Calculate the line integral on the negatively oriented unit circumference (like the hands of the clock) under the $F(x,y)=(e^x+x^2y,e^y-xy^2)$ field
I have thought to do the following to solve this:
$\int_{C'}F\cdot dr=\int_{-C}F\cdot dr=-\int_CF\cdot dr$, where $C$ is a positively oriented curve.
But $\int_CF\cdot dr=\int\int_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA=\int\int_D(-y^2-x^2)dA=-\int\int_D(x^2+y^2)dA=-\int_{0}^{2\pi}\int_{0}^{1}r^3drd\theta=-\int_{0}^{2\pi}1/4d\theta=-2\pi/4=-\pi/2$.
Then $\int_{C'}F\cdot dr=\pi/2$
Is this fine? Thank you.