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Drawing a picture is always a good thing to do for these types of problems.

Start by sketching the graphs of y=8, y=x3, and x=0. We then see that the region to be revolved about the x-axis is in the first quadrant. It is bounded above by the line y=8, on the left by the y-axis, and on the right by the graph of y=x3. Note that the point of intersection in the upper right is (2,8) and can be found by solving the equation 8=x3.

The region is shown, shaded in blue, below:

Now, you generate the solid by revolving this region about the x-axis. The cylindrical shells are generated by revolving a horizontal line segment, shown in green in the diagram, at a fixed y-value about the x-axis.

These line segments "start" at y=0 and "end" at y=8. Thus the integral giving the volume of the solid of revolution is with respect to y and is of the form

∫y=8y=02πry⋅hydy

where ry is the radius of the shell at y and hy is the height of the shell at y.

The height of the shell at y is the length of the line segment at y. Since the length of the line segment at y is the x-coordinate of its right hand endpoint, we have

hy=y1/3.

Keep in mind that we want to write hy in terms of y, since the integral is with respect to y.

The radius of the shell is the height above the x-axis of the line segment: