1 Let $\mathrm{R}$ be the region bounded by the following curves. Find the volume of the solid generated when $R$ is revolved about the $y$-axis.
$$y=x,y=4x,y=16$$
Set up the integral that gives the volume of the solid.
(Type exact answers.)

Step 1 ：Let $\mathrm{R}$ be the region bounded by the following curves. Find the volume of the solid generated when $R$ is revolved about the $y$-axis.

Step 2 ：$$y=x,y=4x,y=16$$

Step 3 ：Set up the integral that gives the volume of the solid. (Type exact answers.)

Step 4 ：The volume of a solid of revolution can be found using the disk method or the shell method. In this case, since we are revolving around the y-axis, the shell method is more appropriate. The shell method formula is:

Step 5 ：$$V=2\pi {\int }_a^{b}r(x)h(x)dx$$

Step 6 ：where $r(x)$ is the distance from the axis of rotation to the shell (in this case, x), and $h(x)$ is the height of the shell (in this case, 4x - x = 3x). The limits of integration a and b are the y-values where the curves intersect, which in this case are 0 and 16.

Step 7 ：So the integral we need to set up is:

Step 8 ：$$V=2\pi {\int }_0^{16}x(3x)dx$$

Step 9 ：$$V=8192\pi$$

Step 10 ：Final Answer: The integral that gives the volume of the solid is $\boxed{{\displaystyle 8192\pi }}$.