Evaluate the following line integral: ∮C[(4x^2+3x+5y)dx+(6x^2+5x+3y)dy] where C is the path around the square with vertices (0,0),(2,0),(2,2) and (0,2).
To evaluate the given line integral, we can make use of Green's theorem, which relates line integrals around a closed curve C to a double integral over the region D enclosed by C.
Green's theorem states that for a vector field F = (P, Q), if its partial derivatives ∂Q/∂x and ∂P/∂y are continuous on a region D in the plane and C is a positively oriented, piecewise-smooth simple closed curve that encloses D, then the line integral of F along C is equal to the double integral of (∂Q/∂x - ∂P/∂y) over D:
∮C (P dx + Q dy) = ∬D (∂Q/∂x - ∂P/∂y) dA
In our case, the vector field F = (P, Q) is given by P = 4x^2 + 3x + 5y and Q = 6x^2 + 5x + 3y. To use Green's theorem, we need to find the partial derivatives of P and Q:
∂P/∂y = 5
∂Q/∂x = 12x + 5
Now, we will compute the double integral (∂Q/∂x - ∂P/∂y) over the region D enclosed by the square with vertices (0,0), (2,0), (2,2), and (0,2).
To compute this double integral, we can break the region D into two triangles. Let's call the triangle with vertices (0,0), (2,0), (0,2) as triangle 1 and the triangle with vertices (0,2), (2,2), (2,0) as triangle 2.
For triangle 1, we have:
∬D1 (∂Q/∂x - ∂P/∂y) dA
= ∫[0,2] ∫[0,2-x] (12x + 5 - 5) dy dx
Simplifying, we get:
∬D1 (∂Q/∂x - ∂P/∂y) dA
= ∫[0,2] (12x + 5 - 5)(2 - x) dx
For triangle 2, we have:
∬D2 (∂Q/∂x - ∂P/∂y) dA
= ∫[0,2] ∫[0,x] (12x + 5 - 5) dy dx
Simplifying, we get:
∬D2 (∂Q/∂x - ∂P/∂y) dA
= ∫[0,2] (12x + 5 - 5)x dx
Now, we can evaluate these two double integrals to find the value of the given line integral.