Question

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

Short answer

Answer: The area of the region enclosed by the curve \(r=\sqrt{\cos \theta}\) is 1 square unit.

Step-by-step solution

Identify the polar equation and analyze the curve

We are given the polar equation \(r = \sqrt{\cos \theta}\). The graph of this equation represents the region we want to find the area of.

Find the bounding curves of the region

Notice that the curve is defined for \(\theta\) within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\), since \(r\) becomes imaginary outside of this range. On the interval \([\frac{\pi}{2}, \frac{3\pi}{2}]\) the radius r remains positive so the curve does not move back to the origin. Thus, the area of the region is enclosed by the curve and the lines \(\theta = -\frac{\pi}{2}\) and \(\theta = \frac{\pi}{2}\).

Determine the limits of integration for \(\theta\)

Since the region is bounded by the lines \(\theta = -\frac{\pi}{2}\) and \(\theta = \frac{\pi}{2}\), our limits of integration for \(\theta\) will be from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Use the polar area formula to calculate the area

The area of a region in polar coordinates is given by the formula: \(A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta\). For our region, we have: \(A = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sqrt{\cos \theta})^2 d\theta = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta d\theta\). Now we can integrate with respect to \(\theta\): \(A = \frac{1}{2} [\sin \theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2}(\sin \frac{\pi}{2} - \sin -\frac{\pi}{2}) = \frac{1}{2}(1 + 1) = 1\). So, the area of the region enclosed by the curve \(r=\sqrt{\cos \theta}\) is 1 square unit.