The area (in sq. units) of the region $(x, y): y^2 \geq 2x$ and $x^2 + y^2 \leq 4x$, $z \geq 0$, $y \geq 0$ is
Step-by-Step Solution
Key Concept: Finding intersection points of two curves and using integration with respect to $x$ to compute enclosed area between parabola and circle.
To find the area between the parabola $y^2 = 2x$ (opens rightward) and circle $x^2 + y^2 = 4x$ (centered at $(2,0)$ with radius $2$), we first find intersection points by solving $x^2 + y^2 = 4x$ and $y^2 = 2x$ simultaneously. Substituting: $x^2 + 2x = 4x$, so $x^2 = 2x$, giving $x = 0$ or $x = 2$. When $x = 0$, $y = 0$ (point $A$); when $x = 2$, $y = \pm 2$ (points $B$ and $C$). The area is computed by integrating from $x = 0$ to $x = 2$: Area $= \left[\int_0^2 \sqrt{4x - x^2}\, dx - \sqrt{2}\int_0^2 \sqrt{x}\, dx\right]$. The first integral evaluates to $[\frac{x-2}{2}\sqrt{4x-x^2} + 2\sin^{-1}(\frac{x-2}{2})]_0^2 - \sqrt{2}\left(\frac{2}{3}\right)[x^{3/2}]_0^2 = [0 - 2\sin^{-1}(-1)] - \sqrt{2}\cdot\frac{2\sqrt{2}}{3} = [\pi] - \frac{4}{3}$. Multiplying by $2$ for symmetry gives Area $= 2[\pi - \frac{4}{3}] = [\pi - \frac{8}{3}]$ sq. units, which equals approximately $\pi - 2.67$ or in the form shown: $\pi - \frac{8}{3}$ sq. units.
Correct Answer: 97