator sections, you are expected to use your calculator to evaluate these definite integrals.

The polar curve $r=\sin 3\theta$ is shown below, where $0\le \theta \le \pi$. Part of the total area enclosed has been shaded.

(a) Find the area of the single shaded loop on the diagram.

(b) Write down the definite integral that represents the area enclosed by the loop below the $x$-axis.

(c) Find the total area enclosed by the curve.

Answer:

(a)

The loop starts and ends when $r=0$ so substitute $r=0$ into $r=\sin 3\theta$

$0=\sin 3\theta$

Solving this equation gives

$\begin{array}{rcl}3\theta & =& 0,\pi ,2\pi ,...\\ \theta & =& 0,\frac{\pi }{3},\frac{2\pi }{3},...\end{array}$

The first two rays are $\theta =0$ and $\theta =\frac{\pi }{3}$ as shown

Substitute $\alpha =0$, $\beta =\frac{\pi }{3}$ and $r=\sin 3\theta$ into the area formula, $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$

$A={\int }_0^{\frac{\pi }{3}}\frac{1}{2}{\left(\sin 3\theta \right)}^{2}d\theta$

Evaluate this definite integral on your calculator

0.261799...

The area of the single shaded loop is 0.262 (to 3 decimal places)

(b)

From part (a), $r=0$ when $\begin{array}{rcl}\theta & =& 0,\frac{\pi }{3},\frac{2\pi }{3},...\end{array}$

The loop in part (a) is between $\theta =0$ and $\theta =\frac{\pi }{3}$, where $r=\sin 3\theta \ge 0$, so $r\ge 0$

The loop in this question is between $\theta =\frac{\pi }{3}$ and $\beta =\frac{2\pi }{3}$ , where $r=\sin 3\theta \le 0$, so $r$ is negative, $r\le 0$, shown below

However, the formula $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$ still works for negative values of $r$ (as $r$ is squared in the integral), so substitute in $\alpha =\frac{\pi }{3}$, $\beta =\frac{2\pi }{3}$ and $r=\sin 3\theta$

$A={\int }_{\frac{\pi }{3}}^{\frac{2\pi }{3}}\frac{1}{2}{\left(\sin 3\theta \right)}^{2}d\theta$

(c)

The total area enclosed by the curve will be the area of 3 loops

Multiply the answer in part (a) by 3

$3\times 0.261799...=0.7853...$

The total area enclosed is 0.785 (to 3 decimal places)