I just need help with two more AP Calculus problems, and then I'll be done for the year. Find volume of region generated by rotating shaded area about the x-axis: y = 3 - (6/x) x = 5 Window: x min: 0 x max: 6 y min: 0 y max: 2 This is what I did for this problem, not completely sure if I did this correctly: y = 3 - (6/x), x = 5 nr^2 pi INTEGRAL(from 2 to 5) [ (3-(6/x))^2 ] =15.122 The second problem: The base of a solid is the region between the line y = 4 and parabola y = x^2. The cross sections perpendicular to x-axis are semicircles, find volume of solid. Thank you for taking the time to read this, OSP

Its been a while, so someone should correct my mistakes but... For the first question, the shape starts at 2 (where y=0) and ends at x=5. Area of a circle is piR^2 so I'm thinking it would be the integral from 2 to 5 of pi((3 - (6/x))^2). You can evaluate that. Some one else check this. The second question is similar.

First one looks right. Second one: PI[integral(-2 to 2) of (x^2/2)^2] I did this all in my head but that should be right. edit: yeah Frankie that's what he had for the 1st one