The two curves (one a U shape, the other an inverted U) intersect at the solution to the equation x^2+2=-x^2+2x+6. This reduces to the quadratic 2x^2-2x-4=0 which becomes x^2-x-2=0. This factorises (x+1)(x-2)=0 so the solutions are -1 and 2, and y=3 and 6. The bounds are limited by the requirement 0<=x<=3. Therefore the bounds for the volume captured by the two curves are 0<=x<=2.

Consider first the volume of rotation of the inverted U, the superior curve, y=-2x^2+2x+6 for 0<=x<=2. Consider the disc width dx standing on its side with radius y. The volume of the disc is (pi)y^2dx=(pi)(-x^2+2x+6)^2dx. A stack of such discs placed on its side is the volume between the curve and the x axis. So we need to integrate (pi)(x^4-4x^3-8x^2+24x+36)dx for 0<=x<=2. That is, (pi)[x^5/5-x^4-8x^3/3+12x^2+36x] for 0<=x<=2.

Now we have to do the same for the other curve y=x^2+2. The volume of the thin disc is (pi)(x^2+2)^2dx=(pi)(x^4+4x^2+4)dx. This integrates as (pi)[x^5/5+4x^3/3+4x] for 0<=x<=2.

Combine the two volumes by subtracting the latter from the former (inferior from the superior) we get (pi)[-x^4-4x^3+12x^2+32x] for 0<=x<=2. So this evaluates to (pi)(-16-32+48+64)=64(pi)=201.06 cubic units.