91Ó°ÊÓ

In order to find the area, we must first understand the geometric meaning of these inequalities. The given inequalities are: 1. \(x^{2}+y^{2} \leq 2 a x\) 2. \(y^{2} \geq a x, x \geq 0\) We must convert these inequalities into equations to find their respective curves. 1. \(x^{2}+y^{2} = 2 a x\) is a circle with center \((a,0)\) and radius \(a\). 2. \(y^{2} = a x\) is a parabola with vertex at the origin and axis parallel to the \(x\)-axis.

To find the area, we must find the region bounded by the curves. For this, we must find the intersection points between the circle and the parabola. We solve the two equations by substitution: \(x^{2} + y^{2} = 2 a x \implies x^{2} - 2 a x + y^2 = 0\) \(y^2 = a x\) We substitute the equation of the parabola into the equation of the circle: \(x^{2} - 2 a x + (a x) = 0\) \(x^{2} - a x = 0\) Thus, \(x(x-a) = 0\). This equation has two solutions: \(x=0\) and \(x=a\). For \(x=0\), we substitute it into the equation of the parabola to get \(y^2 = 0 \implies y=0\). For \(x=a\), we again substitute it into the equation of the parabola to get \(y^2 = a^2 \implies y = \pm a\). Hence, the intersection points are \((0, 0)\), \((a, a)\), and \((a, -a)\).

Now, we will find the area of the region bounded by the circle and the parabola. Note that the region is symmetric about the \(x\)-axis. So, we will find the area of the region in the first quadrant (\(y \geq 0\)) and then multiply it by 2. We set up the integral for the area of the region in the first quadrant as follows: \(Area = 2 \int_{0}^{a} (y_{1} - y_{2}) dx\) where \(y_1\) is the equation of the parabola, and \(y_2\) is the equation of the circle in the first quadrant. We first solve the circle equation for \(y\): \(x^{2}+y^{2} = 2ax \implies y^2 = 2ax - x^2 \implies y = \sqrt{2ax - x^2}\) Now, we can set up the integral as: \(Area = 2 \int_{0}^{a} (\sqrt{ax} - \sqrt{2ax - x^2}) dx\)

To evaluate the integral, we will use substitution: \(u = 2ax - x^2 \implies du = (2a - 2x) dx\) Now the integral becomes: \(Area = 2 \int_{0}^{a} (\sqrt{ax} - \sqrt{u}) \cdot \frac{du}{2a-2x}\) \(Area = 2 \int_{0}^{a^2} (\frac{\sqrt{a^2 - u} - \sqrt{u}}{2a - 2\sqrt{a^2 - u}}) du\) Now, we will integrate using partial fractions decomposition: \(Area = \int_{0}^{a^2} (\frac{2\sqrt{a^2 - u}}{2 a - 2\sqrt{a^2 - u}} - \frac{2\sqrt{u}}{2a - 2\sqrt{a^2 - u}}) du\) After integrating, we get: \(Area = [-\sqrt{a^2 - u} + a\arcsin(\frac{\sqrt{u}}{a})]_{0}^{a^2}\) \(Area = a^2\pi\) Hence, the area of the region bounded by the given inequalities is \(a^2\pi\).