Where $x$ ranges from $0$ to $1$ for $F(x,y) = \dfrac{1}{1+x^{2}} \; \vec{j}$

  • $\begingroup$ 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. $\endgroup$ Dec 19, 2016 at 11:58
  • $\begingroup$ 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. $\endgroup$ Dec 19, 2016 at 12:07
  • $\begingroup$ $$\int_C{\frac{dy}{1+x^2}}=\int_0^1{\frac{2xdx}{1+x^2}}=\ln(1+x^2)|_0^1=\ln2$$ $\endgroup$
    – Daniil
    Dec 19, 2016 at 13:32
  • $\begingroup$ @Daniil Thanks! $\endgroup$ Dec 19, 2016 at 20:46

1 Answer 1


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 !

  • $\begingroup$ Mm... I'll try here and see if I can get the result. $\endgroup$ Dec 19, 2016 at 12:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .