Let \( R \) be the region bounded by the following curves. Find the volume of the solid generated when \( R \) is revolved about the \( y \)-axis. \( y=x, y=4 x, y=16 \) Set up the integral that gives the volume of the solid. 16 \( \int_{0}\left(\frac{15}{16} \pi y^{2}\right) \) dy (Type exact answers.) The volume of the solid is (Type an exact answer.)

To find the volume of the solid generated when the region \( R \) bounded by the curves \( y = x \), \( y = 4x \), and \( y = 16 \) is revolved around the \( y \)-axis, we first need to determine the limits of integration and the outer and inner radii for the washers method. 1. To find the intersection points: - For \( y = x \) and \( y = 16 \): \( x = 16 \). - For \( y = 4x \) and \( y = 16 \): solving \( 16 = 4x \) gives \( x = 4 \). - Therefore, the curves intersect at points \( (4, 16) \) and \( (16, 16) \). 2. Next, we observe the functions: - The outer function is \( y = 4x \) for \( x \) in the range from 0 to 4. - The inner function is \( y = x \). 3. We will express \( x \) in terms of \( y \): - From \( y = x \), we have \( x = y \). - From \( y = 4x \), we have \( x = \frac{y}{4} \). 4. The volume \( V \) can be set up using the formula for volume when rotating around the \( y \)-axis: \[ V = \pi \int_{0}^{16} \left( R^2 - r^2 \right) \, dy \] where \( R = \frac{y}{4} \) (outer radius) and \( r = y \) (inner radius). Putting it together, we get: \[ V = \pi \int_{0}^{16} \left( \left(\frac{y}{4}\right)^2 - (y)^2 \right) dy = \pi \int_{0}^{16} \left( \frac{y^2}{16} - y^2 \right) dy = \pi \int_{0}^{16} \left( -\frac{15y^2}{16} \right) dy \] 5. Calculate the volume: Integrating gives: \[ V = -\frac{15\pi}{16} \cdot \left[ \frac{y^3}{3} \right]_{0}^{16} = -\frac{15\pi}{16} \cdot \left( \frac{16^3}{3} \right) = -\frac{15\pi}{16} \cdot \frac{4096}{3} = -\frac{15 \cdot 4096 \pi}{48} \] Thus, the exact volume is: \[ \frac{15 \cdot 4096 \pi}{48} = 1280\pi \] The integral setup for the volume is: \[ \pi \int_{0}^{16} -\frac{15}{16}y^2 \, dy \] And the volume of the solid is: \[ \frac{1280 \pi}{3} \]