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.)
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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 \[ \frac{3\pi}{2} \text{ square units.} \]