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Consider $$F=\frac {x}{x^3+y^3}dx+\frac{y}{x^3+y^3}dy$$

1) Show that $\int_GF=0$, where $G$ is the arc of a circle or radius $r$ in the first quadrant.

2) Compute the integral of $F$ along the segment connecting $(0,1)$ with $(1,0)$


1) $G$ has the parameterization $y=\sqrt{r^2-x^2}$

$$\int_G\frac {x}{x^3+y^3}dx+\frac{y}{x^3+y^3}dy=\int_0^r\frac {x}{x^3+\sqrt{r^2-x^2}^3}dx+\frac{\sqrt{r^2-x^2}}{x^3+\sqrt{r^2-x^2}^3}\frac{x}{-\sqrt{r^2-x^2}}dx=0$$

2) The segment $I$ has parameterization $y=1-x$. Then

$$\int_I\frac {x}{x^3+y^3}dx+\frac{y}{x^3+y^3}dy=\int_0^1\frac {x}{x^3+(1-x)^3}dx-\frac{(1-x)}{x^3+(1-x)^3}dx=\int_0^1\frac {2x-1}{x^3+(1-x)^3}dx=\int_0^1\frac {2x-1}{3x^2-3x+1}dx$$

How can we proceed? Can we use the residue theorem?

share|cite|improve this question… – enzotib Jul 15 '14 at 12:10
I think the first part can use Green formula. – Shine Jul 15 '14 at 12:23

1)$$F=\frac{1}{x^3+y^3}\frac{1}{2}d(x^2+y^2)$$ In polar coordinates it becomes $$\frac{1}{\cos^3(\theta)+\sin^3(\theta)}\frac{dr^2}{2r^3}=\frac{1}{\cos^3(\theta)+\sin^3(\theta)}\frac{dr}{r^2}$$ You get $0$ just by "integrating" the radial part.

2) If you integrate on the segment between $(0,1)$ and $(1,0)$ you are integrating on the line $y=1-x$: $$F=\frac{1}{2}\frac{d(2x^2-2x+1)}{3x^2-3x+1}=\frac{1}{2}\frac{2}{3}\frac{d(3x^2-3x+1)}{3x^2-3x+1}=\frac{1}{3}d\ln|3x^2-3x+1|\ .$$

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\begin{align*} \frac{2x-1}{3x^2-3x+1}=\frac{1}{3}\frac{6x-3}{3x^2-3x+1}=\frac{1}{3}(\ln|3x^2-3x+1|)' \end{align*}

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