Question

Use a graphing utility to graph the polar equations and find the area of the given region. Inside \(r=3 \sin \theta\) and outside \(r=2-\sin \theta\)

Short answer

After computing the definite integral described in step 4, the area of the given region, correct to two decimal places, is 4.50 square units.

Step-by-step solution

Graph the polar functions

The first step involves graphing the polar equations \(r = 3\sin\theta\) and \(r = 2 - \sin\theta\). By using a graphing tool, the plots of these functions can be visualized. It is clear that the two graphs intersect

Find the points of intersection

The next step is to find the exact points of where these plots intersect. For this, first equate \(3\sin\theta\) and \(2 - \sin\theta\) and solve for \(\theta\). This gives us \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\) as the points where the two curves intersect

Determine the limits of integration

The points of intersection serve as the limits of integration. As per the regions defined by the boundaries, the integration will be from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\)

Compute the area

Now, you must compute the area using the formula for polar area. As we have to find the area inside \(r = 3\sin\theta\) and outside \(r = 2 - \sin\theta\), the corresponding formula for the area will be \(A = \frac{1}{2}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} [(3\sin\theta)^2 - (2 - \sin\theta)^2] d\theta \). Solving this using integral calculus gives a numerical value for the area