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.

  • $\begingroup$ How do you include the positively oriented condition? $\endgroup$ – Rafa Budría Nov 17 '18 at 8:46
  • $\begingroup$ Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así? $\endgroup$ – user402543 Nov 17 '18 at 19:56
  • $\begingroup$ Ah, yes, I didn't see that. Sorry. $\endgroup$ – Rafa Budría Nov 17 '18 at 20:15
  • $\begingroup$ $\pi/2$ is correct. $\endgroup$ – Maxim Dec 2 '18 at 9:29

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