Where $x$ ranges from $0$ to $1$ for $F(x,y) = \dfrac{1}{1+x^{2}} \; \vec{j}$
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$\begingroup$ Can you think of an integral relating work and force? Look at the 'work (physics)' wiki and you will definitely find out how to proceed. $\endgroup$– RumplestillskinDec 19, 2016 at 11:58
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$\begingroup$ Young man, I did not ask the question. I'm just trying to understand and / or find a way to solve. I followed your concern and continue with the same problem. If you can not help me, thank you anyway. $\endgroup$– Felipe MaiaDec 19, 2016 at 12:07
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$\begingroup$ $$\int_C{\frac{dy}{1+x^2}}=\int_0^1{\frac{2xdx}{1+x^2}}=\ln(1+x^2)|_0^1=\ln2$$ $\endgroup$– DaniilDec 19, 2016 at 13:32
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$\begingroup$ @Daniil Thanks! $\endgroup$– Felipe MaiaDec 19, 2016 at 20:46
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1 Answer
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Your curve is given by $c(t)=(t,t^2)$ for $t \in [0,1]$. Then you have to compute
$\int_{c}F(x,y)*d(x,y)=\int_{0}^1 F(c(t)) *c'(t)dt$.
Your turn !
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$\begingroup$ Mm... I'll try here and see if I can get the result. $\endgroup$ Dec 19, 2016 at 12:09