Find the volume of the solid generated by revolving the given region about the line x=16:
y=4-x,y=0,y=2,x=0.
2 answers
Andrew/Isabella/James/Anonymous -- please use the same name for your posts.
The boundary changes, so we need to divide the area into two pieces.
Using discs (washers) of thickness dy,
v = ∫[0,2] π(R^2-r^2) dy
where R=16 and r=14
+ ∫[0,2] π(R^2-r^2) dy
where R=14 and r=16-(4-y)
using shells of thickness dx, we have
v = ∫[0,2] 2πrh dx
where r=16-x and h=2
+ ∫[2,4] 2πrh dx
where r=16-x and h=4-x
Using discs (washers) of thickness dy,
v = ∫[0,2] π(R^2-r^2) dy
where R=16 and r=14
+ ∫[0,2] π(R^2-r^2) dy
where R=14 and r=16-(4-y)
using shells of thickness dx, we have
v = ∫[0,2] 2πrh dx
where r=16-x and h=2
+ ∫[2,4] 2πrh dx
where r=16-x and h=4-x
0 answers