Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integral ∫C7y2xdx+8x2ydy.

Answer:1/2Step-by-step explanation:The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that [tex]\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy[/tex]Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.Mx(x,y) = d/dx 8x²y = 16xyLy(x,y) = d/dy 7y²x = 14xyThus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral[tex] \int\limits^1_0\int\limits^1_0 {2xy} \, dxdy[/tex]We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows[tex]\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2[/tex]We conclude that the line integral is 1/2