Given the bounds $$z=0,\quad x=0,\quad y^2=x^3,\quad y+z=1,\quad z-y=1.$$ I tried finding the volume using the following bounds $$\begin{cases} -1\le y\le 0\\ 0\le x\le y^{2/3}\\ 0\le z\le 1+x^{3/2} \end{cases}$$
Next I calculated the integral which gave me the right answer as in the textbook, but is this a coincidence?
$$V=\int _{-1}^0\int _0^{y^{\frac{2}{3}}}\int _0^{1+x^{\frac{3}{2}}}1dzdxdy=\frac{9}{20}$$
I next wanted to calculate the moment of inertia $I_x$ but this gave me the wrong answer
$$I_x=\frac{20m}{9}\int _{-1}^0\:\int _0^{y^{\frac{2}{3}}}\:\int _0^{1+x^{\frac{3}{2}}}\left(y^2+z^2\right)dzdxdy\:=\frac{50m}{77}.$$
The answer should've been $$I_x=\frac{29m}{77}.$$
It seems when calculating the volume you are missing something. Indeed $y$ can vary from $-1$ to $1$. First let's assume $-1\le y\le 0$. In this case we only need the constraint $z-y=1$. The volume of this part is: $$\int_{-1}^0 \int_{0}^{\sqrt[3] {y^2}}\int_{0}^{1+y}dzdxdy=\frac {9}{40};$$
and for $0\le y\le 1$, we need $y+z=1$. The volume of this part is:
$$\int_{0}^1 \int_{0}^{\sqrt[3] {y^2}}\int_{0}^{1-y}dzdxdy=\frac {9}{40};$$
hence the total volume is $\frac {9}{20}$.
To compute $I_x$, use the same approach. Therefore the answer is:
$$\frac {20}{9}(\int_{-1}^0 \int_{0}^{\sqrt[3] {y^2}}\int_{0}^{1+y}y^2+z^2\ {} dzdxdy \ {} +\int_{0}^1 \int_{0}^{\sqrt[3] {y^2}}\int_{0}^{1-y}\ {y^2+z^2} \ dzdxdy);$$
which is equal to: $$\frac {20}{9}(\frac{261}{3080}+\frac{261}{3080})=\frac{29}{77}.$$