Answer
$\begin{array}{*{20}{c}}
{{\rm{Order}}}&{{\rm{Integral}}{\ }\left( {{\rm}} \right)}\\
{dzdxdy}&{\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{y/2} \mathop \smallint \limits_{z = 0}^{4 - {y^2}} xyz{\rm{d}}z{\rm{d}}x{\rm{d}}y}\\
{dxdydz}&{\mathop \smallint \limits_{z = 0}^4 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - z} } \mathop \smallint \limits_{x = 0}^{y/2} xyz{\rm{d}}x{\rm{d}}y{\rm{d}}z}\\
{dydxdz}&{\mathop \smallint \limits_{z = 0}^4 \mathop \smallint \limits_{x = 0}^{\frac{1}{2}\sqrt {4 - z} } \mathop \smallint \limits_{y = 2x}^{\sqrt {4 - z} } xyz{\rm{d}}y{\rm{d}}x{\rm{d}}z}
\end{array}$
Work Step by Step
1. Iterated integral order: $dzdxdy$
From Example 5 we obtain
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \mathop \smallint \limits_{z = 0}^{4 - {y^2}} xyz{\rm{d}}z{\rm{d}}y{\rm{d}}x$
The order of the integral implies that ${\cal W}$ is a $z$-simple region. The projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$, a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,2x \le y \le 2} \right\}$
Notice that the lower boundary is the line $y=2x$.
Referring to Figure 7 (A), we re-define ${\cal D}$ as a horizontally simple region. Using $y=2x$, the right boundary is the line $x = \frac{y}{2}$. So, the domain description becomes
${{\cal D}_1} = \left\{ {\left( {x,y} \right)|0 \le y \le 2,0 \le x \le \frac{y}{2}} \right\}$
Thus, the triple integral in $dzdxdy$ order:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \left( {\mathop \smallint \limits_{z = 0}^{4 - {y^2}} xyz{\rm{d}}z} \right){\rm{d}}A$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{y/2} \mathop \smallint \limits_{z = 0}^{4 - {y^2}} xyz{\rm{d}}z{\rm{d}}x{\rm{d}}y$
2. Iterated integral order: $dxdydz$
From Example 5 we obtain
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{4 - {y^2}} \mathop \smallint \limits_{x = 0}^{y/2} xyz{\rm{d}}x{\rm{d}}z{\rm{d}}y$
The order of the integral implies that ${\cal W}$ is a $x$-simple region. The projection of ${\cal W}$ onto the $yz$-plane is the domain ${\cal T}$, a vertically simple region defined by
${\cal T} = \left\{ {\left( {y,z} \right)|0 \le y \le 2,0 \le z \le 4 - {y^2}} \right\}$
Notice that the upper boundary is the curve $z = 4 - {y^2}$.
Referring to Figure 7 (B), we re-define ${\cal T}$ as a horizontally simple region. Using $z = 4 - {y^2}$, the right boundary is the curve $y = \sqrt {4 - z} $. So, the domain description becomes
${{\cal T}_1} = \left\{ {\left( {y,z} \right)|0 \le z \le 4,0 \le y \le \sqrt {4 - z} } \right\}$
Thus, the triple integral in $dxdydz$ order:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal T}_1}}^{} \left( {\mathop \smallint \limits_{x = 0}^{y/2} xyz{\rm{d}}z} \right){\rm{d}}A$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{z = 0}^4 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - z} } \mathop \smallint \limits_{x = 0}^{y/2} xyz{\rm{d}}x{\rm{d}}y{\rm{d}}z$
3. Iterated integral order: $dydxdz$
From Example 5 we obtain
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{4 - 4{x^2}} \mathop \smallint \limits_{y = 2x}^{\sqrt {4 - z} } xyz{\rm{d}}y{\rm{d}}z{\rm{d}}x$
The order of the integral implies that ${\cal W}$ is a $y$-simple region. The projection of ${\cal W}$ onto the $xz$-plane is the domain ${\cal S}$, a vertically simple region defined by
${\cal S} = \left\{ {\left( {x,z} \right)|0 \le x \le 1,0 \le z \le 4 - 4{x^2}} \right\}$
Notice that the upper boundary is the curve $z = 4 - 4{x^2}$.
Referring to Figure 7 (C), we re-define ${\cal S}$ as a horizontally simple region. Using $z = 4 - 4{x^2}$, the right boundary is the curve $x = \frac{1}{2}\sqrt {4 - z} $. So, the domain description becomes
${{\cal S}_1} = \left\{ {\left( {x,z} \right)|0 \le z \le 4,0 \le x \le \frac{1}{2}\sqrt {4 - z} } \right\}$
Thus, the triple integral in $dydxdz$ order:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal S}_1}}^{} \left( {\mathop \smallint \limits_{y = 2x}^{\sqrt {4 - z} } xyz{\rm{d}}y} \right){\rm{d}}A$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{z = 0}^4 \mathop \smallint \limits_{x = 0}^{\frac{1}{2}\sqrt {4 - z} } \mathop \smallint \limits_{y = 2x}^{\sqrt {4 - z} } xyz{\rm{d}}y{\rm{d}}x{\rm{d}}z$
In summary:
The triple integral $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V$ is equal to the following iterated integrals:
$\begin{array}{*{20}{c}}
{{\rm{Order}}}&{{\rm{Integral}}{\ }\left( {{\rm{Example{\ }5}}} \right)}\\
{dzdydx}&{\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \mathop \smallint \limits_{z = 0}^{4 - {y^2}} xyz{\rm{d}}z{\rm{d}}y{\rm{d}}x}\\
{dxdzdy}&{\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{4 - {y^2}} \mathop \smallint \limits_{x = 0}^{y/2} xyz{\rm{d}}x{\rm{d}}z{\rm{d}}y}\\
{dydzdx}&{\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{4 - 4{x^2}} \mathop \smallint \limits_{y = 2x}^{\sqrt {4 - z} } xyz{\rm{d}}y{\rm{d}}z{\rm{d}}x}
\end{array}$
$\begin{array}{*{20}{c}}
{{\rm{Order}}}&{{\rm{Integral}}{\ }\left( {{\rm{Exercise{\ }27}}} \right)}\\
{dzdxdy}&{\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{y/2} \mathop \smallint \limits_{z = 0}^{4 - {y^2}} xyz{\rm{d}}z{\rm{d}}x{\rm{d}}y}\\
{dxdydz}&{\mathop \smallint \limits_{z = 0}^4 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - z} } \mathop \smallint \limits_{x = 0}^{y/2} xyz{\rm{d}}x{\rm{d}}y{\rm{d}}z}\\
{dydxdz}&{\mathop \smallint \limits_{z = 0}^4 \mathop \smallint \limits_{x = 0}^{\frac{1}{2}\sqrt {4 - z} } \mathop \smallint \limits_{y = 2x}^{\sqrt {4 - z} } xyz{\rm{d}}y{\rm{d}}x{\rm{d}}z}
\end{array}$
