Find $$\int_C \frac{2x^3+2xy^2-2y}{ x^2+y^2} \, dx+\frac{2y^3+2x^2y+2x}{ x^2+y^2} \, dy$$

Where $C$ is any simple closed loop which contains the origin.

What I figured out

I cannot use the direct version of Green's theorem.

I know that there is another version of greens theorem which is as follows :
$$\oint_{C+S} P \,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA,$$ where $C$ and $S$ are oriented opposite to each other and $D$ is the region enclosed between the two loops.

Using that I found \begin{align*} \int_{C+S} \left( 2x-\frac{2y}{ x^2+y^2}\right) \, dx+\left(2y+\frac{2x}{ x^2+y^2} \right) \, dy &= \iint_D \left[\left(2+\frac{4xy}{(x^2+y^2)^2} \right)-\left(2-\frac{4xy}{(x^2+y^2)^2} \right) \right]dA \\ &=\iint_D \frac{8xy}{(x^2+y^2)^2} dA \end{align*}

The region inside any closed loop can be defined in polar coordinates as $$\{(r, \theta)\in D \mid 0\leq r\leq r(\theta), \, 0\leq \theta \leq 2\pi \}$$

In polar coordinates the integral becomes $$ =\iint_D \frac{8xy}{(x^2+y^2)^2} dA = \int_0^{2\pi}\!\!\int_0^{r(\theta)} \frac{4\sin2\theta}{r} dr\,d\theta $$

Oops, I am again stuck; it's not zero.

  • $\begingroup$ I figured it out, if any one got interested he can ask on it. What i figured out is that the integral i found in last step is zero about any closed loop. I converted this into polar coordinates, if anyone has a better solution, pls contribute $\endgroup$ – Holy cow Jun 13 '14 at 7:27
  • $\begingroup$ i am stuck, can any smart ass help me $\endgroup$ – Holy cow Jun 13 '14 at 7:38
  • $\begingroup$ It's reasonable that you obtain 0: you have to take into account that your 0 is the sum of the integrals on the two components of your boundary. You're result is the sum of the integrations on two contours around the origin with opposite orientations: since the integration does not depend on the simple closed loop you choose the two contribution are equal in modulus but with opposite sign (due to different orientations). $\endgroup$ – Dario Jun 13 '14 at 8:07

You can write your integral as $$2\int_{C} \left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) + \int_{C} \left( 2x\ dx + 2y\ dy\right)\ ,$$ the second form is exact, so its contribution is 0, while for the first for you can do the following. Note that $$\left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) =d\ \arctan\left(\frac{y}{x}\right)\ ,$$ which in polar coordinates can be written as $d\theta$. Your integral is thus just $$-2\int_{C} d\theta = \mp 4\pi$$ where the sign depend on the orientation of the curve you choose: $-$ if it's counterclockwise, $+$ if clockwise.

  • $\begingroup$ i get you idea for thee first part but not for the second part, why is it 0 ! $\endgroup$ – Holy cow Jun 13 '14 at 10:12
  • $\begingroup$ You can see $2x\ dx+2y\ dy$ as $d\left(x^2+y^2\right)$, which in polar coordinates becomes $dr^{2}=2rdr$, so $$\int_{C} \left( 2x\ dx + 2y\ dy\right)=2\int_{C}rdr\ .$$ If you integrate on a circle of fixed radius this integral is 0. $\endgroup$ – Dario Jun 13 '14 at 10:15

Green's theorem fails here because it requires $P$ and $Q$ to be differentiable in the domain $C$. Now here when $$P=\frac{2x^3+2xy^2-2y}{ x^2+y^2}$$, it is not even continuous in $(0,0)$. Since if limit exists at (0,0) then $$\lim_{y \to 0}p(0,y)=\lim_{y \to 0}\frac{-2y}{y^2}$$, which doesn't exist. It is not even bounded.

  • $\begingroup$ I know that green's theorem fails, but the other version of green's theorem works $\endgroup$ – Holy cow Jun 13 '14 at 8:36

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