Question

How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by `y =2/x`, y=0, x(1)=1 , x(2)=6 revolved about the y-axis?

Answer 1

`V=20pi`

Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness `deltax` between the `x`-axis and the curve at some particular `x`-coordinate it would have an area:

`delta A ~~"width" xx "height" = ydeltax = f(x)deltax`

If we then rotated this infinitesimally thin vertical line about `Oy` then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume `delta V` would be given by:

`delta V~~ 2pi xx "radius" xx "thickness" = 2pixdeltaA=2pixf(x)deltax`

If we add up all these infinitesimally thin cylinders then we would get the precise total volume `V` given by:

`V=int_(x=a)^(x=b)2pixf(x) dx`

So for this problem we have:

`V = int_1^6 2pix(2/x) dx`
`\ \ \ = int_1^6 4pi dx`
`\ \ \ = 4pi \ int_1^6 \ dx`
`\ \ \ = 4pi[x]_1^6`
`\ \ \ = 4pi(6-1)`
`\ \ \ = 20pi`