2

$$\int_\gamma \frac{(x^2+y^2-2)\,dx+(4y-x^2-y^2-2) \, dy}{x^2+y^2-2x-2y+2}$$

where $\gamma = 2\sin\left(\frac{\pi x}{2}\right)$ from $(2,0)$ to $(0,0)$.

I think it should be a shortcut to this problem that I cannot see , if that is not the case I will keep trying to simplify it .

Thnaks in advance

Michael Hardy
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  • From my experience these complicated ones you can do it using greens theorem and hopefully get 0. Notice the hopefully – Teh Rod Nov 21 '16 at 01:20
  • Yes but taking the curl of the vector field is not simple. –  Nov 21 '16 at 01:28
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    If this differential form is exact, then the integral is just the difference between the values of the potential at the two points. – Michael Hardy Nov 21 '16 at 02:12

2 Answers2

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Complete the squares in the denominator $(x-1)^2+(y-1)^2$ and change the variables $x-1\mapsto x$ and $y-1\mapsto y$. You get the vector field \begin{align} &\left[\frac{x^2+2x+y^2+2y}{x^2+y^2},\frac{4y-x^2-2x-y^2-2y}{x^2+y^2}\right]=\\ &=\left[1+\frac{2x}{x^2+y^2},-1+\frac{2y}{x^2+y^2}\right]-2\left[\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right]. \end{align} The first term is conservative (with an easy potential), the second term has zero curl (easy to check).

P.S. Actually the second term has a potential too if you restrict you domain e.g. to the plain without the negative $y$-axis, but the potential is less obvious (easier to get in the polar coordinates).

A.Γ.
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  • What causes greens theorem not to work ? i mean i know two criterion now , the curve has to be closed and the vector field must be defined and differentiable in the region am integration over , is there something am missing ? –  Nov 21 '16 at 01:57
  • @Jacksoja Green's theorem works well, of course. The curl is zero, you can check it for the original field with some extra effort. Then due to singularity at $(1,1)$ you have to pick a correct completion of the path - not along the $x$-axis, but along the circle of radius $\sqrt{2}$ centered at $(1,1)$ counterclockwise. – A.Γ. Nov 21 '16 at 02:01
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Hint: Let $I$ denote the integral. Note that the curve can be parameterized using curve using

$$x(t) = t$$ $$y(t) = \sin\left(\frac{\pi x}{2} \right)$$

for $0 \leq t \leq 2$

Thus, separate your integral into two different integrals:

$$\int_\gamma \frac{x^2 + y^2 - 2}{x^2 + y^2 - 2x - 2y + 2} dx \qquad \mathrm{and} \qquad\int_\gamma \frac{4y - x^2 - y^2 - 2}{x^2 + y^2 - 2x - 2y + 2} dy$$

whose sum is equal to $I$.

Meow Mix
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