Calculating the volume limited by $y=z^2$, $z=x^2$, $x=y^2$, $2y=z^2$, $2z=x^2$, $2x=y^2$ I've tried calculating the volume limited by the surfaces $y=z^2$, $z=x^2$, $x=y^2$, $2y=z^2$, $2z=x^2$, $2x=y^2$. I didn't know how to begin so instead I tried solving for $y=z^2$, $z=x^2$, $x=y^2$.
So I checked where they intersect and they can only have values from $0$ to $1$.
I projected over the $xy$-plane and got:
$$\iint_D\left(\int_0^{x^2}\mathrm dz\right)\,\mathrm dx\mathrm dy=\int_0^1\left(\int_0^{\sqrt x}x^2\,\mathrm dy\right)\,\mathrm dx=\frac25.$$
I don't think that I got this right, not really confident in what I did. If what I did is correct, how should I proceed to resolve the original problem now?
 A: $\def\R{\mathbb{R}}\def\e{\mathrm{e}}\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$Assume that the bounded volume is$$
A = \left\{ (x, y, z) \in \R^3 \,\middle|\, \dfrac{y^2}{2} \leqslant x \leqslant y^2,\ \dfrac{z^2}{2} \leqslant y \leqslant z^2,\ \dfrac{x^2}{2} \leqslant z \leqslant x^2\right\}
$$
and make the substitution $(x, y, z) = (\e^u, \e^v, \e^w) =: f(u, v, w)$. Since $\dfrac{∂(x, y, z)}{∂(u, v, w)} = \exp(u + v + w)$ and\begin{align*}
B := f^{-1}(A) &= \{(u, v, w) \in \R^3 \mid 2v - \ln 2 \leqslant u \leqslant 2v,\\
&\peq 2w - \ln 2 \leqslant v \leqslant 2w,\ 2u - \ln 2 \leqslant w \leqslant 2u\},
\end{align*}
then$$
\iiint\limits_A \d x\d y\d z = \iiint\limits_B \left| \frac{∂(x, y, z)}{∂(u, v, w)} \right| \,\d u\d v\d w = \iiint\limits_B \exp(u + v + w) \,\d u\d v\d w.
$$
Now make the substitution $(r, s, t) = (2u - w, 2v - u, 2w - v) =: g(u, v, w)$. Because $r + s + t = u + v + w$, $$
\frac{∂(u, v, w)}{∂(r, s, t)} = \left( \frac{∂(r, s, t)}{∂(u, v, w)} \right)^{-1} = \begin{vmatrix}
2 & 0 & -1\\
-1 & 2 & 0\\
0 & -1 & 2
\end{vmatrix}^{-1} = \frac{1}{7}
$$
and$$
C := g(B) = \{(r, s, t) \in \R^3 \mid 0 \leqslant r, s, t \leqslant \ln 2\},
$$
so\begin{gather*}
\iiint\limits_B \exp(u + v + w) \,\d u\d v\d w = \iiint\limits_C \exp(r + s + t) \left| \frac{∂(u, v, w)}{∂(r, s, t)} \right| \,\d r\d s\d t\\
= \frac{1}{7} \iiint\limits_{0 \leqslant r, s, t \leqslant \ln 2} \e^r \e^s \e^t \,\d r\d s\d t = \frac{1}{7} \left( \int_0^{\ln 2} \e^t \,\d t \right)^3 = \frac{1}{7},
\end{gather*}
i.e. the volume is $\dfrac{1}{7}$.
