The path $L_1,L_2,L_3,L_4$ leads to the integral
$$\int _0^{\sqrt{3}}\int _0^{\sqrt{4-y^2}}3\left(x^2+y^2\right)dxdy$$
which is a little difficult to evaluate
on the other hand the path $A\longrightarrow B\longrightarrow O\longrightarrow A$
will result in an integral which is easier to evaluate (using polar coordinates)
$$\int _0^{\pi /3}\int _0^23r^3drd\theta $$
Spoiler: