Problem 25 - Calculus Early Transcendentals

Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the limaçon \(r=2+\cos \theta\)

Step-by-Step Solution

Answer

Answer: The area of the region enclosed by the limaçon is \(4\pi\) square units.

Step 1: Sketch the polar curve

To sketch the limaçon with equation \(r=2+\cos\theta\), notice that it's an inner loop limaçon because the constant (2) is greater than 1. In general, limaçons have loops when a constant is added or subtracted to the cosine or sine function. Start by plotting some points for different values of \(\theta\) and connect the points smoothly, making sure to create a loop inside (the loop represents the region we want to find the area of).

Step 2: Find points of intersection

The points of intersection (or the endpoints of the region's boundaries) occur when the curve intersects itself. To find these points, set \(r = 2 + \cos\theta\) equal to 0 and solve for \(\theta\). These values are the limits for our integration to find the area of the region. $$ 2 + \cos\theta = 0 \\ \cos\theta = -2 \\ \theta = \pi $$ The limaçon intersects itself when \(\theta = \pi\).

Step 3: Set limits for the integral

In this case, we want to find the area enclosed by the loop, and that enclosed region starts when the curve intersects itself and goes around. As we found in the previous step, the intersection occurs at \(\theta =\pi\). Since the curve is symmetrical, the limits for the integration to find the area are from \(\theta = 0\) to \(\theta = 2\pi\).

Step 4: Apply the polar area formula

To find the area of the region, use the polar area formula: $$ A = \frac{1}{2}\int_{\alpha}^{\beta} [r(\theta)]^2 d\theta $$ In our case, the integral limits are from 0 to \(2\pi\), and \(r(\theta) = 2 + \cos\theta\). Thus, $$ A = \frac{1}{2}\int_{0}^{2\pi}[2 + \cos\theta]^2 d\theta $$

Step 5: Calculate the area

Now, we simply need to evaluate the integral: $$ A = \frac{1}{2}\int_{0}^{2\pi}(4 + 4\cos\theta + \cos^2\theta)d\theta $$ To evaluate the integral more easily, we can rewrite the \(\cos^2\theta\) term as \(\frac{1+\cos(2\theta)}{2}\): $$ A = \frac{1}{2}\int_{0}^{2\pi}(4 + 4\cos\theta + \frac{1+\cos(2\theta)}{2})d\theta $$ Integrate and evaluate: $$ A = \frac{1}{2}\left[4\theta + 4\sin\theta + \frac{1}{2}\theta + \frac{1}{4}\sin(2\theta)\right]_0^{2\pi} $$ $$ A = \frac{1}{2}(8\pi) $$ $$ A = 4\pi $$ So the area of the region enclosed by the limaçon is \(4\pi\) square units.