find the area of the region. interior of r² = 9 sin(2θ)

Explanation:

Step1: Recall area - formula in polar coordinates

The area $A$ of a polar - curve $r = f(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. For the curve $r^{2}=9\sin(2\theta)$, we need to find the appropriate limits of integration. Since $r^{2}\geq0$, we have $\sin(2\theta)\geq0$. The function $y = \sin(2\theta)$ is non - negative when $2k\pi\leq2\theta\leq(2k + 1)\pi$, or $k\pi\leq\theta\leq k\pi+\frac{\pi}{2}$, $k\in\mathbb{Z}$. To find one complete loop of the curve, we can take $k = 0$, so the limits of integration are from $0$ to $\frac{\pi}{2}$.

Step2: Apply the area formula

We know that $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. Substituting $r^{2}=9\sin(2\theta)$, $\alpha = 0$, and $\beta=\frac{\pi}{2}$ into the formula, we get $A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}9\sin(2\theta)d\theta$.

Step3: Integrate $\sin(2\theta)$

Let $u = 2\theta$, then $du=2d\theta$. When $\theta = 0$, $u = 0$; when $\theta=\frac{\pi}{2}$, $u=\pi$. So, $\int\sin(2\theta)d\theta=\frac{1}{2}\int\sin(u)du=-\frac{1}{2}\cos(u)+C=-\frac{1}{2}\cos(2\theta)+C$. Then $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}9\sin(2\theta)d\theta=\frac{9}{2}\left[-\frac{1}{2}\cos(2\theta)\right]_{0}^{\frac{\pi}{2}}$.

Step4: Evaluate the definite integral

$\frac{9}{2}\left[-\frac{1}{2}\cos(2\theta)\right]_{0}^{\frac{\pi}{2}}=\frac{9}{2}\left(-\frac{1}{2}\cos(\pi)+\frac{1}{2}\cos(0)\right)$. Since $\cos(\pi)=-1$ and $\cos(0)=1$, we have $\frac{9}{2}\left(-\frac{1}{2}\times(-1)+\frac{1}{2}\times1\right)=\frac{9}{2}\left(\frac{1}{2}+\frac{1}{2}\right)=\frac{9}{2}$.

Answer:

$\frac{9}{2}$