Yes, another vertex is at (0,2a,0), I don't have a scanner with me to upload the visual. The other vertexes are at (0,a,a), (0,2a,a), (a,a,0), (a,2a,0), (a,a,a), (a,2a,a).
If it's just a scalar is my answer simply the integral at (0,a,0) minus at (0,2a,0) where v = a^3? I think that's what I did.