Find the area of the surface generated by revolving the curve \( y=\sqrt{2 x-x^{2}}, 0.5 \leq x \leq 1.25 \), 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 \( y = \sqrt{2x - x^2} \) from \( x=0.5 \) to \( x=1.25 \) about the \( x \)-axis, we can use the formula for the surface area of revolution: \[ A = 2\pi \int_a^b y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] First, we need to differentiate \( y \) with respect to \( x \): \[ y = (2x - x^2)^{1/2} \] 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 compute \( 1 + \left( \frac{dy}{dx} \right)^2 \): \[ \left( \frac{dy}{dx} \right)^2 = \frac{(1 - x)^2}{2x - x^2} \] \[ 1 + \left( \frac{dy}{dx} \right)^2 = 1 + \frac{(1 - x)^2}{2x - x^2} = \frac{(2x - x^2) + (1 - 2x + x^2)}{2x - x^2} \] \[ = \frac{1}{2x - x^2} \] Thus, we have: \[ \sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \frac{1}{\sqrt{2x - x^2}} \] Now we can put this into our surface area formula. Remember to multiply by \( y \): \[ A = 2\pi \int_{0.5}^{1.25} \sqrt{2x - x^2} \cdot \frac{1}{\sqrt{2x - x^2}} \, dx = 2\pi \int_{0.5}^{1.25} 1 \, dx \] Evaluate the integral: \[ A = 2\pi \left[ x \right]_{0.5}^{1.25} = 2\pi \left( 1.25 - 0.5 \right) = 2\pi \cdot 0.75 = \frac{3\pi}{2} \] Thus, the area of the surface generated is \( \frac{3\pi}{2} \) square units.