#### Answer

The domain is
$$\mathcal{D}=\{(x,y,z)| |x|\leq2,|y|\le3,|z|\le1\}$$
and it is shown on the figure below

#### Work Step by Step

Arguments of every square root have to be nonnegative which gives
$$4-x^2\geq0,\quad 9-y^2\geq0,\quad 1-z^2\geq0.$$
This can be rewritten as
$$x^2\leq4,\quad y^2\leq 9,\quad z^2\leq 1.$$
Taking the square root we get
$$|x|\leq 2,\quad |y|\leq3,\quad |z|\leq1.$$
Finally, this can be rewritten as
$$-2\leq x\leq 2,\quad-3\leq y\leq 3,\quad -1\leq z\leq1,$$
and geometrically, this is a Cuboid bounded by 6 planes
$$x=\pm2,\quad y=\pm3,\quad z=\pm1.$$
So we write for the domain
$$\mathcal{D}=\{(x,y,z)| |x|\leq2,|y|\le3,|z|\le1\}$$
and it is shown on the figure below