Answer
(a) The volume of ${\cal W}$ as a triple integral in the order $dxdzdy$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - {y^2}} \mathop \smallint \limits_{x = - \sqrt y }^{\sqrt y } 1{\rm{d}}x{\rm{d}}z{\rm{d}}y = \frac{{16}}{{21}}$
(b) The volume of ${\cal W}$ as a triple integral in the order $dydzdx$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^4}} \mathop \smallint \limits_{y = {x^2}}^{\sqrt {1 - z} } 1{\rm{d}}y{\rm{d}}z{\rm{d}}x = \frac{{16}}{{21}}$
The two answers agree.
Work Step by Step
Recall from Exercise 31: we have the region ${\cal W}$ bounded by $z = 1 - {y^2}$, $y = {x^2}$, and the plane $z=0$.
(a) To calculate the volume $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V$ as a triple integral in the order $dxdzdy$, we project ${\cal W}$ onto the $yz$-plane to obtain the domain ${\cal T}$. On the $yz$-plane, ${\cal T}$ is bounded below by $z=0$ and bounded above by $z = 1 - {y^2}$. The left and right boundaries of ${\cal T}$ are $y=0$ and $y=1$, respectively. Thus, ${\cal T}$ is a vertically simple region given by
${\cal T} = \left\{ {\left( {y,z} \right)|0 \le y \le 1,0 \le z \le 1 - {y^2}} \right\}$
Referring to the figure attached, we see that ${\cal W}$ is located on both positive and negative $x$-axis. Using $y = {x^2}$, we obtain $x = \pm \sqrt y $. Thus, we can describe it as a $x$-simple region bounded below by $x = - \sqrt y $ and bounded above by $x = \sqrt y $.
Thus, the region description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le y \le 1,0 \le z \le 1 - {y^2}, - \sqrt y \le x \le \sqrt y } \right\}$
The volume of ${\cal W}$ as a triple integral in the order $dxdzdy$ is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - {y^2}} \mathop \smallint \limits_{x = - \sqrt y }^{\sqrt y } 1{\rm{d}}x{\rm{d}}z{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - {y^2}} \left( {x|_{ - \sqrt y }^{\sqrt y }} \right){\rm{d}}z{\rm{d}}y$
$ = 2\mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - {y^2}} \sqrt y {\rm{d}}z{\rm{d}}y$
$ = 2\mathop \smallint \limits_{y = 0}^1 \sqrt y \left( {z|_0^{1 - {y^2}}} \right){\rm{d}}y$
$ = 2\mathop \smallint \limits_{y = 0}^1 \sqrt y \left( {1 - {y^2}} \right){\rm{d}}y$
$ = 2\mathop \smallint \limits_{y = 0}^1 \left( {{y^{1/2}} - {y^{5/2}}} \right){\rm{d}}y$
$ = 2\left( {\frac{2}{3}{y^{3/2}} - \frac{2}{7}{y^{7/2}}} \right)|_0^1$
$ = 2\left( {\frac{2}{3} - \frac{2}{7}} \right) = \frac{{16}}{{21}}$
The volume of ${\cal W}$ as a triple integral in the order $dxdzdy$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - {y^2}} \mathop \smallint \limits_{x = - \sqrt y }^{\sqrt y } 1{\rm{d}}x{\rm{d}}z{\rm{d}}y = \frac{{16}}{{21}}$
(b) To calculate the volume $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V$ as a triple integral in the order $dydzdx$, we project ${\cal W}$ onto the $xz$-plane to obtain the domain ${\cal S}$. We need to find the boundary curve of ${\cal S}$. A point $P$ on this curve is the projection of a point $Q$ on the intersection of the two surfaces, namely $z = 1 - {y^2}$ and $y = {x^2}$. Since $Q$ lies in the intersection curve, we have
$z = 1 - {y^2} = 1 - {x^4}$
Thus, the boundary curve of ${\cal S}$ is $z = 1 - {x^4}$. We see from the figure attached, we can describe ${\cal S}$ as a vertically simple region, where
${\cal S} = \left\{ {\left( {x,z} \right)| - 1 \le x \le 1,0 \le z \le 1 - {x^4}} \right\}$
We see that ${\cal W}$ is a $y$-simple region bounded below by $y = {x^2}$ and bounded above by $z = 1 - {y^2}$. Thus, ${\cal W}$ consists of all points lying between $y = {x^2}$ and $y = \sqrt {1 - z} $.
Thus, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)| - 1 \le x \le 1,0 \le z \le 1 - {x^4},{x^2} \le y \le \sqrt {1 - z} } \right\}$
The volume of ${\cal W}$ as a triple integral in the order $dydzdx$ is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^4}} \mathop \smallint \limits_{y = {x^2}}^{\sqrt {1 - z} } 1{\rm{d}}y{\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^4}} \left( {y|_{{x^2}}^{\sqrt {1 - z} }} \right){\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^4}} \left( {\sqrt {1 - z} - {x^2}} \right){\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {\left( { - \frac{2}{3}{{\left( {1 - z} \right)}^{3/2}} - {x^2}z} \right)|_0^{1 - {x^4}}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( { - \frac{2}{3}{x^6} - {x^2}\left( {1 - {x^4}} \right) + \frac{2}{3}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {\frac{1}{3}{x^6} - {x^2} + \frac{2}{3}} \right){\rm{d}}x$
$ = \left( {\frac{1}{{21}}{x^7} - \frac{1}{3}{x^3} + \frac{2}{3}x} \right)|_{ - 1}^1$
$ = \frac{1}{{21}} - \frac{1}{3} + \frac{2}{3} + \frac{1}{{21}} - \frac{1}{3} + \frac{2}{3} = \frac{{16}}{{21}}$
The volume of ${\cal W}$ as a triple integral in the order $dydzdx$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^4}} \mathop \smallint \limits_{y = {x^2}}^{\sqrt {1 - z} } 1{\rm{d}}y{\rm{d}}z{\rm{d}}x = \frac{{16}}{{21}}$
The two answers agree.
