Let $\P$ be a probability measure on $\R^2$ with density $\r$ and $X,Y$ denote the projections $(x,y)\mapsto x$ and $(x,y)\mapsto y$ respectively. If $\int\r(x,y)\,dx > 0$, then for all measurable $f:\R^2\rar[0,\infty]$ : $$ \E(f|Y=y) =\frac{\int f(x,y)\r(x,y)\,dx}{\int\r(x,y)\,dx}, $$ where $\E(f|Y=y)\in\R$ is a convenient notation for the value of $\E(f|Y)\colon=\E(f|\s(Y))$ on the set $[Y=y]$. If $\P$ is the uniform distribution on $(a,b)\times(c,d)$, then $$ \E(f|X=x)=\frac1{d-c}\int_c^d f(x,y)\,dy\quad\mbox{and}\quad \E(f|Y=y)=\frac1{b-a}\int_a^b f(x,y)\,dx~. $$
Define $h(y)$ by the right hand side of the above formula. Obviously $h(Y)$ is $\s(Y)$-measurable. Now for any $A\in\s(Y)$ there is some $B\in\B$, such that $A=[Y\in B]$ and therefore we get by the transformation theorem of measure theory: \begin{eqnarray*} \E(h(Y);A) &=&\int_B\int_\R h(y)\r(x,y)\,dx\,dy\\ &=&\int_B\int_\R\Big( \int f(z,y)\r(z,y)\,dz\Big/\int\r(z,y)\,dz\Big)\r(x,y)\,dx\,dy\\ &=&\int_B\int_\R f(z,y)\r(z,y)\,dz\,dy =\E(f(X,Y)I_B(Y)) =\E(f(X,Y);A)~. \end{eqnarray*}