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(C) The given curves are $x^2 + y^2 = 8$ (a circle with center $(0,0)$ and radius $2\sqrt{2}$) and $y^2 = 2x$ (a parabola opening to the right).To fin

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$2\pi + \frac{4}{3}$

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(C) The given curves are $x^2 + y^2 = 8$ (a circle with center $(0,0)$ and radius $2\sqrt{2}$) and $y^2 = 2x$ (a parabola opening to the right).To find the intersection points,substitute $y^2 = 2x$ into the circle equation: $x^2 + 2x - 8 = 0$.$(x+4)(x-2) = 0$,so $x = 2$ (since $x \ge 0$ for the parabola).At $x = 2$,$y^2 = 4$,so $y = \pm 2$.The area of the smaller portion is symmetric about the $x$-axis,so we calculate $2 	imes \int_{0}^{2} \sqrt{2x} \, dx + 2 	imes \int_{2}^{2\sqrt{2}} \sqrt{8 - x^2} \, dx$.First integral: $2 \int_{0}^{2} \sqrt{2} x^{1/2} \, dx = 2\sqrt{2} [\frac{2}{3} x^{3/2}]_{0}^{2} = 2\sqrt{2} 	imes \frac{2}{3} 	imes 2\sqrt{2} = \frac{16}{3}$.Second integral: $2 \int_{2}^{2\sqrt{2}} \sqrt{(2\sqrt{2})^2 - x^2} \, dx = 2 [\frac{x}{2} \sqrt{8-x^2} + \frac{8}{2} \sin^{-1}(\frac{x}{2\sqrt{2}})]_{2}^{2\sqrt{2}}$.$= 2 [ (0 + 4 \sin^{-1}(1)) - (1 \sqrt{4} + 4 \sin^{-1}(\frac{1}{\sqrt{2}})) ] = 2 [ 4(\frac{\pi}{2}) - (2 + 4(\frac{\pi}{4})) ] = 2 [ 2\pi - 2 - \pi ] = 2\pi - 4$.Total area = $\frac{16}{3} + 2\pi - 4 = 2\pi + \frac{4}{3}$.

The area of the smaller portion between the curves $x^2 + y^2 = 8$ and $y^2 = 2x$ is

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