91Ó°ÊÓ

To make a sketch of the region and its bounding curves, we should plot both equations in polar coordinates. 1. \(r = \sqrt{\cos 2 \theta}\) This is a rose curve with 4 petals. As we are interested in the first quadrant, we will only focus on the right lobe of the curve in the first quadrant. 2. \(r = \frac{1}{\sqrt{2}}\) This is a circle with a radius of \(\frac{1}{\sqrt{2}}\). The center of the circle is at the origin, and it extends into all quadrants.

We need to determine where the right lobe of \(r=\sqrt{\cos 2 \theta}\) intersects the circle \(r=1 / \sqrt{2}\) in the first quadrant. To do this, we can set the two equations equal to each other and solve for theta: $$\sqrt{\cos 2 \theta} = \frac{1}{\sqrt{2}}$$ Squaring both sides we get: $$\cos 2 \theta = \frac{1}{2}$$ Now solve for \(\theta\) in the first quadrant: $$2 \theta = \frac{\pi}{3}\Rightarrow \theta = \frac{\pi}{6}$$

Now that we have the intersection points, we can find the area of the region by integrating the difference between their polar equations with respect to \(\theta\): Bounded Area \(= \frac{1}{2}\int_{0}^{\frac{\pi}{6}}( r_1^2 - r_2^2) d\theta\), where \(r_1(\theta) = \sqrt{\cos 2 \theta}\) and \(r_2(\theta) = \frac{1}{\sqrt{2}}\). Calculating the integral: $$\frac{1}{2}\int_{0}^{\frac{\pi}{6}}(\cos 2 \theta - \frac{1}{2}) d\theta = \frac{1}{2}\left[\frac{1}{2}\sin 2 \theta - \frac{1}{2}\theta\right]_{0}^{\frac{\pi}{6}}$$ Now plug in the limits of integration and subtract: $$\frac{1}{2}\left[\frac{1}{2}(1) - \frac{1}{2}(\frac{\pi}{6})\right] \approx 0.0898$$ The area of the region inside the right lobe of \(r=\sqrt{\cos 2 \theta}\) and inside the circle \(r=1 / \sqrt{2}\) in the first quadrant is approximately 0.0898 square units.