Describe a subset of $R^{3}$ and modify the order of an iterated triple integral I have come to a problem in a multivariate calculus book that I am having trouble solving. The problem goes :
If $D \subseteq \mathbb{R}^{3}$ and :
\begin{equation}
\int\int\int_{D} d(x,y,z) = \int_{-1}^{1} \left[ \int_{x^{2}}^{1} \left( \int_{0}^{1-y} dz\right)dy\right]dx
\end{equation}
then describe $D$. Rewrite the triple integral as an iterated integral in which $dx$, $dy$, and
$dz$ appear in each of the following orders (i.) $dz,dx,dy$, (ii.) $dx,dy,dz$, (iii.) $dx,dz,dy$, (iv.) $dy,dz,dx$, (v.) $dy,dx,dz$.
I'm not sure how to approach this problem. I was able to do previous problems in the book that required changing the order of 2D iterated integrals, but in the 3D case I am having more trouble.
Can someone help with this ?
Edit :
After reading the comments, I have come up with a possible solution. The solution is below :
We have for the cross section perpendicular to the z-axis at $z = 0$ :

Now suppose we have $E \subset \mathbb{R}^{3}$ s.t. :
\begin{equation}
\int\int\int_{E} d(x,y,z) = \int_{-1}^{1} \left[ \int_{x^{2}}^{1} \left( \int_{0}^{1} dz \right) dy \right] dx
\end{equation}
Then $E$ is a parabolic cylinder with height $1$ and base at the plane corresponding to $z = 0$ :
\begin{equation}
E = \{ (x,y,z) \in \mathbb{R}^{3} \; : \; x \in [-1,1] \text{ and } y \in [x^{2},1] \text{ and } z \in [0,1] \}
\end{equation}
$D$ is a subset of $E$. Define plane $P$ :
\begin{equation}
P = \{ (x,y,z) \in \mathbb{R}^{3} \; : \; y + z = 1 \}
\end{equation}
Now define region of $\mathbb{R}^{3}$ below $P$ :
\begin{equation}
M = \{ (x,y,z) \in \mathbb{R}^{3} \; : \; y + z \leq 1 \}
\end{equation}
We see :
\begin{equation}
D = \{ (x,y,z) \in \mathbb{R}^{3} \; : \; (x,y,z) \in E \bigcap M \}
\end{equation}
We see :
\begin{equation}
y + z \leq 1 \Leftrightarrow y \leq 1 - z
\end{equation}
We see :
\begin{equation}
y = x^{2} \Rightarrow x = \pm \sqrt{y}
\end{equation}
So for plane $z = z_{0}$ we have the cross section of $D$ :

So you can peform the integration by adding up slices like those shown above. In this case the order is $[dy,dx,dz]$ or $[dx,dy,dz]$ I believe.
Let :
\begin{equation}
I = \int\int\int_{D} d(x,y,z)
\end{equation}
Then I think we have for $dy,dx,dz$ :
\begin{equation}
I = \int_{0}^{1} \left[ \int_{-\sqrt{1-z}}^{\sqrt{1-z}} \left( \int_{0}^{x^{2}} dy \right) dx \right] dz \; \checkmark
\end{equation}
and :
\begin{equation}
I = \int_{0}^{1} \left[ \int_{0}^{1-z} \left( \int_{-\sqrt{y}}^{\sqrt{y}} dx \right) dy \right] dz \; \checkmark
\end{equation}
Now still need $[dz,dx,dy]$, $[dx,dz,dy]$, and $[dy,dz,dx]$.
For $[dz,dx,dy]$ and $[dx,dz,dy]$ we will need to sum cross sections that are perpendicular to the y-axis.
We can draw the cross-section of $D$ at $y = y_{0}$ as :

So we have :
\begin{equation}
I = \int_{0}^{1} \left[ \int_{-\sqrt{y}}^{\sqrt{y}} \left( \int_{0}^{1-x^{2}} dz \right) dx \right] dy \; \checkmark
\end{equation}
and :
\begin{equation}
I = \int_{0}^{1} \left[ \int_{0}^{1-y} \left( \int_{-\sqrt{1-z}}^{\sqrt{1-z}} dx \right) dz \right] dy \; \checkmark
\end{equation}
Now only need $[dy,dz,dx]$. We see :
\begin{align}
y \in [x^{2},1] \text{ and } z \in [0,1-y]  & \Rightarrow z \in [1-1,1-x^{2}]\\
      & \Rightarrow z \in [0,1-x^{2}]
\end{align}
We see :
\begin{align}
z = 1-x^{2}  & \Leftrightarrow z - 1 = -x^{2} \\
  & \Leftrightarrow 1 - z = x^{2}
\end{align}
So :
\begin{equation}
y \in [x^{2},1] \Rightarrow y \in [1-z,1]
\end{equation}
and :
\begin{equation}
I = \int_{-1}^{1} \left[ \int_{0}^{1-x^{2}} \left( \int_{1-z}^{1} dy \right) dz \right] dx \; \checkmark
\end{equation}
 A: I have marked errors in $\color {red} {\textbf {red}}$ and my corrections are in $\color {blue} {\textbf {blue}}$. If you need explanation on any of them, please ask.
i) order $dz \ dx \ dy$
Your integral setup -
\begin{equation}
I = \int_{0}^{1} \left[ \int_{-\sqrt{y}}^{\sqrt{y}} \left( \int_{0}^{\color{red}{\bf{1-x^2}}} dz \right) dx \right] dy
\end{equation}
Correct integral -
\begin{equation}
I = \int_{0}^{1} \left[ \int_{-\sqrt{y}}^{\sqrt{y}} \left( \int_{0}^{\color{blue}{\bf{1-y}}} dz \right) dx \right] dy
\end{equation}
ii) order $dx \ dy \ dz$ - your working is correct
iii) order $dx \ dz \ dy$
Your integral setup -
\begin{equation}
I = \int_{0}^{1} \left[ \int_{0}^{1-y} \left( \int_{\color {red} {\bf {-\sqrt{1-z}}}}^{\color {red} {\bf{\sqrt{1-z}}}} dx \right) dz \right] dy
\end{equation}
Correct integral -
\begin{equation}
I = \int_{0}^{1} \left[ \int_{0}^{1-y} \left( \int_{\color {blue} {\bf {-\sqrt{y}}}}^{\color {blue} {\bf{\sqrt{y}}}} dx \right) dz \right] dy
\end{equation}
iv) order $dy \ dz \ dx$
Your integral setup -
\begin{equation}
I = \int_{-1}^{1} \left[ \int_{0}^{1-x^{2}} \left( \int_{\color {red} {\bf {1-z}}}^{\color {red} {\bf 1}} dy \right) dz \right] dx
\end{equation}
Correct integral -
\begin{equation}
I = \int_{-1}^{1} \left[ \int_{0}^{1-x^{2}} \left(\int_{\color {blue} {\bf {x^2}}}^{\color {blue} {\bf {1-z}}} dy \right) dz \right] dx
\end{equation}
v) order $dy \ dx \ dz$
Your integral setup -
\begin{equation}
I = \int_{0}^{1} \left[ \int_{-\sqrt{1-z}}^{\sqrt{1-z}} \left( \int_{\color {red} {\bf 0}}^{\color {red} {\bf {x^2}}} dy \right) dx \right] dz
\end{equation}
Correct integral -
\begin{equation}
I = \int_{0}^{1} \left[ \int_{-\sqrt{1-z}}^{\sqrt{1-z}} \left( \int_{\color {blue} {\bf {x^2}}}^{\color {blue} {\bf {1-z}}} dy \right) dx \right] dz
\end{equation}
