Calculate boundary of two domains $D_1 =\{ -1 \le x \le 1-\sqrt{y^2+z^2}\}$
$D_2 =\{ x^2+y^2+z^2\le 8,z\ge 2\}$
Is it correct to say that the boundary of $D_1$ is : 
$B(D_1) =\{ (x-1)^2=y^2+z^2,(x+1)^2=y^2+z^2\}$
and the boundary of $D_2$ : 
$B(D_2) =\{z\ge 2 ,x^2+y^2+z^2=8 \} \cup \{x^2+y^2=4\}$
?
Especially the last one. I know for sure that the boundary is $x^2+y^2=4$, but i'm not sure that the boundary is also the 'cover' of the shpere($x^2+y^2+z^2=8$)

$C= \{x \in \mathbb R^3 ; x^2+y^2=4\}$ is not a subset of $B(D_2)$. $C$ is the right circular cylinder of axis $Oz$ and with radius $2$. $C$ is unbounded while $D_2$ is bounded. What is true is that 
$$B(D_2) = \{z\ge 2 ,x^2+y^2+z^2=8 \} \cup \{z=2, x^2+y^2 \le 4\}.$$
Your $B(D_1)$ isn't correct either. To "compute" the boundary, use the fact that one of the inequality has to be an equality and look at what it implies for the second inequality.
