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Field : $$\int_\ell \frac{-1}{1+(y-x)^2}\,dx + \frac{1}{1+(y-x)^2}\,dy$$

Find the path from point $(0,0)$ to $(1,2)$ along the ellipse $(x-1)^2 +(y/2)^2 =1$.

I thought of checking the green formula because there are no undefined points. I get the answer zero, which mean (on a closed loop, with solid inner area) that the path doesnt matter.

I pick an Easy road to $(1,2)$ with $y= 2x$. I get the answer $13/3$ But the answer is $\pi/4$.

Regards Oskar

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Aard
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1 Answers1

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The vector field $\vec{F} = \left(\frac{-1}{1+(y-x)^2}, \frac{1}{1+(y-x)^2}\right)$ is conservative, so the value of the integral is independent of path.

By following the path $y=2x$, we get the integral: $$\int_0^1 \frac{1}{1+x^2}dx = \arctan(1)-\arctan(0) = \frac{\pi}{4}$$

Your mistake is probably found where you substituted the path into the integral.

apnorton
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