Question 1158831
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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:

{{{ dV[1] = (2*pi*r*h) dx = 2*pi*x*(1/4)x^3*dx}}}

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:

{{{  V[1] = int(2*pi*x*(1/4)x^3, dx, 0,2) }}}
{{{  V[1] = (pi/2)*(1/5)*x^5 }}} evaluated at x=2 minus value at x=0

{{{ V[1] = (pi/10)(32-0) }}}
{{{ V[1] = (32pi/10) = 16pi/5 }}}  

And the 2nd section... ( x=2 to x=4 ):

{{{ dV[2] = (2*pi*r*h) dx = 2*pi*x*(4-x) * dx }}}
{{{ V[2] = int(2*pi*x*(4-x), dx, 2, 4) }}}
{{{ V[2] = 2pi*(2x^2-(1/3)x^3) }}} evaluated at x=4 minus value at x=2
{{{ V[2] = 32pi/3 }}}


{{{ V = V[1]+V[2] = 16pi/5 + 32pi/3 = 48pi/15 + 160pi/15 = highlight(208pi/15)}}}   (approx. 43.56 cubic units)

I think this is right. Its late though, and for me its always dangerous doing math when tired...