1285. Find the areas of two figures bounded by the parabola $y^{2}=2 x$ and the circle $y^{2}=4 x-x^{2}$.
Solution. Transforming the equation of the circle, we get

$$
(x-2)^{2}+y^{2}=4
$$

from which it follows that the center of the circle lies at point $C(2 ; 0)$ and its radius $R=2$ (Fig. 80).

Solving the system of equations

$$
\left\{\begin{array}{l}
y^{2}=4 x-x^{2} \\
y^{2}=2 x
\end{array}\right.
$$

we obtain the following points

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-247.jpg?height=466&width=596&top_left_y=864&top_left_x=528)

Fig. 80 intersection points of the given curves: $O(0 ; \quad 0), \quad A(2 ; \quad 2), \quad B(2 ;-2)$.

From Fig. 80, it is clear that the area of the shaded figure $S=2 \int_{0}^{2}\left(\sqrt{4 x-x^{2}}-\sqrt{2 x}\right) d x=2\left(\int_{0}^{2} \sqrt{4 x-x^{2}} d x-\sqrt{2} \int_{0}^{2} \sqrt{x} d x\right)$. The first integral on the right side of the equation is equal to $\frac{1}{4}$ of the area of the circle, i.e., it is equal to $\pi$, therefore

$$
S=2\left(\pi--\left.\frac{2 \sqrt{2}}{3} x^{\frac{3}{2}}\right|_{0} ^{2}\right)=2\left(\pi-\frac{8}{3}\right)
$$

The area of the unshaded figure

$$
S_{1}=\pi R^{2}-S=4 \pi-2\left(\pi-\frac{8}{3}\right)=2\left(\pi+\frac{8}{3}\right)
$$