Use the functions:
y=x+1, y=0, x=0, x=2
Find the volume when rotated around y=3
Find the volume when rotated around y=-1
PI[tex]\int[/tex] R2dx or perhaps PI[tex]\int[/tex] R2 - r2dx
The Attempt at a Solution
i don't think that this is a particularly hard question, but I just can't seem to get the right answer. (perhaps I'm making a silly mistake)
So first I graphed the equations, all in the first quadrant, vertical line at x=2. bounded region looks like a trapezoid.
PI[tex]\int[/tex] (from 0-2) of [3-(x+1)]2 dx
however the answer is 46pi/3 and I get 8pi/3 <-- very obviously wrong.
is this a washer method problem?
for the next, PI[tex]\int[/tex] (from 0-2) of [1+(x+1)]2 dx
i get 32pi/3 and the answer is 50pi/3