Question Use the disk method to find the volume of the solid of revolution bounded by the \( x \)-axis and the graphs of \( f(x)=3 \sqrt{6 \sin x}, x=\frac{2 \pi}{3} \), and \( x=\pi \) rotated about the \( x \)-axis. Enter your answer in terms of \( \pi \).

To find the volume of the solid of revolution using the disk method, we start by identifying the bounds of integration which are \( x = \frac{2\pi}{3} \) and \( x = \pi \). The volume \( V \) is given by the integral: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] Substituting our function \( f(x) = 3 \sqrt{6 \sin x} \) into the formula gives: \[ V = \pi \int_{\frac{2\pi}{3}}^{\pi} [3 \sqrt{6 \sin x}]^2 \, dx \] This simplifies to: \[ V = \pi \int_{\frac{2\pi}{3}}^{\pi} 54 \sin x \, dx \] Now we compute the integral: \[ V = 54\pi \int_{\frac{2\pi}{3}}^{\pi} \sin x \, dx \] The integral of \( \sin x \) is \( -\cos x \), so we evaluate the definite integral from \( \frac{2\pi}{3} \) to \( \pi \): \[ V = 54\pi \left[-\cos x\right]_{\frac{2\pi}{3}}^{\pi} \] Calculating the boundaries: \[ = 54\pi \left[(-\cos(\pi)) - (-\cos(\frac{2\pi}{3}))\right] \] \[ = 54\pi [1 - \left(-\frac{1}{2}\right)] \] \[ = 54\pi \left(\frac{3}{2}\right) \] \[ = 81\pi \] Therefore, the volume of the solid of revolution is: \[ \boxed{81\pi} \]