Given the polar curves $r^2=18\cdot \sin \left(2\theta \right)$ and $r=3$, what is the area of the region that lies inside both curves?

• A. $\frac{9\pi }{2}$
• B. $\frac{18\pi }{5}$
• C. $\frac{27\pi }{4}$
• D. $9\pi$

Answer: (A)

Verified step by step guidance

1. Identify the given polar curves: the first curve is given by \(r^{2} = 18 \cdot \sin(2\theta)\), and the second curve is the circle \(r = 3\).
2. Rewrite the first curve to express \(r\) explicitly: since \(r^{2} = 18 \sin(2\theta)\), we have \(r = \sqrt{18 \sin(2\theta)}\). Note that \(r\) is real only where \(\sin(2\theta) \geq 0\).
3. Find the points of intersection between the two curves by setting \(r\) equal: \$3 = \sqrt{18 \sin(2\theta)}\(. Square both sides to get \)9 = 18 \sin(2\theta)\(, which simplifies to \)\sin(2\theta) = \frac{1}{2}\(. Solve for \)\theta\( in the interval \)[0, 2\pi)$ to find the limits of integration.
4. Determine the region inside both curves: this corresponds to values of \(\theta\) where \(r\) from the first curve is less than or equal to 3, and \(r\) from the second curve is constant at 3. The overlapping region is bounded by the smaller radius at each \(\theta\) between the intersection points.
5. Set up the integral for the area of the overlapping region using the formula for area in polar coordinates: \(A = \frac{1}{2} \int_{\alpha}^{\beta} \min(r_1(\theta), r_2(\theta))^{2} \, d\theta\). Here, split the integral if necessary to account for which curve is inside. Evaluate the integral over the appropriate \(\theta\) intervals to find the total area.