The graph of this function resembles a parabola on its side with vertex pointing to the right (positive) intercepting the y axis at 0 and 4. We can find the vertex by differentiating the function: dx/dy=8y-3y^2. The derivative is zero at y=0 and 8/3. Therefore the vertex is x=4(8/3)^2-(8/3)^3=256/9-512/27=256/27, making the vertex (256/27,8/3).
Now consider rotation about the x axis. We will get a type of parabolic torus or doughnut which can be split into two parts: the upper part from x=0 to 256/27, while y=4 is decreasing to 8/3; the lower part from x=0 to 256/27 and y=0 to 8/3. The upper and lower parts have different integrals that are each "washers" of thickness dx. The upper washer has an outer radius of y and an inner radius of 8/3; the lower washer has an outer radius of 8/3 and an inner radius of y.
The integrals are: upper volume VU=(pi)integral(y^2-(8/3)^2)dx) and lower volume VL=(pi)integral((8/3)^2-y^2)dx). The complete volume of rotation is VU+VL. The limits of VU and VL are x=0 to 256/27, and y=4 (VU) to 8/3 and 0 (VL) to 8/3. We can replace dx by (8y-3y^2)dy allowing us to integrate with respect to y.
VU=(pi)integral(y^2-(64/9))(8y-3y^2)dy) and VL=-VU. The limits of y for VU and VL are different: [4,8/3] for VU and [0,8/3] for VL.
VU=(pi)integral(8y^3-3y^4-512y/9+64y^2/3)dy)=(pi)(2y^4-3y^5/5-256y^2/9+64y^3/9)[4,8/3]=-148.2728+321.6991=173.426.
VL=-(pi)(2y^4-3y^5/5-256y^2/9+64y^3/9)[0,8/3]=148.2728.
V=VU+VL=321.70.