jmc

For $q(x)$ to be defined, the quantities under both radicals must be nonnegative, and the denominator must be nonzero. Thus we must have $x\ge 0$ and $1-x^2>0$. The solution to the second inequality is $|x|<1$, so both inequalities are satisfied precisely when $x$ is in the interval $\boxed{[0,1)}$.