Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = 6.374$
Work Step by Step
We have a solid region ${\cal W}$ above $z = {x^2} + {y^2}$ and below $z=5$, and bounded by $y=0$ and $y=1$. So, it is a $z$-simple region bounded below by the surface $z = {x^2} + {y^2}$ and bounded above by the plane $z=5$.
The projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$, which is shown in the figure attached. The left and right boundaries of ${\cal D}$ are the arcs of the circle of radius $\sqrt 5 $. So, we describe ${\cal D}$ as a horizontally simple region, defined by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 1, - \sqrt {5 - {y^2}} \le x \le \sqrt {5 - {y^2}} } \right\}$
Thus, the triple integral is equal to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = {x^2} + {y^2}}^5 y{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = - \sqrt {5 - {y^2}} }^{\sqrt {5 - {y^2}} } y\left( {z|_{{x^2} + {y^2}}^5} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = - \sqrt {5 - {y^2}} }^{\sqrt {5 - {y^2}} } y\left( {5 - {x^2} - {y^2}} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = - \sqrt {5 - {y^2}} }^{\sqrt {5 - {y^2}} } \left( {5y - {x^2}y - {y^3}} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \left( {\left( {5xy - \frac{1}{3}{x^3}y - x{y^3}} \right)|_{ - \sqrt {5 - {y^2}} }^{\sqrt {5 - {y^2}} }} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \left( {10y\sqrt {5 - {y^2}} - \frac{2}{3}y{{\left( {5 - {y^2}} \right)}^{3/2}} - 2{y^3}\sqrt {5 - {y^2}} } \right){\rm{d}}y$
You need to find the antiderivatives of the integrands above and we list the results below:
$\begin{array}{*{20}{c}}
{{\rm{Integrand}}}&{10y\sqrt {5 - {y^2}} }&{ - \frac{2}{3}y{{\left( {5 - {y^2}} \right)}^{3/2}}}&{ - 2{y^3}\sqrt {5 - {y^2}} }\\
{{\rm{Antiderivative}}}&{ - \frac{{10}}{3}{{\left( {5 - {y^2}} \right)}^{3/2}}}&{\frac{2}{{15}}{{\left( {5 - {y^2}} \right)}^{5/2}}}&{\frac{2}{{15}}{{\left( {5 - {y^2}} \right)}^{3/2}}\left( {10 + 3{y^2}} \right)}
\end{array}$
So,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V$
$ = \left( { - \frac{{10}}{3}{{\left( {5 - {y^2}} \right)}^{3/2}} + \frac{2}{{15}}{{\left( {5 - {y^2}} \right)}^{5/2}} + \frac{2}{{15}}{{\left( {5 - {y^2}} \right)}^{3/2}}\left( {10 + 3{y^2}} \right)} \right)|_0^1$
$ = - \frac{{10}}{3}\cdot8 + \frac{2}{{15}}\cdot32 + \frac{2}{{15}}\cdot8\cdot13 + \frac{{10}}{3}\cdot5\sqrt 5 - \frac{2}{{15}}\cdot25\sqrt 5 - \frac{2}{{15}}\cdot5\sqrt 5 \cdot10$
$ = - \frac{{80}}{3} + \frac{{64}}{{15}} + \frac{{208}}{{15}} + \frac{{50}}{3}\sqrt 5 - \frac{{50}}{{15}}\sqrt 5 - \frac{{100}}{{15}}\sqrt 5 $
$ = - \frac{{128}}{{15}} + \frac{{20\sqrt 5 }}{3} \simeq 6.374$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = 6.374$.
