Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V $
$= \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_{z = \sqrt {{x^2} + {y^2}} }^1 f\left( {x,y,z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$
Work Step by Step
We have the solid region given by
${\cal W} = \left\{ {\left( {x,y,z} \right):\sqrt {{x^2} + {y^2}} \le z \le 1} \right\}$
The description of ${\cal W}$ implies that this is a $z$-simple region bounded below by $z = \sqrt {{x^2} + {y^2}} $ and bounded above by $z=1$.
To express $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V$ as an iterated integral in the order $dzdydx$ we project ${\cal W}$ onto the $xy$-plane to obtain the domain ${\cal D}$. From Figure 16 and the figure attached, we see that ${\cal D}$ is a disk of radius $1$, that is, ${x^2} + {y^2} = 1$. Notice that ${\cal D}$ is both vertically simple and horizontally simple region. To have integral with $dzdydx$ order we choose to describe ${\cal D}$ as a vertically simple region:
${\cal D} = \left\{ {\left( {x,y} \right)| - 1 \le x \le 1, - \sqrt {1 - {x^2}} \le y \le \sqrt {1 - {x^2}} } \right\}$
Thus, the triple integral equals to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V $
$= \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_{z = \sqrt {{x^2} + {y^2}} }^1 f\left( {x,y,z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$
