Question

How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region `y = sqrt(x`), `y = 0`, `y = 12 - x` rotated about the x axis?

Answer 1

Volume = `(99pi)/2`

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)2pi \ x \ f(x) \ dx`

Similarly if we rotate about `Ox` instead of `Oy` then we get the volume as:

`V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy`

So for this problem we have:

We need the point of intersection for the bounds of integration;

`12-x = sqrt(x) => (12-x)^2 =x`
`:. 144-24x+x^2=x`
`:. x^2 - 25x + 144 =0`
`:. (x-9)(x-16) =0 => x=9,16`

And our solution is `x=9`, (as `x=16` is the solution for `-sqrt(x)`)

`x=9 => y=12-x = 3`

So the bounding curves `y=sqrt(x)` and `y=12-x` meet at `(9,3)`

And we also need `x=g(y)`, rather than `y=f(x)`

`y=12-x => x=12-y`
`y = sqrt(x) \ \ \ \ \ => x=y^2`

Then the required volume is given by:

`V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy`
`\ \ \= 2pi int_(y=0)^(y=3) \ y{(12-y) - (y^2)} \ dy`
`\ \ \= 2pi int_(y=0)^(y=3) \ y(12-y - y^2) \ dy`
`\ \ \= 2pi int_(y=0)^(y=3) \ 12y-y^2 - y^3) \ dy`
`\ \ \= 2pi [ 6y^2 - 1/3y^3 - 1/4y^4 ]_0^3`
`\ \ \= 2pi (54-9-81/4)`
`\ \ \= (99pi)/2`