Answer
(a) ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le x} \right\}$
(b) ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le \sqrt {1 - {x^2}} } \right\}$
Work Step by Step
(a) We have $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^x \mathop \smallint \limits_0^{{x^2} + {y^2}} f\left( {x,y,z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$.
From the order of the integral we obtain the domain ${\cal W}$:
$0 \le x \le 1$, ${\ \ \ }$ $0 \le y \le x$, ${\ \ \ }$ $0 \le z \le {x^2} + {y^2}$
Thus, the projection of ${\cal W}$ onto the $xy$-plane is described by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le x} \right\}$
(b) We have $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_2^4 f\left( {x,y,z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$.
From the order of the integral we obtain the domain ${\cal W}$:
$0 \le x \le 1$, ${\ \ \ }$ $0 \le y \le \sqrt {1 - {x^2}} $, ${\ \ \ }$ $2 \le z \le 4$
Thus, the projection of ${\cal W}$ onto the $xy$-plane is described by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le \sqrt {1 - {x^2}} } \right\}$
