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=f(x)=3x-x^2` and x axis revolved about the x=-1?

Answer 1

`V=(45pi)/2 units^3`

Explanation:

Imagine a cylindrical shell as a rectangular prism with width `f(x)`, length `2pir`, and thickness `deltax`. The reason for the length being `2pir` is that if we unravel a cylindrical shell into a rectangular prism, the length corresponds to the circumference of the circular cross-section of the original cylinder.

The volume of the cylindrical shell is width x length x thickness (height)

Now width `=f(x)=3x-x^2`

`:.` Volume of shell`=2pir(3x-x^2)deltax`

If we sketch the parabola `y=3x-x^2` we can see that the region bound by this parabola and the x-axis resides in the domain `0<=x<=3`

Also, the radius of the cylindrical shell is considered to be the distance from the axis of revolution `x=-1` and the edge of the shell (i.e. at a position `x` within the domain `0<=x<=3`)

Hence, r`=1+x`

`:.`Volume of shell`=2pi(1+x)(3x-x^2)deltax`

i.e. the change in volume, `deltav=2pi(1+x)(3x-x^2)deltax`

To find the volume, we limit the thickness of the shells, and find their summation within the domain `0<=x<=3`

`V=lim_(deltax->0)sum_(x=0)^3(2pi(1+x)(3x-x^2)deltax)`

`=2piint_0^3(1+x)(3x-x^2)dx`

`=2piint_0^3(3x+2x^2-x^3)dx`

`=2pi[3/2x^2+2/3x^3-1/4x^4]_0^3`

`=2pi[45/4]`

`=(45pi)/2 units^3`