Let \( R \) be the region bounded by the following curves. Find the volume of the solid generated when \( R \) is revolved about the \( x \)-axis. \( y=11 \sqrt{\sin x}, y=11, x=0 \) The volume of the solid is \( \square \) cubic units. (Type an exact answer.)

To find the volume of the solid generated when the region \( R \) bounded by the curves \( y = 11 \sqrt{\sin x} \), \( y = 11 \), and \( x = 0 \) is revolved about the \( x \)-axis, we can set up the integral using the disk method. First, we need to determine the points of intersection of the two curves \( y = 11 \sqrt{\sin x} \) and \( y = 11 \): 1. Setting \( 11 \sqrt{\sin x} = 11 \): \[ \sqrt{\sin x} = 1 \implies \sin x = 1 \] The solution for \( \sin x = 1 \) in the interval from \(0\) to \(\frac{\pi}{2}\) is \( x = \frac{\pi}{2} \). Next, we find the volume \( V \) when the region is revolved about the \( x \)-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \): The volume \( V \) can be calculated by the integral: \[ V = \pi \int_{0}^{\frac{\pi}{2}} (f(x)^2 - g(x)^2) \, dx \] where \( f(x) = 11 \) and \( g(x) = 11 \sqrt{\sin x} \). Thus, the volume becomes: \[ V = \pi \int_{0}^{\frac{\pi}{2}} (11^2 - (11 \sqrt{\sin x})^2) \, dx = \pi \int_{0}^{\frac{\pi}{2}} (121 - 121 \sin x) \, dx \] \[ = \pi \cdot 121 \int_{0}^{\frac{\pi}{2}} (1 - \sin x) \, dx \] Now, we calculate the integral: \[ \int (1 - \sin x) \, dx = x - (-\cos x) = x + \cos x. \] Thus, evaluating from \( 0 \) to \( \frac{\pi}{2} \): \[ \left[ x + \cos x \right]_{0}^{\frac{\pi}{2}} = \left( \frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right) \right) - \left( 0 + \cos(0) \right) = \frac{\pi}{2} + 0 - 1 = \frac{\pi}{2} - 1. \] Therefore, substituting back into the volume equation: \[ V = \pi \cdot 121 \left(\frac{\pi}{2} - 1 \right) = \frac{121\pi^2}{2} - 121\pi. \] Thus, the final volume of the solid is: \[ V = \frac{121\pi^2}{2} - 121\pi \quad \text{cubic units.} \]