For 10C, my choice of parameterisation $\mathbf{r} (x,y) = ( x, y, z(x, y))$ fails to effect the right answer, but that of user ellya does function.
Yet for 9C, the parameterisation $\mathbf{r} (x,y) = ( x, y, z(x, y))$ does function. Nonetheless, in my first attempt, I didn't realise its legitimacy and had thought that it also failed. But in reality, the failure was due to my algebraic errors. I did compute the right answer by virtue of the parameterisation: $\mathbf{r} (\theta, z) = ( \sqrt{z} \cos \theta, \sqrt{z} \sin \theta, z)$ for all $0 \le \theta \le 2\pi, 1/9 \le z \le 1$. But this is much harder than need be; it wastes more time and effort and complicates the calculations.
$1.$ Why do we need to parameterise for these questions? How would we determine that we must?
$2.$ When does the parameterisation $\mathbf{r} (x,y) = ( x, y, z(x, y))$ function? When doesn't it?
we must parametrize because if we integrate over a paraboloid or a cyclinder we have restrictions that cannot be expressed in form $(x,y,f(x,y))$.
For instance the unit cylinder needs $x^2+y^2= 1$ if we try to write $(x,y,z)$ on the domain $[0,1]\times[0,1]\times[0,1]$ we end up including point that do not satisfy $x^2+y^2= 1$ for instance the point $(1,1,0)$, but if we put:
$(\cos\theta,\sin\theta,z)$, the $x$ and $y$ values are now forced to satisfy $x^2+y^2=1$, since $x^2+y^2=cos^2\theta+\sin^2\theta=1$.
We need to parametrize if our domain is non rectangular.