Question

the polar curve r = 2 + 3 sin θ is graphed for 0 ≤ θ ≤ 2π. 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 π/2 < θ < π. c)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 $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. For the inner - loop, $\alpha=\pi+\arcsin(\frac{2}{3})$ and $\beta = 2\pi-\arcsin(\frac{2}{3})$, and $r = 2 + 3\sin\theta$. So the area of the inner - loop is $A=\frac{1}{2}\int_{\pi+\arcsin(\frac{2}{3})}^{2\pi-\arcsin(\frac{2}{3})}(2 + 3\sin\theta)^{2}d\theta$.

Step2: Find horizontal tangent points

We know that $x=r\cos\theta=(2 + 3\sin\theta)\cos\theta=2\cos\theta+3\sin\theta\cos\theta$ and $y=r\sin\theta=(2 + 3\sin\theta)\sin\theta=2\sin\theta+3\sin^{2}\theta$.
Differentiate $x$ and $y$ with respect to $\theta$:
$\frac{dx}{d\theta}=-2\sin\theta+3\cos^{2}\theta - 3\sin^{2}\theta=-2\sin\theta+3\cos2\theta$
$\frac{dy}{d\theta}=2\cos\theta + 6\sin\theta\cos\theta=2\cos\theta(1 + 3\sin\theta)$
For a horizontal tangent, $\frac{dy}{d\theta}=0$ and $\frac{dx}{d\theta}
eq0$.
From $\frac{dy}{d\theta}=0$, we have $\cos\theta=0$ or $\sin\theta=-\frac{1}{3}$.
If $\cos\theta=0$, then $\theta=\frac{\pi}{2}$ or $\theta=\frac{3\pi}{2}$. But we want $\frac{\pi}{2}<\theta<\pi$, so we consider $\sin\theta=-\frac{1}{3}$. Then $\cos\theta=-\sqrt{1-\sin^{2}\theta}=-\frac{2\sqrt{2}}{3}$ (since $\frac{\pi}{2}<\theta<\pi$).
$r=2+3\sin\theta=2 - 1 = 1$
$x=r\cos\theta=1\times(-\frac{2\sqrt{2}}{3})=-\frac{2\sqrt{2}}{3}$
$y=r\sin\theta=1\times(-\frac{1}{3})=-\frac{1}{3}$

Step3: Set up arc - length integral

The arc - length 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 $\frac{dr}{d\theta}=3\cos\theta$.
To find the limits for the outer - loop, we note that the outer - loop is traced out for $\theta$ values where $r\geq0$. The outer - loop is traced for $\theta\in[2\pi-\arcsin(\frac{2}{3}),\pi+\arcsin(\frac{2}{3})]$.
So the arc - length integral is $L=\int_{2\pi-\arcsin(\frac{2}{3})}^{\pi+\arcsin(\frac{2}{3})}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta$.

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) $(-\frac{2\sqrt{2}}{3},-\frac{1}{3})$
c) $\int_{2\pi-\arcsin(\frac{2}{3})}^{\pi+\arcsin(\frac{2}{3})}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta$