The easiest way seems to be to take the integration variable to be $x= |\frac{z_1+z_2}{2}|$ and then integrate over the position of the point furthest from the origin.:
This gives $$ \operatorname{exp\_abs}(2) = \frac{\int_0^1 \mathrm{d}x \, x^2 \int_x^1 \mathrm{d}y \sqrt{1-y^2} }{\int_0^1 \mathrm{d}x \, x \int_x^1 \mathrm{d}y \sqrt{1-y^2} } = \frac{2}{3}\frac{\int_0^1 \mathrm{d}y\, y^3\sqrt{1-y^2} }{\int_0^1 \mathrm{d}y \,y^2 \sqrt{1-y^2} } = \frac{64}{45\pi}.$$
The easiest way seems to be to take the integration variable to be $x= |\frac{z_1+z_2}{2}|$ and then integrate over the position of the point furthest from the origin. This gives $$ \operatorname{exp\_abs}(2) = \frac{\int_0^1 \mathrm{d}x \, x^2 \int_x^1 \mathrm{d}y \sqrt{1-y^2} }{\int_0^1 \mathrm{d}x \, x \int_x^1 \mathrm{d}y \sqrt{1-y^2} } = \frac{2}{3}\frac{\int_0^1 \mathrm{d}y\, y^3\sqrt{1-y^2} }{\int_0^1 \mathrm{d}y \,y^2 \sqrt{1-y^2} } = \frac{64}{45\pi}.$$
The easiest way seems to be to take the integration variable to be $x= |\frac{z_1+z_2}{2}|$ and then integrate over the position of the point furthest from the origin:
This gives $$ \operatorname{exp\_abs}(2) = \frac{\int_0^1 \mathrm{d}x \, x^2 \int_x^1 \mathrm{d}y \sqrt{1-y^2} }{\int_0^1 \mathrm{d}x \, x \int_x^1 \mathrm{d}y \sqrt{1-y^2} } = \frac{2}{3}\frac{\int_0^1 \mathrm{d}y\, y^3\sqrt{1-y^2} }{\int_0^1 \mathrm{d}y \,y^2 \sqrt{1-y^2} } = \frac{64}{45\pi}.$$
