Issues evaluating a Fourier transform on a restriction.

Question

The F.T. of the function $R^3\rightarrow R \ -U(x,y,z)\ $ is: $$g(k_{1},k_{2},k_{3})=\iiint_{R^{3}} U(x,y,z)e^{-i(k_1 x+k_2 y+ k_3 z)}dxdydz$$ By putting $z=0$ in the previous equation, you obtain: $$g(k_{1},k_{2},k_{3})=\iiint_{R^{3}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdydz=$$ $$=\iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy\int _{R}dz$$ Now,I don't know how to deal with the last integral $\int _{R}dz$. Since previously I have put z=0 then it should be dz=0 and also $\int _{R}dz =0 $. However in this case it results $g(k_{1},k_{2},k_{3})$=0 which doesn't make sense. I'd say that the right solution should be: $$\iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$$ That comes if $\int _{R}dz=1$. But I don't know how to proof it, someone knows? Thanks.