Finding double integral of this region using polar coordinates? So I've got the region $R$ like in the image below, and need to find the double integral $$\iint\limits_{R}\frac{1}{4}\sqrt{2x^2+2y^2}dA$$ over that region, given $a=8$ and $c=5$.

So here's what I did:
Assuming that the shaded area in the 2nd (top left) quadrant is identical to that in the fourth (bottom right), and using polar coordinates, the region $R$ would be $$R=\{(r,\theta)|0 \leq r \leq 8, -\arcsin\left(\frac{5}{8}\right) \leq \theta \leq \arcsin\left(\frac{5}{8}\right)+\pi/2\}$$
which gives the integral $$\int\limits_{0}^8\int\limits_{-\arcsin(5/8)}^{\arcsin(5/8)+\pi/2}\frac{1}{4}\sqrt{2}r\cdot r\mathrm d\theta\mathrm dr .$$Using integration methods, I got
$$\frac{\sqrt{2}}{4}\int\limits_{0}^8 r^2\theta\Big|_{-\arcsin(5/8)}^{\arcsin(5/8)+\pi/2} \mathrm dr$$
$$\frac{\sqrt{2}}{4}\int\limits_{0}^8 r^2(2\arcsin(5/8)+\pi/2) \mathrm dr$$
$$\frac{\sqrt{2}}{4}\cdot (2\arcsin(5/8)+\pi/2)\cdot\frac{r^3}{3} \Big|_0^8$$
$$\frac{\sqrt{2}}{4}\cdot (2\arcsin(5/8)+\pi/2)\cdot\frac{512}{3}\approx176.3$$
Apparently, this is not the correct answer. Could someone please point out what's wrong with my calculation?
UPDATE: turns out the mistake was in the upper bound for my inner integral (the $\mathrm d\theta$'s) integral: it should be $\arccos(\frac{5}{8})+\pi/2$ instead of $\arcsin(\frac{5}{8})+\pi/2$
 A: $$\iint\limits_{R}\frac{1}{4}\sqrt{2x^2+2y^2}dA=$$
$$\frac {\sqrt 2}{4} \int _\alpha ^{\alpha +\pi } \int_0^8 r^2 dr d\theta =$$
$$\frac {\sqrt 2}{4} \frac {8^3}{3}\pi $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Lets $\ds{\delta \equiv \arccos\pars{5 \over 8}}\,,\ x = 8\cos\pars{\theta}\,,\ y = 8\sin\pars{\theta}$:

\begin{align}
&\iint_{\large\mathbb{R}^{2}}{1 \over 4}\root{2x^{2} + 2y^{2}}
\bracks{\root{x^{2} + y^{2}} < 8}
\bracks{-\,{\pi \over 2} + \delta < \theta < {\pi \over 2} + \delta }\,\mrm{dS}
\\[5mm] = &\
{1 \over 4}\int_{-\pi/2 + \delta}^{\pi/2 + \delta}\int_{0}^{8}\pars{\root{2}r}r
\,\dd r\,\dd\theta =
{\root{2} \over 4}\int_{-\pi/2 + \delta}^{\pi/2 + \delta}{1 \over 3}\,8^{3}\,\dd\theta =
\bbx{{128\root{2} \over 3}\,\pi} \approx  189.5630
\end{align}
