Question

Each of the integral expressions that follow represents the area of a region in the plane bounded by a function expressed in polar coordinates. Use the ideas from this section and from Chapter 9 to sketch the regions, and then evaluate each integral

$\frac{1}{2}{\int }_0^{\mathrm{\pi }}{\cos }^{2}3\theta d\theta$

Short answer

The area is$\frac{\mathrm{\pi }}{4}$square units.

Step-by-step solution

Step 1. Given information

Integral:

$\frac{1}{2}{\int }_0^{\mathrm{\pi }}{\cos }^{2}3\theta d\theta$

Step 2. Plot the curve

Since the area of a function $r=f\left(\theta \right)$is $\frac{1}{2}{\int }_0^{\mathrm{\pi }}{r}^{2}d\theta$

So by comparing the given integral we get,

$r=\cos 3\theta$

So the graph of this curve is:

Step 3. Evaluate integral

$$\begin{gathered} \frac{1}{2}{\int }_0^{\mathrm{\pi }}{\cos }^{2}3\theta d\theta \\ =\frac{1}{2}{\int }_0^{\mathrm{\pi }}{\cos }^{2}3\theta d\theta \\ =\frac{1}{2}{\int }_0^{\mathrm{\pi }}\left(\frac{1+\cos 6\theta }{2}\right)d\theta \\ =\frac{1}{4}{\int }_0^{\mathrm{\pi }}\left(1+\cos 6\theta \right)d\theta \\ =\frac{1}{4}{\left[\theta +\frac{sin6\theta }{6}\right]}_0^{\pi } \\ =\frac{1}{4}\left[\mathrm{\pi }+\frac{\sin 6\mathrm{\pi }}{6}-0\right] \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$