Question

Use a graphing utility to graph the polar equations and find the area of the given region. Common interior of \(r=4 \sin \theta\) and \(r=2\)

Short answer

The area of the common interior of the graphs of the given polar equations is \(2\pi\) square units.

Step-by-step solution

Graph the Equations

Start by graphing the polar equations using a graphing utility. The graph of \(r=4 \sin \theta\) is a circle with radius 2 centered at (0,2) while that of \(r=2\) is a circle with radius 2 centered at the origin.

Find the Intersection Points

Next, determine the points where the graphs intersect. These are the values of \(\theta\) for which the two equations are equal, i.e. when \(4 \sin \theta = 2\). Solve this to get \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Calculate the Area

Now calculate the area of the common interior. This is given by \(\frac{1}{2}\) multiplied by the integral from \(\theta_1\) to \(\theta_2\) of \((f(\theta))^2 - (g(\theta))^2 d\theta\), where f and g are the given polar equations. In this case, it would be \(\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} ((4 \sin \theta)^2 - 2^2)d\theta \). Solve the integral to find the area.