Set up a triple integral that represents the volume of part of the ball x^2 + (y-1)^2 + z^2 </= 1 that is outside the cone rho = pi/6 and inside the cone rho = pi/3

In the answer, the dTheta part of the triple integral goes from 0 to pi. My question is how do you find that? Normally theta goes from 0 to 2pi.