= Solution
Evaluate the <two-body Lorentz-invariant phase space> in the parent's rest frame, $p=(m,\mathbf0)$. The spatial <Dirac delta function> sets $\mathbf q_2=-\mathbf q_1$. For massless daughters $E_1=E_2=q$, so
$$
d\Pi_2=\frac1{(2\pi)^2}\frac{q^2dq\,d\Omega}{4q^2}\delta(m-2q)
=\frac{d\Omega}{32\pi^2}.
$$
The factor $1/2$ in the last step is the derivative of the delta-function argument. Substituting into the <decay width> gives, for labelled or distinct daughters,
$$
\boxed{\Gamma=\frac1{64\pi^2m}\int d\Omega\,|\mathcal A|^2.}
$$
The angular domain is one sphere: $0\le\varphi<2\pi$ and $-1\le\cos\theta\le1$, equivalently $0\le\theta\le\pi$. The printed negative-to-positive polar-angle interval is not a second independent coverage of the sphere.
For identical daughters, integration over the full labelled two-body <phase space> counts each physical final state twice. One must therefore include an additional factor $1/2!$, or equivalently integrate over an unordered half-domain. This distinction is necessary for the two identical <photons> in <6/d/solution>.
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