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. $\endgroup$ – Rumplestillskin Dec 19 '16 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. $\endgroup$ – Felipe Maia Dec 19 '16 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 '16 at 13:32
  • $\begingroup$ @Daniil Thanks! $\endgroup$ – Felipe Maia Dec 19 '16 at 20:46

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. $\endgroup$ – Felipe Maia Dec 19 '16 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.