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, follow these steps: 1. **Identify the Function and Bounds:** - The function given is \( f(x) = 3 \sqrt{6 \sin x} \). - The region is bounded by \( x = \frac{2\pi}{3} \) and \( x = \pi \) along the \( x \)-axis. 2. **Set Up the Disk Method Integral:** - When revolving around the \( x \)-axis, the volume \( V \) is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] - Plugging in the given function and bounds: \[ V = \pi \int_{\frac{2\pi}{3}}^{\pi} \left(3 \sqrt{6 \sin x}\right)^2 \, dx \] 3. **Simplify the Integrand:** \[ \left(3 \sqrt{6 \sin x}\right)^2 = 9 \times 6 \sin x = 54 \sin x \] So the integral becomes: \[ V = 54\pi \int_{\frac{2\pi}{3}}^{\pi} \sin x \, dx \] 4. **Evaluate the Integral:** \[ \int \sin x \, dx = -\cos x \] Applying the limits: \[ V = 54\pi \left[ -\cos x \right]_{\frac{2\pi}{3}}^{\pi} = 54\pi \left( -\cos \pi + \cos \frac{2\pi}{3} \right) \] \[ \cos \pi = -1 \quad \text{and} \quad \cos \frac{2\pi}{3} = -\frac{1}{2} \] Substituting these values: \[ V = 54\pi \left( -(-1) + \left(-\frac{1}{2}\right) \right) = 54\pi \left( 1 - \frac{1}{2} \right) = 54\pi \times \frac{1}{2} = 27\pi \] **Final Answer:** \[ 27\pi \]