Hi, I have the following HW question and I'm a little confused:
Find the volume of the solid generated by revolving the region bounded by y=1/sqrt(x) and y=0 for 1 <= x <= 2 about the line y=-1.
So far I have the integral
2pi (-1-y)(1-1/y^2) dy from -1 to 1, but that's not right. Can someone please help me?
The region of interest is just a small patch above the x-axis. y ranges from 0 to 1, but the right boundary changes at 1/√2. The volume consists of a cylindrical ring of thickness 1/√2 plus a shape with curved edge.
using shells of thickness dy
v = ∫2πrh dy
where r=y+1 and h=x-1 (from 1/√2 to 1)
and h=1 (from 0 to 1/√2)
v = ∫[0,1/√2] 2π(y+1)(1) dy
+ ∫[1/√2,1] 2π(y+1)(1/y^2-1) dy
= π(1/2 + √2) + π(-9/2+3√2+ln2)
= π(-4+42+ln2)
That's pretty strange. We can check it using discs of thickness dx:
v = ∫π(R^2-r^2) dx
where R=1+y and r=1
v = ∫[1,2] π((1+1/√x)^2-1^2) dx
= π(-4+4√2+ln2)
I guess we're good to go.