The height of the equation is $z= x^3 +4y$ and it is bounded by $Y = x^3$ and $Y = 2x$, I have drawn the bounded region below. As you can see, for the left region if we take the double integration of the type 1 area it is bounded from $2x$ to $x^3$, and the right region is bounded from $x^3$ to $2x$, how do I solve this?
Note that both the areas (Area in $1$st quadrant and $3$rd quadrant) are similar. We can find any of the areas and multiply by $2$ to find the desired area.
For the area in $1$st quadrant, consider vertical elemental strips with length $dx$ going from $0$ to $\sqrt{2}$ and for height, the upper limit is given by $y=2x$ and the lower limit by $y = x^3$.
Area in first quadrant is given by, $$A = \int_0^{\sqrt{2}}\int_{x^3}^{2x} (x^3 + 4y) \ dy dx $$
Can you take it from here? (Solving this double integral is not a hard task tbh).