Find the volume generated when the region bounded by \( f(x)=\frac{8}{\sqrt{x}} \) for \( 1 \leq x \leq 8 \) is rotated through \( 2 \pi \) about the \( x \)-axis. A \( \quad \pi \int_{1}^{8} \frac{64}{\sqrt{x}} d x \) B \( \pi \int_{1}^{8} \frac{8}{x} d x \) C \( \pi \int_{1}^{8} \frac{64}{x} d x \) D \( \pi \int_{1}^{8} \frac{8}{\sqrt{x}} d x \)

We are given the function \[ f(x)= \frac{8}{\sqrt{x}}, \] defined on the interval \(1 \leq x \leq 8\). The volume \( V \) of the solid formed by rotating this region about the \( x \)-axis is given by the formula \[ V = \pi \int_{a}^{b} \left[f(x)\right]^2 dx. \] Substitute \( f(x) \) into the formula: \[ V = \pi \int_{1}^{8} \left(\frac{8}{\sqrt{x}}\right)^2 dx. \] Square the function: \[ \left(\frac{8}{\sqrt{x}}\right)^2 = \frac{64}{x}. \] Thus, the volume becomes: \[ V = \pi \int_{1}^{8} \frac{64}{x} \, dx. \] Comparing this expression with the given options, it matches option C: \[ \pi \int_{1}^{8} \frac{64}{x} \, dx. \]