Problem 27 - Calculus Early Transcendentals

Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside one leaf of \(r=\cos 3 \theta\)

Step-by-Step Solution

Answer

Based on the given polar equation \(r = \cos{3\theta}\), the area of the region inside one leaf can be calculated using the limits of integration \(\alpha = \frac{\pi}{6}\) and \(\beta = \frac{\pi}{2}\) and the formula for area in polar coordinates, which resulted in an area of \(\frac{\pi}{12}\).

Step 1: Sketch the polar curve

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.

Step 2: Determine the limits of integration

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.

Step 3: Find the area of the region inside one leaf

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 \(\frac{\pi}{12}\).