Let $R=\{(x,y,z):3x^2 \leq z \leq 3, \ -z \leq y \leq z\}$ and compute the volume $\text{Volume}(R).$
Plotting the first inequality:
and since $y$ varies between $-z$ and $z$, the region is just a cylindric paraboloid bounded by the parabola $z=3x^2,$ the planes $z=3$, $y=-z$ and $y=z$. Now it remains to get the bounds for the trippleintegral.
Clearly $ -1\leq x \leq 1$ and it's already given that $-z \leq y \leq z$. But It's also given that (and it's evident from the figure above) that $3x^2 \leq z \leq 3$. So
$$I=\iiint_R1 \ dzdydz=\int_{-1}^{1}\int_{-z}^{z}\int_{3x^2}^{3}1 \ dzdydx= 8z.$$
Now I got a my answer in terms of $z$, but this is ok since I don't know the bounds explicitly for $y$.
Am I on the right path?
You can't "decompose" the integral like this with those boundaries for the variables. When you do such "decomposition" of the integral you are actually first projecting the region onto the $xy$-plane and afterwards on $x$-line and so on. When you do the projection onto the $xy$-plane it can't be dependent on $z$.
To get around this you can express $y$ in terms of $x$ using the inequality $3x^2 \le z \le 3$. However a better solution would be to project the region onto the $xz$-plane first and afterwards onto the $x$-line. Then the solution would be:
$$\iiint_R1 \ dzdydz=\int_{-1}^{1}\int_{3x^2}^{3}\int_{-z}^{z}1 \ dydzdx$$
Now just evaluate the integral, which shouldn't be that hard.