Interval for area bounded by $r = 1 + 3 \sin \theta$

Notice how $\arcsin(-\frac{1}{3}) = - \arcsin(\frac{1}{3})$, so your answer now looks like

$$ \frac{11 \pi}{4} + \frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2 \\ $$

That means your area is greater than the answer in your book by:

$$ 2 \left(\frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2\right) $$

This might indicate you are calculating the area of the outer loop whereas your book is calculating the inner loop. If you choose the interval $[\frac{3 \pi}{2}, 2 \pi - \arcsin(\frac{1}{3})]$ to calculate the half as you did before, you get:

$$ \begin{eqnarray} A &=& 2 \times \frac{1}{2} \int_{\frac{3 \pi}{2}}^{2 \pi - \arcsin \frac{1}{3}} (1 + 3 \sin \theta)^2 \, \textrm{d}\theta \\ &=& \left[\frac{11 \theta}{2} - 6 \cos \theta - \frac{9 \sin(2 \theta)}{4} \right]_{\frac{3 \pi}{2}}^{2 \pi - \arcsin \frac{1}{3}} \\ &=& \frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2 \\ \end{eqnarray} $$

This seems to agree with the answer in your book.