Evaluating $\iint dx\,dy$ over the region bounded by $y^2=x$ and $x^2+y^2=2x$ in the first quadrant 
Identify the region bounded by the curves $y^2=x$ and $x^2+y^2=2x$, that lies in the first quadrant and evaluate $\iint dx\,dy$ over this region.

In my book the solution is like:

$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}} \, dx \, dy\\
&=\int_{x=0}^1 \big[y\big]_{\sqrt x}^{\sqrt{2x-x^2}}\,dx\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&{\begin{aligned}\\
=\int_0^1\sqrt{1-x^2}\,dx-\int_0^1\sqrt{x}&\,dx\text{(applying} \int_0^af(x)\,dx=&\int_0^af(a-x)\,dx \text{ in the first part)}\\
\end{aligned}\\}\\
&=\left[\frac{\sqrt{1-x^2}}{2}+\sin^{-1}x\right]_0^1-\left[\frac{x^{\frac{3}{2}}}{\frac32}\right]_0^1\\
&=\frac{\pi}{2}-\frac12-\frac23(1-0)\\
&=\frac{\pi}{2}-\frac76\\
\end{align}\\
$$
And I did it like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}}dx\,dy\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&=\int_0^1\sqrt{1-(x-1)^2}\,dx-\int_0^1\sqrt{x}\,dx\\
&{\begin{aligned}\\
=&\left[\frac{x-1}{2}\sqrt{1-(x-1)^2}+\frac12\sin^{-1}(x-1)\right]_0^1&-\left[\frac23x^{\frac32}\right]_0^1\\
\end{aligned}\\}\\
&=-\frac{\pi}{4}-\frac23\\
\end{align}\\
$$
Which one is correct?

$$\int\limits_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx =\\=
\int_\limits0^1\sqrt{2x-x^2}\,dx-\int\limits_0^1\sqrt{x}\,dx = \frac{\pi}{4} - \frac{2}{3}
$$
For first one:
$$\int_\limits0^1\sqrt{2x-x^2}\,dx = \int_\limits0^1\sqrt{1-(x-1)^2}\,dx=\\=\int\limits_{-1}^{0}\sqrt{1-y^2}\,dy=\int\limits_{-\frac{\pi}{2}}^{0}\cos^2t\,dt = \int\limits_{-\frac{\pi}{2}}^{0}\frac{1+\cos 2t}{2}\,dt=\\=\frac{1}{2}\left(\int\limits_{-\frac{\pi}{2}}^{0}\,dt+ \int\limits_{-\frac{\pi}{2}}^{0}\cos 2t\,dt \right)  =\frac{1}{2}\left(\frac{\pi}{2} +0\right)= \frac{\pi}{4}
$$
