Polar coordinates are generally pretty robust. You can mess with $\theta$ all you want and things will still work out ok.
The transformation: $ \begin{array}{ccc} x & = & r \cos(\theta) \\ y & = & r \sin(\theta) \end{array} $ satisfies the necessary assumptions so the change of variables formula can be applied as long as $r>0$ and $\theta$ ranges over an interval of length $2\pi$. So one can use $[0,2\pi)$, $[-\pi,\pi)$, $[-\pi/2,3\pi/2)$, $[12345+\sqrt{3},12345+\sqrt{3}+2\pi)$, or $[a,a+2\pi)$ for any $a\in\mathbb{R}$. The only thing to watch out for is $r$. Remember that $r$ is the determinant of the Jacobian of this transformation, so the change of variables formula (for double integrals) tells us to insert the absolute value of the Jacobian -- which is $r$ since $r>0$. On the other hand if (for some really odd unexplained reason) you choose to let $r$ range over all real numbers and $\theta$ range over an interval of length $\pi$, you'll need to use "$|r|$" in the integral.
By the way, the same reasoning applies to triple integrals. When converting to cylindrical coordinates, you can let $\theta$ range over any interval of length $2\pi$ and all is fine (just don't mess with "r>0").
For spherical coordinates, if you use the "same" $\theta$ as in cylindrical coordinates you have $ \begin{array}{ccc} x & = & \rho\cos(\theta)\sin(\phi) \\ y & = & \rho\sin(\theta)\sin(\phi) \\ z & = & \rho\cos(\phi) \end{array} $ The standard domain for the spherical coordinates is $\rho>0$, $0 \leq \theta < 2\pi$, and $0 \leq \phi < \pi$. The determinant of the Jacobian of this transform is then $\rho^2\sin(\phi)$. Since $\sin(\phi)\geq 0$ when $0 \leq \phi < \pi$, there is no need for an absolute value when using the change of variables formula. This remains true if we change the interval $\theta$ ranges over to something else of the form $[a,a+2\pi)$. However, notice if we wanted to let $\phi$ range over $[0,2\pi)$ and $\theta$ over $[0,\pi)$, we would still have a valid change of coordinates, but our Jacobian determinant $\rho^2\sin(\phi)$ would now need absolute values since sine could take on negative values.
Long story short: Don't worry about changing $\theta$'s interval.