Setting up the triple integral in spherical coordinates in a different order 
Let D be the region bounded below by the plane $z=0$, above by the sphere $x^2+y^2+z^2 = 4$ and on the sides by the cylinder $x^2+y^2 = 1$
  Set up the triple integral in spherical coordinates that give the volume of D using the following order of integration:
  $\mathrm{d\phi d\rho d\theta}$

I am able to set the integral easily for the normal order of integration in spherical coordinates but am facing lot of trouble in setting the integral for the given order.
So could anyone help with it?
I tried with the range of $\phi$ as $[0,\pi/6]$ and then $[\pi/6,\pi/2]$, $\theta$ will obviously vary from 0 to $2\pi$ but I am not able to decide the limits of $\rho$ in either case.
 A: 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 $\rho$).
