Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

y = 6x
y = 18
x = 0

V = Int(2pi*r*h dx)[0,3]

where
r = x
h = 18-y

V = 2pi*Int(x*(18-6x) dx)[0,3]
= 2pi * Int(18x - 6x^2 dx)[0,3]
= 2pi*(9x^2 - 2x^3)[0,3]
= 2pi*(27-18)
= 18pi

To use the shell method, we need to set up and evaluate an integral that represents the volume of the solid generated by revolving the plane region about the y-axis.

First, let's visualize the given plane region. We have the lines y = 6x, y = 18, and x = 0.

The region is bounded by the x-axis on the left (x = 0), the line y = 6x on the right, and the line y = 18 from above.

To find the limits of integration, we need to determine the x-values where the lines intersect.

Setting y = 6x equal to y = 18, we get:

6x = 18
x = 18/6
x = 3

Therefore, the limits of integration for x are from 0 to 3.

Now, let's consider a vertical slice of the solid at an x-value x (between 0 and 3). We'll call this width Δx.

The height of this vertical slice can be determined by looking at the functions that bound the region: y = 6x and y = 18.

The height is given by the difference between the two functions:

Height = (y = 18) - (y = 6x)
Height = 18 - 6x

The circumference of this vertical slice is given by the perimeter of the curve traced by rotating the line segment about the y-axis.

The circumference is given by 2π times the radius, which is the x-value:

Circumference = 2πx

Therefore, the volume of this thin slice is ∆V = Height * Circumference * ∆x.

To find the total volume, we need to sum the volumes of all the thin slices.

This results in the integral formula:

V = ∫[from x=0 to x=3] (2πx * (18 - 6x)) dx

Now, let's evaluate this integral to find the volume:

V = ∫[from x=0 to x=3] (2πx * (18 - 6x)) dx

To evaluate this integral, we can use the power rule and the constant rule of integration.

Let's simplify the integrand:

V = ∫[from x=0 to x=3] (36πx - 12πx^2) dx

Using the power rule of integration, we get:

V = [(18πx^2) - (4πx^3/3)] from x=0 to x=3

Evaluating this expression, we get:

V = [(18π(3)^2) - (4π(3)^3/3)] - [(18π(0)^2) - (4π(0)^3/3)]

V = [162π - 108π/3] - [0 - 0]

V = 162π - 36π

V = 126π

Therefore, the volume of the solid generated by revolving the plane region about the y-axis is 126π units^3.