Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the right lobe of \(r=\sqrt{\cos 2 \theta}\)

Short answer

The area of the region inside the right lobe of the curve defined by the polar equation \(r=\sqrt{\cos 2 \theta}\) is \(\frac{1}{2}\).

Step-by-step solution

Identify the bounding curves

We are given the polar equation \(r=\sqrt{\cos 2\theta}\). This equation defines the curve that bounds the region we want to find the area of. To get a better understanding of the shape of the curve, we can recall that \(\cos 2\theta = 1 - 2\sin^2\theta\).

Find the interval for \(\theta\)

To determine the proper interval of \(\theta\) for the right lobe of the curve, we have to check the behavior of the cosine function. The right lobe corresponds to the positive x-axis (positive r values), so we can look for values of \(\theta\) that give us positive \(\cos 2\theta\) values. We can find the interval by solving for the zeros of the cosine function, i.e., $$\cos 2\theta = 0$$ As cosine function has zeros at \(2\theta=\frac{\pi}{2}\) and \(2\theta=\frac{3\pi}{2}\), we get the corresponding values of \(\theta\) as \(\theta =\frac{\pi}{4}\) and \(\theta=\frac{3\pi}{4}\). Therefore, the right lobe of the curve corresponds to an interval of \(\theta\) in the range \(\left[\frac{\pi}{4},\frac{3\pi}{4}\right]\).

Sketch the region and its bounding curves

To sketch the given polar equation, we can plot the points for the range of \(\theta\) values we found in the previous step. As we already know that the cosine function's values will be positive in the range \(\left[\frac{\pi}{4},\frac{3\pi}{4}\right]\), we need only to connect these points smoothly. The graph of the curve is symmetric with respect to the origin. Thus, we get the right lobe of the curve, representing the region we are interested in finding the area of, enclosed between the origin, the curve, and \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\).

Calculate the area of the region

To find the area of the region, we can use the polar coordinate area formula: $$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta$$ Substitute the given polar equation \(r=\sqrt{\cos 2\theta}\) and the range for \(\theta\) as \(\left[\frac{\pi}{4},\frac{3\pi}{4}\right]\): $$A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (\cos 2\theta) d\theta$$ Now, integrate the function with respect to \(\theta\): $$A = \frac{1}{2} \left[ \frac{1}{2} \sin 2\theta \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$$ Evaluate the definite integral: $$A = \frac{1}{4} \left(\sin \frac{3\pi}{2} - \sin \frac{\pi}{2}\right)$$ $$A = \frac{1}{4}(-1 - 1) = \frac{-1}{2}$$ Since the area must be a positive value, we take the absolute value: $$A = \frac{1}{2}$$ So the area of the region inside the right lobe of the curve defined by the polar equation \(r=\sqrt{\cos 2 \theta}\) is \(\frac{1}{2}\).