The region described can be divided into two sections. The first section is from x=0 to x=2 and is the area under y=(1/4)x^3. Note that y=-x+4 and y=(1/4)x^3 intersect at (2,2). The second section is from x=2 to x=4 and is simply the area under y=-x+4.
Using concentric cylinders...
Each cylinder has an infinitesimal volume that is equal to 2pi*r*h*dx. If it helps, you can envision one such cylinder as cut and rolled out to form a rectangle. That rectangle has height equal to that of the function 'h' ((1/4)x^3, for the first section) and 'r' is just x so the length is 2*pi*x. Finally, for the 3rd dimension, the thickness is dx.
For the first section then:
And by integration, we can add all these infinitesimal volumes, allowing for the varying heights of the cylinders as f(x) changes over the limits of integration:
evaluated at x=2 minus value at x=0
And the 2nd section... ( x=2 to x=4 ):
evaluated at x=4 minus value at x=2
(approx. 43.56 cubic units)
I think this is right. Its late though, and for me its always dangerous doing math when tired...
