y ranges from -a to 0 so we are talking about the lower half plane in the xy- plane.
For each y, x ranges from -\sqrt{a^2- y^2} to 0. x= -\sqrt{a^2- y^2} gives x^2+ y^2= a^2, the circle with center (0,0) and radius a. Since x is the negative square root and we already know that we are in the lower half plane (y< 0), we are looking at the portion of that circle in the third quadrant.
For each (x,y), z ranges from a- \sqrt{a^2- x^2- y^2} to a. z= a- \sqrt{a^2- x^2- y^2} gives x^2+ y^2+ (z-a)^2= a^2, the sphere with center at (0,0,a) and radius a. Since z goes up to a, we are in the lower half sphere. That is, we are looking at the part of the sphere x^2+ y^2+ (z-a)^2= a^2 where x and y are negative and z< a.
I would be inclined to introduce new coordinates: x'= x, y'= y, z'= z- a. In those coordinates, the region we are integrating over is the "seventh octant" in which x, y, and z are all negative. To get that sphere, let \rho go from 0 to a, \theta from \pi to 3\pi/2, and \phi from \pi/2 to \pi. The integrand is still "1" and the "differential of volume" is still \rho sin^2(\phi)d\phi d\theta d\rho. In fact, because the integrand is "1", this integral is just the volume of the region: 1/8 of the volume of a sphere of radius a and so is (1/8)(4/3)\pi a^3= (1/6)\pi a^3. You can use that as a check.
