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?
 A: Clearly, your area cannot be negative, so your result is immediately incorrect.
The system $$x = y^2 \\ x^2 + y^2 = 2x$$ is readily solved by substitution.  We have $$\begin{align}
0 &= x^2 + y^2 - 2x \\
&= x^2 + x - 2x \\
&= x^2 - x \\
&= x(x-1).
\end{align}$$  Hence $x \in \{0, 1\}$ and the full solution set is $$(x,y) \in \{(0,0), (1, -1), (1, 1)\}.$$    In the first quadrant, the area of interest may be expressed as
$$\begin{align}
\int_{x = 0}^1 \int_{y = \sqrt{x}}^\sqrt{2x-x^2} \, dy \, dx
&= \int_{x=0}^1 \sqrt{2x - x^2} - \sqrt{x} \, dx \\
&= \int_{x=0}^1 \sqrt{1 - (1-x)^2} - \int_{x=0}^1 \sqrt{x} \, dx \\
&= \int_{u=0}^1 \sqrt{1-u^2} \, du - \left[\frac{2}{3}x^{3/2}\right]_{x=0}^1 \\
&= \int_{\theta = 0}^{\pi/2} \sqrt{1 - \sin^2 \theta} \cos \theta \, d\theta - \frac{2}{3} \\
&= \int_{\theta = 0}^{\pi/2} \cos^2 \theta \, d \theta - \frac{2}{3} \\
&= \int_{\theta = 0}^{\pi/2} \frac{1 + \cos 2\theta}{2} \, d\theta - \frac{2}{3} \\
&= \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\theta = 0}^{\pi/2} - \frac{2}{3} \\
&= \left(\frac{\pi}{4} + 0 - 0 + 0\right) - \frac{2}{3} \\
&= \frac{\pi}{4} - \frac{2}{3}.
\end{align}$$  This step-by-step calculation should resolve all doubt.  This is because for a fixed $x \in [0,1]$, we note $$y = \sqrt{x} \le \sqrt{2x-x^2}.$$  Alternatively, we may change the order of integration, but this requires us to solve the equation for the circle in terms of $x$.  We can do this by completing the square:  $x^2 - 2x + y^2 = 0$ implies $$1-y^2 = x^2 - 2x + 1 = (x-1)^2,$$ hence $$x = 1 \pm \sqrt{1-y^2},$$ and we choose the negative root because we require $x < 1$.  Therefore, the area can be expressed as $$\int_{y=0}^1 \int_{x=1 - \sqrt{1-y^2}}^{y^2} \, dx \, dy.$$  Both integrals evaluate to $$\frac{\pi}{4} - \frac{2}{3}.$$
As has already been noted, the figure is misleading because the point $(1,1)$ lies directly above the center of the circle at $(1,0)$.

We can also check our solution by noting that the desired area is equal to the area under a parabola $y = x^2$ on $x \in [0,1]$, minus the area of a unit square from which a quarter of a unit circle has been cut out; i.e., this is simply $$\int_{x=0}^1 x^2 \, dx - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2}{3}.$$
A: $$\int\limits_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx =\\=
\int_\limits0^1\sqrt{2x-x^2}\,dx-\int\limits_0^1\sqrt{x}\,dx = \frac{\pi}{4} - \frac{2}{3}
$$
For first one:
$$\int_\limits0^1\sqrt{2x-x^2}\,dx = \int_\limits0^1\sqrt{1-(x-1)^2}\,dx=\\=\int\limits_{-1}^{0}\sqrt{1-y^2}\,dy=\int\limits_{-\frac{\pi}{2}}^{0}\cos^2t\,dt = \int\limits_{-\frac{\pi}{2}}^{0}\frac{1+\cos 2t}{2}\,dt=\\=\frac{1}{2}\left(\int\limits_{-\frac{\pi}{2}}^{0}\,dt+ \int\limits_{-\frac{\pi}{2}}^{0}\cos 2t\,dt \right)  =\frac{1}{2}\left(\frac{\pi}{2} +0\right)= \frac{\pi}{4}
$$
