Interval for area bounded by $r = 1 + 3 \sin \theta$ I'm trying to calculate the area of the region bounded by one loop of the graph for the equation
$$
r = 1 + 3 \sin \theta
$$
I first plot the graph as a limaçon with a maximum outer loop at $(4, \frac{\pi}{2})$ and a minimum inner loop at $(-2, -\frac{3 \pi}{2})$. I then note the graph is symmetric with respect to the $\frac{\pi}{2}$ axis and the zero for the right half is at $\theta = \arcsin(-\frac{1}{3})$.
So, I chose the interval $[\arcsin(-\frac{1}{3}),\frac{\pi}{2}]$ to calculate the area which can then be multiplied by $2$ for the other half. The problem is that the answer in the book seems to use $\arcsin(\frac{1}{3})$ instead, note the change of sign.
Just to make sure I'm not misunderstanding where I went wrong, I get the answer
$$
\frac{11 \pi}{4} - \frac{11}{2} \arcsin(-\frac{1}{3}) + 3 \sqrt 2
$$
Whereas the book gets
$$
\frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2
$$
It's a subtle change of sign but I'd really like to understand where I went wrong.

$$
\begin{align}
\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12r^2\,\mathrm{d}\theta
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12(1+3\sin(\theta))^2\,\mathrm{d}\theta\\
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\sin^2(\theta)\right)\mathrm{d}\theta\\
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\left(\frac{1-\cos(2\theta)}2\right)\right)\mathrm{d}\theta\\
&=\left[\frac{11}4\theta-3\cos(\theta)-\frac98\sin(2\theta)\right]_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\\
&=\frac{11}4\left(\pi+2\arcsin\left(\frac13\right)\right)+3\sqrt2
\end{align}
$$
This is the area of the outer loop

The sum of both loops is
$$
\begin{align}
&\int_0^{2\pi}\frac12\left(1+6\sin(\theta)+9\left(\frac{1-\cos(2\theta)}2\right)\right)\mathrm{d}\theta\\
&=\left[\frac{11}4\theta-3\cos(\theta)-\frac98\sin(2\theta)\right]_0^{2\pi}\\
&=\frac{11}2\pi
\end{align}
$$
so the area of the inner loop is
$$
\frac{11}4\left(\pi-2\arcsin\left(\frac13\right)\right)-3\sqrt2
$$
