Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the rose \(r=4 \cos 2 \theta\) and outside the circle \(r=2\)

Short answer

Answer: The area of the region is \(\frac{\pi}{4}\).

Step-by-step solution

Sketch the curves

Start by sketching the given curves using polar coordinates: 1. The rose curve: \(r = 4\cos 2\theta\) 2. The circle: \(r = 2\) First, let's analyze the rose curve. Since the equation is \(r = 4\cos 2\theta\), it has even symmetry. This means that it will have 4 petals. The maximum length of the petal is when cosine is equal to 1: \(r=4\). For the circle, the equation is \(r = 2\). This indicates that it's a circle centered at the origin with a radius of 2.

Find points of intersection

Next, determine the points where the curves intersect. For this, we need to solve the system of equations: \(r = 4\cos 2\theta\) and \(r = 2\) Set these two equations equal to each other: \(2 = 4\cos 2\theta\) Divide by 4: \(\frac{1}{2} = \cos 2\theta\) Now, we can solve for \(\theta\): \(2\theta = \arccos\frac{1}{2}\) \(\theta = \frac{\arccos\frac{1}{2}}{2}\) From this result, we find two intersection points since the rose curve has even symmetry: \(\theta_1 = \frac{\pi}{6}\) and \(\theta_2 = \frac{5\pi}{6}\).

Compute the area using integration

Now we can compute the area of the region using integration. To obtain the correct area, we must subtract the area inside the circle and outside the rose curve from the area of the rose curve. We can do this using the polar area formula: \(A = \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_{1}^{2}-r_{2}^{2})d\theta\) Here, \(r_1 = 4\cos 2\theta\) and \(r_2 = 2\). Plug in these values and the limits of integration we found in Step 2: \(A = \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\cos^2 2\theta - 4)d\theta\) Now, we use the double-angle formula: \(\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}\): \(A = \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\cdot\frac{1 + \cos 4\theta}{2} - 4)d\theta\) Next, we integrate and evaluate the definite integral: \(A = \frac{1}{2}\left[\theta - \frac{1}{4}\sin 4\theta\right]_{\pi/6}^{5\pi/6}\) \(A = \frac{1}{2}\left(\frac{4\pi}{6} - \frac{1}{4}\sin \frac{20\pi}{6} - \frac{\pi}{6} + \frac{1}{4}\sin \frac{4\pi}{6}\right)\) \(A = \frac{1}{2}\left(\frac{3\pi}{6}\right)\) \(A = \frac{3\pi}{12}\) \(A = \frac{\pi}{4}\) The area of the region inside the rose curve and outside the circle is \(\frac{\pi}{4}\).