Quick visualization question about solids of revolution?

What do the the cross sectional slices look like? Like the segment of a not quite circular pizza pie.

What are the expressions for the area of fthe cross sectional slice at any point x in the interval of the rotation?

Area=Int[1 to 8](9-x)dx-Int[1 to 8](8/x)dx

What is the shape formed when the area between the functions xy=8 and x+y=9 is rotated around the x-axis? The outside surface is a truncated cone with an inner surface bulged like that pizza segment.

ok Destin--- in case you utilize the shell technique you will be waiting to apply x= 0 and x = 3 as your limits of integration through fact the functionality is non-bring about any respect factors in that variety. V = 2pi*Int(x*(a million/(x^2+a million))dx = 2pi*Int(x/(x^2+a million))dx from x=0 to x=3. you will could do a substitution u = x^2+a million du = 2xdx . this components V = pi*Int(du/u) from u = a million to u = 10. you will be waiting to do something.