91Ó°ÊÓ

To sketch the polar curve \(r = \cos{3\theta}\), examine the points (r, θ) in polar coordinates as θ ranges from \(0\) to \(2\pi\). Observe that the function has three peaks and valleys, creating a three-leaf shape. This is because each time \(\theta\) moves through an interval of length \(\frac{\pi}{3}\), the argument of the cosine function changes by \(\pi\). Since we are only interested in one leaf, we will focus on the "top" leaf when the graph is plotted on paper or a computer screen. The curve forms one full leaf as \(\theta\) ranges from \(0\) to \(\frac{2\pi}{3}\) pyplot's linspace.

To find the area of the region inside one leaf, we must determine the limits of integration. We do this by finding the values of \(\theta\) where \(r = 0\), or when the curve intersects itself. This occurs when \(\cos{3\theta} = 0\). Solving for \(\theta\), we find \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{\pi}{2}\), where the curve begins and ends, respectively.

Use the formula for area in polar coordinates: \(A = \frac{1}{2}\int_{\alpha}^{\beta} r(\theta)^2 d\theta\), where \(\alpha = \frac{\pi}{6}\) and \(\beta = \frac{\pi}{2}\). Thus, the area of the region inside one leaf is: $$ A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\cos{3\theta})^2 d\theta $$ Now integrate: $$ A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \left(\frac{1 + \cos{6\theta}}{2}\right) d\theta $$ Break the integral into two parts: $$ A = \frac{1}{2}\left(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{2} d\theta + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{2}\cos{6\theta} d\theta\right) $$ Evaluate each of these integrals separately: $$ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{2} d\theta = \frac{1}{2} \cdot \frac{\pi}{3} $$ $$ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{2}\cos{6\theta} d\theta = \frac{1}{12}\left(\sin{(6\cdot{\pi}{2})}-\sin(6\cdot{\pi}{6})\right) = 0 $$ Combine the results: $$ A = \frac{1}{2}\left(\frac{1}{2} \cdot \frac{\pi}{3} + 0\right) = \frac{\pi}{12} $$ The area of the region inside one leaf is \(\frac{\pi}{12}\).