91Ó°ÊÓ

To find the point(s) of intersection between the circle and the parabola, we replace \(x\) from the parabola equation into the circle equation: \(x^{2}+y^{2}=2 \Rightarrow y^{2} + y^{2} = 2 \Rightarrow 2y^2 = 2 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1\) Find the corresponding \(x\) values for each \(y\) value: When \(y=1\), \(x=y^{2}\Rightarrow x=1\) When \(y=-1\), \(x=y^{2}\Rightarrow x=1\) So, the points of intersection are \((1,1)\) and \((1,-1)\).

We will use the method of integration to find the area of the region common to both. To do this, we need first to express the given equation in terms of the same variable, in this case, \(x\). Rearrange the parabola equation \(y^{2}=x\) to have \(y\) in terms of \(x\) as follows: \(y = \sqrt{x}\) (upper half of the parabola) \(y = -\sqrt{x}\) (lower half of the parabola) We also express the circle equation \(x^{2}+y^{2}=2\) as \(y\) in terms of \(x\): \(y = \sqrt{2-x^{2}}\) \(y = -\sqrt{2-x^{2}}\) For the area enclosed by these curves, we will integrate the difference between the functions for the circle and parabola for both the upper and lower half and sum the results: \(A = \int_{0}^{1}[\sqrt{2-x^{2}}-\sqrt{x}] dx + \int_{0}^{1}[-\sqrt{2-x^{2}}-(-\sqrt{x})] dx\)

Now we can find the separate integrals and add their results to find the area. \(\int_{0}^{1}[\sqrt{2-x^{2}}-\sqrt{x}] dx = \frac{2\pi}{3} - \frac{4}{3}\) \(\int_{0}^{1}[-\sqrt{2-x^{2}}+\sqrt{x}] dx = - (\frac{2\pi}{3} - \frac{4}{3})\) Adding the integration results: \(A = (\frac{2\pi}{3} - \frac{4}{3}) + (- (\frac{2\pi}{3} - \frac{4}{3})) = \frac{4}{3} - \frac{2\pi}{3} + \frac{4}{3} - \frac{2\pi}{3} = \frac{8}{3} - \frac{4\pi}{3}\) So, the area of the region common to the circle and the parabola is \(A = \frac{8}{3} - \frac{4\pi}{3}\).