Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line.

y = x3
y = 0
x = 3

a)the line x = 4

To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 4, we will use the disk method.

The disk method involves integrating the area of a series of disks that make up the solid. Each disk has a radius and a thickness, which will be used to calculate its volume.

First, let's find the points of intersection between the graphs of the equations. The graphs are y = x^3, y = 0, and x = 3. The intersection points are (3, 0) and (3, 27).

To set up the integral for the disk method, we need to determine the radius and height of each disk. The radius of each disk is the distance between the line of revolution (x = 4) and the function y = x^3. The height of each disk is the thickness of the disk, which we will call Δx.

Since we are revolving the region about the line x = 4, the radius of each disk can be calculated as |4 - x^3|. And the thickness of each disk, Δx, is simply dx.

The integral for the disk method is given by the following formula:

V = ∫[a, b] π(radius)^2 Δx

In this case, we integrate from x = 3 to x = 4, since 4 is the line of revolution.

V = ∫[3, 4] π(4 - x^3)^2 dx

Now, we can simplify and solve the integral to find the volume of the solid:

V = π∫[3, 4] (16 - 8x^3 + x^6) dx

Integrating the function, we get:

V = π[16x - 2x^4 + (1/7)x^7] from 3 to 4

Plugging in the limits of integration, we have:

V = π[(16(4) - 2(4^4) + (1/7)(4^7)) - (16(3) - 2(3^4) + (1/7)(3^7))]

V = π[(64 - 2(256) + (1/7)(16384)) - (48 - 2(81) + (1/7)(2187))]

V = π[(64 - 512 + 2340.5714) - (48 - 162 + 312.4286)]

V = π[1892.5714 - 597.5714]

V ≈ 4892.138 cubic units

Therefore, the volume of the solid generated by revolving the region about the line x = 4 is approximately 4892.138 cubic units.