posted by Liz .
Find the volume of the solid generated by the region in the first quadrant bounded above by the 3x+y=6, below by the x-axis, and on the left by the y-axis, about the line x= -2.
If we include the area out to x = -2, we have
v = ∫[0,6] πr^2 dy
where r = 2+x = 2+(6-y)/3 = 4-y/3
v = π∫[0,6](4-y/3)^2 dy = 56π
But we have to subtract out the interior cylinder of radius 2 and height 6, or 24π, leaving us with just 32π generated by the rotating triangle.