The region bounded by \( y=2+\sin x, y=0, x=0 \) and \( 2 \pi \) is revolved about the \( y \)-axis. Find the volume that results. Hint: \[ \int x \sin x d x=\sin x-x \cos x+C \]

To find the volume of the solid formed by revolving the region bounded by \( y = 2 + \sin x \), \( y = 0 \), \( x = 0 \), and \( x = 2\pi \) about the \( y \)-axis, we can use the method of cylindrical shells. The formula for the volume \( V \) when using cylindrical shells is given by: \[ V = 2\pi \int_{a}^{b} x f(x) \, dx \] where \( f(x) \) is the height of the shell at position \( x \), and \( [a, b] \) is the interval over which we are integrating. In our case, the height of the shell is given by \( f(x) = 2 + \sin x \), and we are integrating from \( x = 0 \) to \( x = 2\pi \). Thus, we have: \[ V = 2\pi \int_{0}^{2\pi} x (2 + \sin x) \, dx \] We can split this integral into two parts: \[ V = 2\pi \left( \int_{0}^{2\pi} 2x \, dx + \int_{0}^{2\pi} x \sin x \, dx \right) \] First, we calculate the integral \( \int_{0}^{2\pi} 2x \, dx \): \[ \int_{0}^{2\pi} 2x \, dx = 2 \left[ \frac{x^2}{2} \right]_{0}^{2\pi} = \left[ x^2 \right]_{0}^{2\pi} = (2\pi)^2 - 0^2 = 4\pi^2 \] Next, we calculate the integral \( \int_{0}^{2\pi} x \sin x \, dx \). We can use integration by parts, where we let: - \( u = x \) and \( dv = \sin x \, dx \) - Then, \( du = dx \) and \( v = -\cos x \) Using integration by parts: \[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C \] Now we evaluate \( \int_{0}^{2\pi} x \sin x \, dx \): \[ \int_{0}^{2\pi} x \sin x \, dx = \left[ -x \cos x + \sin x \right]_{0}^{2\pi} \] Calculating at the bounds: At \( x = 2\pi \): \[ -x \cos x + \sin x = -2\pi \cos(2\pi) + \sin(2\pi) = -2\pi(1) + 0 = -2\pi \] At \( x = 0 \): \[ -x \cos x + \sin x = -0 \cdot \cos(0) + \sin(0) = 0 \] Thus, \[ \int_{0}^{2\pi} x \sin x \, dx = -2\pi - 0 = -2\pi \] Now we can combine our results: \[ V = 2\pi \left( 4\pi^2 - 2\pi \right) = 2\pi (4\pi^2 - 2\pi) = 2\pi \cdot 2\pi (2\pi - 1) = 4\pi^2 (2\pi - 1) \] Thus, the volume of the solid is: \[ \boxed{4\pi^2 (2\pi - 1)} \]