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.
\[
y=x, y=4 x, y=16
\]
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 ：\[y=x, y=4 x, y=16\]

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 \(\boxed{8192\pi}\).