Consider the region $A = \{(s,t) \in \mathbb{R}^2; s \geq 0, t \geq 0, s^2+t^2 \leq 1 \}$. Be $X = (X_1, X_2)$ a random vector that is uniformly distributed on $A$.
I want to compute $E(X_1\mid X_2=t)$.
We have $f_{X_1,X_2}(s,t)=\frac{4}{\pi}$ and $$f_{X_1}(s)=\int_0^{\sqrt{1-s^2}} \frac{4}{\pi} \, dt = \frac{4}{\pi} \sqrt{1-s^2}. $$ Then $$E(X_1) = \int_0^{\sqrt{1-t^2}} s f_{X_1}(s) \, ds = - \frac{4t^3-4}{3 \pi}.$$
How can I now determine the conditional expectation?
The conditional expectation is given by \begin{align} \mathbb E[X_1\mid X_2=t] &= \int_0^{\sqrt{1-t^2}} s f_{s\mid t}(s\mid t)\ \mathsf ds, \end{align} where $$f_{X_1\mid X_2}(s\mid t) = \frac{f_{X_1,X_2}(s,t)}{f_{X_2}(t)}. $$ By symmetry, $f_{X_2}(t) = \frac 4\pi\sqrt{1-t^2}$. So we compute $$ \mathbb E[X_1\mid X_2=t] = \int_0^{\sqrt{1-t^2}} s\cdot \frac{4/\pi}{4/\pi\sqrt{1-t^2}}\ \mathsf ds = \frac1{\sqrt{1-t^2}}\int_0^{\sqrt{1-t^2}}s\ \mathsf ds = \frac12 \sqrt{1-t^2}. $$