Answer
(a) the limits of integration are
$ - 1 \le z \le 2$, ${\ \ \ }$ $0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le r \le 2$
Thus, the triple integral:
$\mathop \smallint \limits_{z = - 1}^2 \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 f\left( {r,\theta ,z} \right)r{\rm{d}}r{\rm{d}}\theta {\rm{d}}z$
(b) the limits of integration are
$ - 2 \le z \le 0$, ${\ \ \ }$ $0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le r \le \sqrt {4 - {z^2}} $
Thus, the triple integral:
$\mathop \smallint \limits_{z = - 2}^0 \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt {4 - {z^2}} } f\left( {r,\theta ,z} \right)r{\rm{d}}r{\rm{d}}\theta {\rm{d}}z$
Work Step by Step
We have a triple integral in cylindrical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} f\left( {r,\theta ,z} \right)r{\rm{d}}r{\rm{d}}\theta {\rm{d}}z$
a) The region ${\cal W}$ is defined by ${\cal W} = \left\{ {\left( {x,y,z} \right)|{x^2} + {y^2} \le 4, - 1 \le z \le 2} \right\}$.
In cylindrical coordinates, we have $0 \le \theta \le 2\pi $ and ${r^2} = {x^2} + {y^2}$. So, $r \le 2$.
Since we need the integration to be in the order ${\rm{d}}r{\rm{d}}\theta {\rm{d}}z$, so the limits of integration are
$ - 1 \le z \le 2$, ${\ \ \ }$ $0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le r \le 2$
Thus, the triple integral:
$\mathop \smallint \limits_{z = - 1}^2 \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 f\left( {r,\theta ,z} \right)r{\rm{d}}r{\rm{d}}\theta {\rm{d}}z$
(b) The lower hemisphere of the ball of radius $2$, centered at origin is described by
${\cal W} = \left\{ {\left( {x,y,z} \right)|{x^2} + {y^2} + {z^2} \le 4, - 2 \le z \le 0} \right\}$
In cylindrical coordinates, we have $0 \le \theta \le 2\pi $ and ${r^2} = {x^2} + {y^2}$. The hemisphere becomes ${r^2} + {z^2} \le 4$.
Since we need the integration to be in the order ${\rm{d}}r{\rm{d}}\theta {\rm{d}}z$, so the limits of integration are
$ - 2 \le z \le 0$, ${\ \ \ }$ $0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le r \le \sqrt {4 - {z^2}} $
Thus, the triple integral:
$\mathop \smallint \limits_{z = - 2}^0 \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt {4 - {z^2}} } f\left( {r,\theta ,z} \right)r{\rm{d}}r{\rm{d}}\theta {\rm{d}}z$
