Thanks a lot for the answer. It looks quite close to the rough idea I would like to have as an answer.

Thanks a lot for the answer. It looks quite close to the rough idea I would like to have as an answer.

@Markiyan Hirnyk 
I've already presented the surface, namely $\sqrt((\zeta-x)^2+(\eta-y)^2)\le(t-\tau)$.

@Markiyan Hirnyk 
you can also that notation...

@Markiyan Hirnyk 
Because, you can consider the double integral as an integral over the surface $\sqrt((\zeta-x)^2+(\eta-y)^2)\le(t-\tau)$

Thank you for your answer. It is good to know.

Thank you for your answer. It is good to know.