$$F=(xy^2, y+x)$$ Integrate the $(\bigtriangledown \times F) \cdot k$ over the region of the first eightant that is bounded of the curves $y=x^2$ and $y=x$. $$$$ Green's Theorem: $$\int \int_R {(\bigtriangledown \times F) \cdot k}dxdy=\oint_S{F}d \sigma=\int_S{Mdx+Ndy}=\int_{S_1}{Mdx+Ndy}+\int_{S_2}{Mdx+Ndy}=\int_{S_1}{xy^2dx+(y+x)dy}+\int_{S_2}{xy^2dx+(y+x)dy}$$
$S_1: y=x$ $$\int_0^1{(x^3+2x)}dx=\frac{5}{4}$$ $S_2: y=x^2$ $$\int_0^1{(x^5+(x^2+x)2x)dx}=\frac{4}{3}$$ So $$\int \int_R {(\bigtriangledown \times F) \cdot k}dxdy=-\frac{5}{4}-\frac{4}{3}=-\frac{31}{12}$$
But in my textbook it is as followed: $$\int \int_R {(\bigtriangledown \times F) \cdot k}dxdy=\frac{4}{3}-\frac{5}{4}=\frac{1}{12}$$ But why do we take these signs???
Integrals are done over oriented lines, surfaces, or other objects. An integral done $\int_0^1$ is different from an integral done $\int_1^0$. That's why they drew arrows on the picture, to tell you how each segment of curve is oriented. You oriented $S_1$ correctly, but you oriented $S_2$ backwards. Try fixing that sign and trying again.