What's the intuition for a marginal expectation on a uniform triangular domain?

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Very often I've seen the basic density $$f(x,y)=2,\quad \text{for} \ x+y<1 \text{ where}\ x\in (0,1),\ y\in (0,1) $$

and it's relatively standard to show that $E(X)=E(Y)=\frac{1}{3}$.

I was wondering, what is the intuition for this result?

(I don't mean an intuition for the marginal of $X$ and then calculating the expectation, but a way to see possible visually why this might be the case)

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The point $\langle X,Y\rangle$ is uniformly distributed over the triangle $\triangle\langle 0,0\rangle\langle 0,1\rangle\langle 1,0\rangle$.

This triangle's centroid is $\langle \tfrac 13,\tfrac 13\rangle$ - which is one third of the vector sum of the vertices.

This is the centre of gravity of a such a shape with a uniform density.

That is all.