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?

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#