The inequalities describing the region in spherical coordinates are
$$
0 \le \rho \le 2
,\quad
0 \le \rho \sin\phi \le 1
,\quad
0 \le \phi \le \pi/2
,\quad
0 \le \theta \le 2\pi
.
$$
When you're doing the $\rho$ integral first (innermost), what you call the “usual order”, you rewrite the inequalities for $\rho$ and $\phi$ as
$$
0 \le \rho \le \min(2,1/\sin\phi)
,\qquad
0 \le \phi \le \pi/2
,
$$
and set up the bounds for the integrals accordingly (split into cases depending on whether $1/\sin\phi$ is less than or greater than 2), as in your comment above:
$$
\int_{\phi=0}^{\pi/6}
\int_{\rho=0}^2
+
\int_{\phi=\pi/6}^{\pi/2}
\int_{\rho=0}^{1/\sin\phi}
.
$$
To do the $\phi$ integral first instead, write the inequalities for $\rho$ and $\phi$ as
$$
0 \le \rho \le 2
,\quad
0 \le \sin\phi \le 1/\rho
,\quad
0 \le \phi \le \pi/2
.
$$
Here you split into cases depending on whether $1/\rho$ is less than or greater than 1 (since $\sin \phi \le 1$ automatically, which renders the middle inequality inactive if $1/\rho>1$). Like this:
$$
\int_{\rho=0}^1 \int_{\phi=0}^{\pi/2}
+
\int_{\rho=1}^2 \int_{\phi=0}^{\arcsin(1/\rho)}
.
$$
Geometrically, when you're doing the $\phi$ integral first, you are integrating over a circular arc in space (with a fixed radius $\rho$ and a fixed longitude $\theta$), starting on the $z$ axis, and stopping either when you hit the $xy$ plane or when you hit the cylinder (whichever happens first, and this depends on the value of $\rho$).
