I think the idea is to have on one side and on the remaining three sides. The problem you state gives on the top boundary. The remaining cases would be to have on the botton, right and left boundaries, respectively.
Can anyone help me think of the three similar cases I need to examine, I was thinking 0<x<pi/2 0<y<pi/2, 0<x<pi 0<y<pi/2, 0<x<pi/2 0<y<pi, with the same boundaries as those parts of the original square, but it doesn't really work for me, any help would be greatly appreciated!