No, not really. Just because the region you are integrating in is a sphere doesn't meant you should expect pi to show up. If you think about any integral in either polar, cylindrical, or spherical coordinates, if theta shows up in the integrand as the argument of a trig function, then evaluation ion the d(theta) integration will cause he angle (which is where the pi would come from, of course) converted into whatever the resulting trig function results in.
example in polar coordinates for simplicity:\[\int_0^1\int_0^{\pi}rdrd\theta=\frac\pi2\](with no integrand, this integral just represents the area of the region of integration; a circle with radius 1)
However if we wanted to integrate \(y=r\sin\theta\) *over* that region of a half-circle we get\[\int_0^1\int_0^\pi r^2\sin\theta drd\theta=\frac23\]Notice the \(\pi\) got swallowed up by the fact that it was evaluated as the argument of a trig function, which is very common an in no way unusual in these situations.