Answer
The volume of the solid: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{{59}}{{30}}$.
Work Step by Step
We have a solid region ${\cal W}$ bounded by $y = {x^2}$, $x = {y^2}$, $z=x+y+5$, and $z=0$.
Referring to the figure attached, ${\cal W}$ is a $z$-simple region. So, the volume is equal to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = 0}^{x + y + 5} {\rm{d}}z} \right){\rm{d}}A$,
where ${\cal D}$ is the projection of ${\cal W}$ onto the $xy$-plane.
To evaluate the integral over ${\cal D}$, we must find the boundary of ${\cal D}$.
Step 1. Find the boundary of ${\cal D}$
We find the intersection of the curves $y = {x^2}$ and $x = {y^2}$ by solving the equation
$y = {x^2} = \sqrt x $
Squaring both sides gives ${x^4} = x$. So,
${x^4} - x = 0$
$x\left( {{x^3} - 1} \right) = 0$
$x = 0$, ${\ \ \ }$ $x = 1$
Step 2. Express ${\cal D}$ as a simple domain
Notice that ${\cal D}$ is both vertically and horizontally simple region. We choose to describe ${\cal D}$ as a vertically simple region. Thus,
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,{x^2} \le y \le \sqrt x } \right\}$
Step 3. Write the triple integral as an iterated integral and evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = 0}^{x + y + 5} {\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = {x^2}}^{\sqrt x } \left( {\mathop \smallint \limits_{z = 0}^{x + y + 5} {\rm{d}}z} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = {x^2}}^{\sqrt x } \left( {z|_0^{x + y + 5}} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = {x^2}}^{\sqrt x } \left( {x + y + 5} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {\left( {xy + \frac{1}{2}{y^2} + 5y} \right)|_{{x^2}}^{\sqrt x }} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {{x^{3/2}} + \frac{1}{2}x + 5{x^{1/2}} - {x^3} - \frac{1}{2}{x^4} - 5{x^2}} \right){\rm{d}}x$
$ = \left( {\frac{2}{5}{x^{5/2}} + \frac{1}{4}{x^2} + \frac{{10}}{3}{x^{3/2}} - \frac{1}{4}{x^4} - \frac{1}{{10}}{x^5} - \frac{5}{3}{x^3}} \right)|_0^1$
$ = \frac{2}{5} + \frac{1}{4} + \frac{{10}}{3} - \frac{1}{4} - \frac{1}{{10}} - \frac{5}{3} = \frac{{59}}{{30}}$
Thus, the volume of the solid: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{{59}}{{30}}$.
