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 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

First, convert $x=r\cos\theta=(2 + 3\sin\theta)\cos\theta=2\cos\theta+3\sin\theta\cos\theta$ and $y=r\sin\theta=(2 + 3\sin\theta)\sin\theta=2\sin\theta+3\sin^{2}\theta$. Then, find $\frac{dy}{d\theta}=2\cos\theta + 6\sin\theta\cos\theta=\cos\theta(2 + 6\sin\theta)$ and $\frac{dx}{d\theta}=-2\sin\theta+3\cos^{2}\theta - 3\sin^{2}\theta=-2\sin\theta+3(\cos^{2}\theta-\sin^{2}\theta)$. For a horizontal tangent, $\frac{dy}{d\theta}=0$ and $\frac{dx}{d\theta}
eq0$. Since $\frac{dy}{d\theta}=\cos\theta(2 + 6\sin\theta)=0$ and $\frac{\pi}{2}<\theta<\pi$, if $\cos\theta = 0$, $\theta=\frac{\pi}{2}$ (not in the domain), if $2+6\sin\theta = 0$, $\sin\theta=-\frac{1}{3}$, then $\cos\theta=-\frac{2\sqrt{2}}{3}$. Substitute into $x$ and $y$: $x=(2 + 3(-\frac{1}{3}))(-\frac{2\sqrt{2}}{3})=-\frac{2\sqrt{2}}{3}$ and $y=(2 + 3(-\frac{1}{3}))(-\frac{1}{3})=-\frac{1}{3}$.

Step3: Set up arc - length integral

The arc - length of a polar curve 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. When $r$ is non - negative, we know that $r = 2+3\sin\theta\geq0$. The outer - loop is traced for $\theta$ values such that $r\geq0$. The limits can be found by considering the non - negativity of $r$. The outer - loop is traced for $\theta\in[0,\pi+\arcsin(\frac{2}{3})]\cup[2\pi-\arcsin(\frac{2}{3}),2\pi]$. We can use the integral $L=\int_{0}^{\pi+\arcsin(\frac{2}{3})}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta+\int_{2\pi-\arcsin(\frac{2}{3})}^{2\pi}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta$.

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) $\int_{0}^{\pi+\arcsin(\frac{2}{3})}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta+\int_{2\pi-\arcsin(\frac{2}{3})}^{2\pi}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta$