A disk will have radius extending from the graph to the line of rotation. If the line of rotation is y= 8, then the radius is R(x)= |y- 8|= |f(x)- 8|. For B, the non-shaded area, since it goes up to the line y= 8, the volume would be

where the upper limit, is the value of x where the graph of y= f(x) crosses the line y= 8. That is, .

For (a) where there is unshaded area between the shaded area and the line of rotation, you cannot use the "disk method" directly. What you can do is use the "washer" method. A washer has an outer radius of an inner radius of and so its area is the area between those two radii, . That is really the same as doing two separate "disk" calculations, to find an "outer volume" and "inner volume" and subtracting:

For a problem like the one you show, where one edge is just y= 0, that outer volume is just the volume of a cylinder- which here would be and, again, h is the value of x where the graph crosses the line of rotation, f(x)= 8.