Question

the polar curve $r = 2 + 3sin\theta$ is graphed for $0leq\thetaleq2pi$.
a) find the area enclosed by the inner loop of the curve.
b)determine the cartesian coordinates of the point on the curve where the tangent line is horizontal, for $\frac{pi}{2}<\thetac)set up, but do not evaluate, an integral expression for the arc length of the outer loop.

Explanation:

Step1: Find limits for inner - loop area

To find the limits for the inner - loop of $r = 2+3\sin\theta$, set $r = 0$. So, $2 + 3\sin\theta=0$, then $\sin\theta=-\frac{2}{3}$. The solutions in $[0,2\pi]$ are $\theta_1=\pi+\arcsin(\frac{2}{3})$ and $\theta_2 = 2\pi-\arcsin(\frac{2}{3})$. The area $A$ of a polar curve $r = f(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. Here, $\alpha=\pi+\arcsin(\frac{2}{3})$ and $\beta = 2\pi-\arcsin(\frac{2}{3})$, and $r = 2 + 3\sin\theta$. So, $A=\frac{1}{2}\int_{\pi+\arcsin(\frac{2}{3})}^{2\pi-\arcsin(\frac{2}{3})}(2 + 3\sin\theta)^{2}d\theta$.

Step2: Find horizontal - tangent points in Cartesian coordinates

First, convert $r = 2+3\sin\theta$ to Cartesian coordinates using $x = r\cos\theta=(2 + 3\sin\theta)\cos\theta$ and $y = r\sin\theta=(2 + 3\sin\theta)\sin\theta$. Differentiate $y$ with respect to $\theta$: $\frac{dy}{d\theta}=(2 + 3\sin\theta)\cos\theta+3\cos\theta\sin\theta$. Set $\frac{dy}{d\theta}=0$ for horizontal tangent.
\[

$$\begin{align*} (2 + 3\sin\theta)\cos\theta+3\cos\theta\sin\theta&=0\\ \cos\theta(2 + 6\sin\theta)&=0 \end{align*}$$

\]
Since $\frac{\pi}{2}<\theta<\pi$, $\cos\theta
eq0$. Then $2 + 6\sin\theta=0$, so $\sin\theta=-\frac{1}{3}$. $\cos\theta=-\sqrt{1-\sin^{2}\theta}=-\frac{2\sqrt{2}}{3}$. $r = 2+3\sin\theta=2 - 1=1$. $x = r\cos\theta=-\frac{2\sqrt{2}}{3}$ and $y = r\sin\theta=-\frac{1}{3}$.

Step3: Set up arc - length integral for outer - loop

The arc - length $L$ of a polar curve $r = f(\theta)$ is given by $L=\int_{\alpha}^{\beta}\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}d\theta$. First, find $\frac{dr}{d\theta}=3\cos\theta$. For the outer - loop, we need to find the appropriate limits. The curve $r = 2+3\sin\theta$ starts and ends a full outer - loop. We can use $\alpha = 0$ and $\beta = 2\pi$. So, $L=\int_{0}^{2\pi}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta$.

Answer:

a) $\frac{1}{2}\int_{\pi+\arcsin(\frac{2}{3})}^{2\pi-\arcsin(\frac{2}{3})}(2 + 3\sin\theta)^{2}d\theta$
b) $x =-\frac{2\sqrt{2}}{3},y =-\frac{1}{3}$
c) $\int_{0}^{2\pi}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta$