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I have the 1-form $$dz=2xy\, dx+(x^{2}+2y)\, dy$$ And I want to integrate it from $(x_{1},y_{1})$, to $(x_{2},y_{2})$.

If I'm not drunk, checking mixed partials, I find that $dz$ is an exact differential. BUT, when I want to calculate explicitly the integral

$$\int_{\sigma}dz$$ where $\sigma$ is

i) $(x_{1},y_{1})\to(x_{2},y_{1})\to(x_{2},y_{2})$

ii) $(x_{1},y_{1})\to(x_{1},y_{2})\to(x_{2},y_{2})$

I find that the integral has a different value if I take the trajectory ii) instead of i). Why this happens?

Calculating the integral for i)

$\int dz=\int_{x_{1}}^{x_{2}}2xy_{1}\, dx+\int_{y_{1}}^{y_{2}}(x_{2}^{2}+2y)\, dy=y_{1}(x_{2}^{2}-x_{1}^{2})+x_{2}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}$

Calculating the integral for ii)

$\int dz=\int_{y_{1}}^{y_{2}}(x_{1}^{2}+2y)\, dy+\int_{x_{1}}^{x_{2}}2xy_{2}\, dx=x_{1}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}+y_{2}(x_{2}^{2}-x_{1}^{2})$

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You tried to compare the results of the two computations before fully simplifying; there are cancelling factors in the first result that do not appear in the second and vice versa which threw you off. –  anon May 20 '13 at 7:18
    
Yes you are right. Thanks! @anon –  Anuar May 20 '13 at 7:19

3 Answers 3

up vote 2 down vote accepted

$$\begin{array}{lr} & \color{Purple}{y_1x_2^2}-\color{Blue}{y_1x_1^2}+\color{Magenta}{x_2^2y_2}-\color{Purple}{x_2^2y_1}+\color{Green}{y_2^2-y_1^2} \\ & = \color{Teal}{x_1^2y_2}-\color{Blue}{x_1^2y_1}+\color{Green}{y_2^2-y_1^2}+\color{Magenta}{y_2x_2^2}-\color{Teal}{y_2x_1^2} \end{array}$$

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Very good answer. I reward your patience. –  Anuar May 20 '13 at 7:22

Actually $dz$ is an exact differential: $$z = x^2y + y^2. $$ And integral does not depend on path. I think you lost something when you integrate. I just can't understand how do you integrate on these path.

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1  
I link the points of i) and ii) with straight lines. Then I integrate over the trajectory dividing the integrals in 2 parts, one part where $x$ stays constant, and the other part when $y$ stays fixed. –  Anuar May 20 '13 at 7:10

The differential is exact, and you got the same answer. Just do some algebra and you will get $$ -x_{1}^{2}y_1+x_{2}^{2}y_{2}+y_{2}^{2}-y_{1}^{2} $$

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Completely agree. Thanks! –  Anuar May 20 '13 at 7:20

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