To solve this problem, we will use the Green's theorem. According to the Green's Theorem, the line integral over a curve C of (Pdx + Qdy) is equal to the double integral over the region D enclosed by C of (dQ/dx - dP/dy) dA. Here, we have the functions P(x, y) = 2y²x and Q(x, y) = 3x²y.
Firstly, we calculate the partial derivatives of P and Q. The derivative of Q with respect to x, denoted as dQ/dx, is the derivative of 3x²y with respect to x, which results in 6xy.
Secondly, we calculate the partial derivative of P with respect to y, denoted as dP/dy, that is, the derivative of 2y²x with respect to y, which results in 4xy.
Next, we calculate the difference between the two derivatives to get the integrand of the double integral. The result is 6xy - 4xy, which simplifies to 2xy.
Now we want to compute the double integral of 2xy over the region D enclosed by the square C. The square C is defined by the limits 0 and 1 for both x and y. Therefore, our double integral is ∫ from 0 to 1 ∫ from 0 to 1 (2xy) dx dy.
When we calculate this double integral over the region [0,1] x [0,1], we get the result 1/2.
Therefore, by Green's Theorem, the line integral over the square C of the function 2y²xdx+3x²ydy equals 1/2.