Volume of y=2x(2+x) from [0,2] rotated around y = 4, -2 and x = 4 using washer/ cylinder method?

I am having trouble finding the volume of this shape rotated around each of those axes, can anyone show me how to do this sort of integration?Volume of y=2x(2+x) from [0,2] rotated around y = 4, -2 and x = 4 using washer/ cylinder method?
I have created a graph for you that will take care of the axis is y = 4 case. To see this graph, which you must do or my solution won't make any sense, click on the following link:





http://i369.photobucket.com/albums/oo133鈥?/a>








(1) By Washers





Integrate 蟺 * ((2x / (2 + x)) - 4)^2 as x goes from 0 to 2





For the moment I'm just going to work with ((2x / (2 + x)) - 4)^2





((2x - 4 * (x + 2))^2) / (x + 2)^2





((2x - 4x - 8)^2 / (x + 2)^2 = (-8 - 2x)^2 / (x + 2)^2 =





((-(8 + 2x + -4 + 4))^2) / (x + 2)^2 = [(4 + 2x + 4) / (x + 2)]^2 =





[ (4 / (x + 2)^2) + 2^2]





Now 鈭?((4 / (x + 2))^2 + 4 dx = 鈭?((16 * (x + 2)^(-2) + 4 dx =





16 * (-2) * (x + 2)^(-3) + 4x = (-32 / (x + 2)^3) + 4x





For x = 2 this evaluates to: (-32 / 64 + 8) = (-1/2 + 8) = (15/2)





For x = 0 this evaluates to (-32 / 8) + 0 = - 4





So, not forgetting about the 蟺, we have





蟺 * ((15/2) - (-4)) = 蟺 * ((15/2) + 8) = 31蟺/2.....%26lt;%26lt;%26lt;%26lt;%26lt;....Answer by washers





To be continued

worms