Answer
$-30$
Work Step by Step
Green's Theorem states that: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$
Here, we have $f=\ln x+y ; g =-x^2$
Now, $\dfrac{\partial g}{\partial x} =-2x$ and $\dfrac{\partial f}{\partial y}=1$
Therefore,
$ \iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\int_{1}^3\int_{1}^{4} (-2x-1) \ dy \ dx \\=\int_{1}^3[y(2x-1)]_1^4 \ dx\\=\int_1^3 3(-2x-1) \ dx \\=3[-x^2-x)]_1^3\\=3(-3^2-3)-3(-1-1) \\=-30$
