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}
