Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the inner loop of \(r=\cos \theta-\frac{1}{2}\)

Short answer

Answer: The area of the region inside the inner loop is \(\frac{\pi}{6}\) square units.

Step-by-step solution

Determine the range of \(\theta\) for the inner loop

To find the range for \(\theta\) that corresponds to the inner loop, we look for when the value of \(r\) becomes zero. Let's set \(r\) equal to zero and solve for \(\theta\): $$r = \cos\theta - \frac{1}{2} = 0$$ $$\cos\theta = \frac{1}{2}$$ $$\theta = \pm\frac{\pi}{3}$$ The inner loop will form between \(\theta=-\frac{\pi}{3}\) and \(\theta=\frac{\pi}{3}\).

Sketch the curve and identify the bounding curves

Now, let's plot the curve for \(r = \cos\theta - \frac{1}{2}\) in the given range. The curve will look like a distorted circle with an inner loop between \(\theta=-\frac{\pi}{3}\) and \(\theta=\frac{\pi}{3}\). The bounding curves are the radial lines \(\theta=-\frac{\pi}{3}\) and \(\theta=\frac{\pi}{3}\), as well as the curve given by \(r = \cos\theta - \frac{1}{2}\).

Calculate the area of the enclosed region

The area of a region enclosed by a polar curve is given by: $$A=\frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta$$ In this case, our polar function is \(r=\cos\theta-\frac{1}{2}\), and the limits of integration are \(\alpha=-\frac{\pi}{3}\) and \(\beta=\frac{\pi}{3}\). Plugging in these values, we get: $$A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (\cos\theta-\frac{1}{2})^2 d\theta$$ Apply the binomic formula and integrate: $$A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (\cos^2\theta-2\cos\theta\frac{1}{2}+\frac{1}{4}) d\theta$$ $$A=\!-\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (\cos^2\theta\!-\cos\theta\!+\frac{1}{4}) d\theta$$ Now, use the identity \(\cos^2\theta=\frac{1+\cos2\theta}{2}\) and integrate term by term: $$A=\!-\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (\frac{1+\cos2\theta}{2}\!-\cos\theta\!+\frac{1}{4}) d\theta=-\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (1+\cos2\theta-2\cos\theta+\frac{1}{2}) d\theta$$ $$A=-\frac{1}{2}\left[\theta+\frac{1}{2}\sin{2\theta}-\sin\theta+\frac{1}{2}\theta\right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}$$ Substitute the limits of integration and simplify to find the area: $$A=-\frac{1}{2}\left[\frac{2\pi}{3}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\frac{\pi}{3}\right]=-\frac{1}{2}\left[\frac{\pi}{3}\right]=\boxed{\frac{\pi}{6}}$$ The area of the region inside the inner loop of \(r = \cos\theta-\frac{1}{2}\) is \(\frac{\pi}{6}\) square units.