Question

Find the areas of the following regions. The region inside the outer loop but outside the inner loop of the limaçon \(r=3-6 \sin \theta\)

Short answer

Answer: The area of the region is \(\frac{33\pi}{4}\).

Step-by-step solution

Determine the limits of integration

For the limits of integration, we need to determine when the inner and outer loops intersect. We set \(r=0\) to find the intersection points: \(r(\theta) = 3-6\sin\theta=0\) \(\sin\theta=\frac{1}{2}\) \(\theta=\frac{\pi}{6}, \frac{5\pi}{6}\) These are the intersection points between the inner and outer loops when \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{5\pi}{6}\). Our limits of integration will be from \(\theta=\frac{\pi}{6}\) to \(\theta=\frac{5\pi}{6}\).

Compute the area of the outer loop

To find the area of the outer loop we use the formula for the area of a polar curve: \(A = \frac{1}{2}\int_{\alpha}^{\beta}[r(\theta)]^2\,d\theta\) For the outer loop, the limits of integration are \(\alpha=\frac{\pi}{6}\) and \(\beta=\frac{5\pi}{6}\).Plug in the given equation into the formula: \(A_{outer} = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(3-6\sin\theta)^2\,d\theta\)

Compute the area of the inner loop

The intersection points are the same as in step 1. To find the area of the inner loop we will use the same formula, but this time the limits of integration are from \(\alpha=0\) to \(\beta=\frac{\pi}{6}\): \(A_{inner}=\frac{1}{2}\int_0^{\frac{\pi}{6}}(3-6\sin\theta)^2\,d\theta\)

Compute the net area of the region

To find the net area between the outer and inner loops, we need to subtract the area of the inner loop from the area of the outer loop: \(A_{net} = A_{outer} - A_{inner} = \frac{1}{2}\left(\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(3-6\sin\theta)^2\,d\theta - \int_0^{\frac{\pi}{6}}(3-6\sin\theta)^2\,d\theta\right)\) Calculating the integrals, we get: \(A_{net} = \frac{1}{2}\left(9\pi - 6\int_0^{\frac{\pi}{6}}(\sin\theta)^2\,d\theta \right)\) \(A_{net}= \frac{1}{2}\left(9\pi - 6\left[\frac{1}{2}\theta - \frac{1}{4}\sin2\theta\right]_0^{\frac{\pi}{6}}\right)\) Finally, we plug in the limits and simplify to find the net area: \(A_{net}= \frac{1}{2}\left(9\pi -6\frac{\pi}{12}\right)\) \(A_{net}= \frac{33\pi}{4}\) So, the area of the region inside the outer loop but outside the inner loop of the limaçon is \(\frac{33\pi}{4}\).