how to compute E[X] given X+Y>1, X, Y iid ~ uniform (0, 1)

Question

Given $X, Y$ iid $\sim \operatorname{uniform} (0, 1)$

1. one way is $F(x\mid X+Y>1) = P(X<x\mid X+Y>1) = \frac{ P(1-Y<X<x)}{ P(X+Y>1)} = 2 (x-1+y),$ this is incorrect.

2. or, $E[X\mid X+Y>1] = \int_0^1 \int_{1-x}^1 x \,dy \,dx = \frac{1}{3} $, doesn't look right, since E[X | X+Y>1] should be larger than $\frac{1}{2}$.

3. Thanks for Henry's comment! This seems right, $$ E[X\mid X+Y>1] = \frac{E[X; X+ Y > 1]}{P(X+Y>1)} = \frac{\int_0^1 \int_{1-x}^1 x \,dy \,dx}{ P(X+Y > 1) } = \frac{2}{3} $$

And similar way to compute $E[XY]$ given $X + Y > 1$. I think this is resolved.