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?
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 $$\int_{x=0}^1 x^2 \, dx - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2}{3}.$$