Given $$z=0$$ $$x=0$$ $$y^2=x^3$$ $$y+z=1$$ $$z-y=1$$
I need to find the volume $V$ and mass of inertia $I_x$ but I have not really a clue of how to choose my bounds for my triple integrals
I was thinking
$$-1\le y\le 1$$ $$0\le z\le 1+y$$ $$0\le x\le y^{\frac{2}{3}}$$
Your parameterization is slightly incorrect. The upper bound of $z$ should instead be the shortest vertical distance between the plane $z=0$ and either plane $z=1-y$ and $z=1+y$, which is captured by $z=\min\{1-y,1+y\}$. The subsequent volume integral (taking advantage of symmetry about $y=0$) would be
$$\begin{align*}V &= \int_{-1}^1 \int_0^{\min\{1-y,1+y\}} \int_0^{y^{2/3}} dx \, dz \, dy \\[1ex] &= \int_0^1 \int_0^{1-y} \int_0^{y^{2/3}} dx \, dz \, dy + \int_{-1}^0 \int_0^{1+y} \int_0^{y^{2/3}} dx \, dz \, dy \\[1ex] &= 2 \int_0^1 \int_0^{1-y} \int_0^{y^{2/3}} dx \, dz \, dy\end{align*}$$