Use Green's Theorem to evaluate the line integral. Orient the curve counterclockwise unless otherwise indicated. ?C(lnx+y)dx−x2dy where C is the rectangle with vertices (1, 1), (3, 1), (1, 4), and (3, 4)​

Respuesta :

By Green's theorem,

[tex]\displaystyle \int_C P(x,y)\,dx + Q(x,y)\,dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \, dA[/tex]

where [tex]D[/tex] is the interior of [tex]C[/tex]. It's easy to see that

[tex]D = \left\{(x,y) ~:~ 1 \le x \le 3 \text{ and } 1 \le y \le 4\right\}[/tex]

Now,

[tex]\dfrac{\partial Q}{\partial x} = \dfrac{\partial}{\partial x}(-x^2) = -2x[/tex]

[tex]\dfrac{\partial P}{\partial y} = \dfrac{\partial}{\partial y}(\ln(x) + y) = 1[/tex]

so that the line integral reduces to

[tex]\displaystyle \int_C (\ln(x)+y)\,dx - x^2\,dy = -\int_1^3 \int_1^4 (2x+1) \, dy\,dx = \boxed{-30}[/tex]