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(16-x^2)` and the x axis rotated about the x axis?

Answer 1

Volume `= (256pi)/3 \ "unit"^3`

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:
graph{(y-sqrt(16-x^2))=0 [-8, 8, -2, 5]}

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

`sqrt(16-x^2) = 0 => x =+- 4`, and the range of `y` is `y in [-4,4]`

By symmetry we can restrict `y in [0,4]` and double the result.

`y=sqrt(16-x^2) => y^2=16-x^2 => x=sqrt(16-y^2)`

The Volume of Revolution about `Ox` is given by:

`1/2V = int_(y=a)^(y=b)2pi \ y \ g(y) \ dy`
`\ \ \ \ \ \ = int_(y=0)^(y=4)2pi \ y \ sqrt(16-y^2) \ dy`
`\ \ \ \ \ \ = 2pi \ int_0^4 y \ sqrt(16-y^2) \ dy`

We can use a substitution to perform this integral;

Let `u=16-y^2 => (du)/(dy)= -2y`
When `{ (y=-4), (y=4) :} => { (u=16), (u=0) :}`

Substituting into the integral we get:

`1/2V = 2pi \ int_16^0 \ sqrt(u) (-1/2)\ du`
`\ \ \ \ \ \ = pi \ int_0^16 \ sqrt(u) \ du`
`\ \ \ \ \ \ = pi \ [u^(3/2)/(3/2)]_0^16`
`\ \ \ \ \ \ = 2/3pi \ [u^(3/2)]_0^16`
`\ \ \ \ \ \ = 2/3pi \ (16^(3/2) - 0)`
`\ \ \ \ \ \ = (128pi)/3`
`:. V = (256pi)/3`

Note:
The volume is that of a sphere of radius `4`, so we can validate the volume as `V=4/3pir^3 = (256pi)/3`