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Chapter 11: Problem 47

Find the areas of the following regions. The region common to the circles \(r=2 \sin \theta\) and \(r=1\)

Short Answer

Expert verified
Answer: The approximate area of the region common to the circles \(r=2\sin\theta\) and \(r=1\) is 1.720 square units.

Step by step solution

01

Find the points of intersection between the two circles

To find the points of intersection, we need to find where \(r=2\sin\theta\) and \(r=1\) overlap. Let's equate the two equations and solve for \(\theta\): $$ 2\sin\theta = 1 \\ \sin\theta = \frac{1}{2} $$ This is true when \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\). Now that we have found the points of intersection, we can proceed in calculating the area of the overlapping region.
02

Set up the integral for the overlapping area

We can calculate the area of the overlap between the circles by using the polar coordinate equation for the area of a region: $$ \frac{1}{2}\int_{\alpha}^{\beta}(r_1^2-r_2^2)d\theta $$ In our case, \(r_1 = 2\sin\theta\), \(r_2 = 1\), \(\alpha = \frac{\pi}{6}\), and \(\beta = \frac{5\pi}{6}\). Plugging these values into the polar coordinate equation for the area of a region, we get: $$ A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(2\sin\theta)^2 - (1^2)d\theta $$
03

Evaluate the integral

First, simplify the integrand: $$ A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(4\sin^2\theta - 1)d\theta $$ Now use the double angle formula to further simplify: $$ \sin^2\theta = \frac{1 - \cos(2\theta)}{2} $$ Plug it into the equation of the area we get $$ A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(4\cdot\frac{1 - \cos(2\theta)}{2} - 1)d\theta $$ $$ A = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(2 - 2\cos(2\theta) - 1)d\theta $$ Now evaluate the integral: $$ A = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(1 - 2\cos(2\theta))d\theta \\ A = \Big[\theta - \sin(2\theta)\Big]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} $$
04

Calculate the final area

Now we need to substitute the limits of integration: $$ A = \Big[\frac{5\pi}{6} - \sin(2\cdot\frac{5\pi}{6})\Big] - \Big[\frac{\pi}{6} - \sin(2\cdot\frac{\pi}{6})\Big] $$ $$ A = \frac{4\pi}{6} - \left(\sin\frac{5\pi}{3} - \sin\frac{\pi}{3}\right) = \frac{2\pi}{3} - (-\frac{1}{2} + \frac{\sqrt{3}}{2}) $$ Now simplify the expression $$ A = \frac{2\pi}{3} +\frac{1}{2} - \frac{\sqrt{3}}{2} \approx 1.720 $$ The area of the region common to the circles \(r=2\sin\theta\) and \(r=1\) is approximately 1.720 square units.

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