Given the polar curves $r=3\sin \left(\theta \right)$ and $r=3\cos \left(\theta \right)$, what is the area of the region that lies inside both curves?

• A. $9$
• B. $\frac{9}{8}$
• C. $\frac{9}{4}$
• D. $\frac{9}{2}$

Answer: (B)

Verified step by step guidance

1. Identify the two polar curves given: \(r = 3 \sin(\theta)\) and \(r = 3 \cos(\theta)\). These represent circles in polar coordinates.
2. Find the points of intersection by setting the two equations equal: \$3 \sin(\theta) = 3 \cos(\theta)\(, which simplifies to \)\sin(\theta) = \cos(\theta)\(. Solve for \)\theta$ to find the angles where the curves intersect.
3. Determine the region inside both curves. Since both are circles centered on the axes, the overlapping region is symmetric and lies between the intersection angles found in the previous step.
4. Set up the integral for the area of the overlapping region. The area inside a polar curve \(r(\theta)\) from \(\alpha\) to \(\beta\) is given by \(\frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 \, d\theta\). For the overlapping region, integrate the minimum of the two \(r\) values over the appropriate interval.
5. Calculate the area by integrating \(\frac{1}{2} (3 \sin(\theta))^2\) and \(\frac{1}{2} (3 \cos(\theta))^2\) over their respective intervals determined by the intersection points, then sum these areas to find the total area inside both curves.