Let R be the region bounded by the curves y=0 and y=x^2+x between x=0 and x=1. Compute the volume of the solid of revolution obtained when R is rotated about the axis y=-1.
Disc method: integral of pi*r^2 = volume from the bounds
The Attempt at a Solution
The bounds are obvious = 0 to 1.
The hard part is accounting for the empty area inbetween as we rotate around y=-1 rather then y=0.
What I did to solve the problem was moved the volume up 1 to the origin. I then solved the entire volume of the problem from 0 to 1 using the radius of x^2+x+1 (the +1 to account the entire area).
My answer here was 37pi/10.
I then went back and took the volume of the squared middle region which had radius of 1 and then subtracted that from my first answer.
My final answer was 27pi/10.
Can anyone confirm my train of thought/answer?