Finding double integral of this region using polar coordinates?

Question

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 $\arcsin(\frac{5}{8})+\pi/2$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

Answer 2

$$\iint\limits_{R}\frac{1}{4}\sqrt{2x^2+2y^2}dA=$$

$$\frac {\sqrt 2}{4} \int _\alpha ^{\alpha +\pi } \int_0^8 r^2 dr d\theta =$$

$$\frac {\sqrt 2}{4} \frac {8^3}{3}\pi $$