Find the area of the surface generated by revolving the curve \( y=\sqrt{2 x-x^{2}}, 0.25 \leq x \leq 1 \), about the \( x \)-axis. The area of the surface generated is \( \square \) square units. (Type an exact answer, using \( \pi \) as needed.)

To find the area of the surface generated by revolving the curve \( y = \sqrt{2x - x^2} \) around the \( x \)-axis, we can use the formula for the surface area of revolution: \[ S = 2\pi \int_a^b y \sqrt{\left(\frac{dy}{dx}\right)^2 + 1} \, dx \] First, we need to compute \( \frac{dy}{dx} \): Given \( y = \sqrt{2x - x^2} \), we can differentiate it using the chain rule: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{2x - x^2}}(2 - 2x) = \frac{1 - x}{\sqrt{2x - x^2}} \] Next, we need to simplify \( \sqrt{\left(\frac{dy}{dx}\right)^2 + 1} \): \[ \left(\frac{dy}{dx}\right)^2 = \left(\frac{1 - x}{\sqrt{2x - x^2}}\right)^2 = \frac{(1 - x)^2}{2x - x^2} \] So, \[ \left(\frac{dy}{dx}\right)^2 + 1 = \frac{(1 - x)^2}{2x - x^2} + 1 = \frac{(1 - x)^2 + (2x - x^2)}{2x - x^2} = \frac{1 - 2x + x^2 + 2x - x^2}{2x - x^2} = \frac{1}{2x - x^2} \] Thus, \[ \sqrt{\left(\frac{dy}{dx}\right)^2 + 1} = \frac{1}{\sqrt{2x - x^2}} \] Now we can substitute this into our surface area integral: \[ S = 2\pi \int_{0.25}^{1} y \sqrt{\left(\frac{dy}{dx}\right)^2 + 1} \, dx = 2\pi \int_{0.25}^{1} \sqrt{2x - x^2} \cdot \frac{1}{\sqrt{2x - x^2}} \, dx = 2\pi \int_{0.25}^{1} 1 \, dx \] Now we compute the integral: \[ S = 2\pi \left[ x \right]_{0.25}^{1} = 2\pi \left(1 - 0.25\right) = 2\pi \cdot 0.75 = \frac{3\pi}{2} \] Therefore, the area of the surface generated is \[ \frac{3\pi}{2} \text{ square units.} \]