The question is $$ \int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy, \quad \gamma:y=\sqrt{x}, \ \ 0\leq x\leq 1$$ I've tried to solve it like this: $$\int_{0}^{1}\sin(\sqrt{t}-t) +t+ \sin(t-\sqrt{t}) \ \frac{1}{2\sqrt{t}}\ dt \quad\text{with} \ \ \gamma:r(t)=(t,\sqrt{t}) $$

How should i proceed from here ? Any suggestion would be great, thanks

  • $\begingroup$ I think there are some errors: We have $dx\neq dy$ since $dx=dt$ but $dy=\frac{1}{2\sqrt{t}}dt$. And where is the $2xy$ part? $\endgroup$ – Fakemistake Jan 13 at 18:34
  • $\begingroup$ $2xy$ is with $dx$ or $dy$? $\endgroup$ – Math Lover Jan 13 at 18:35
  • $\begingroup$ It seems that you have a typo in your integral. $\endgroup$ – user9464 Jan 13 at 19:17
  • $\begingroup$ I was unclear sorry, $Q=2xy+\sin(x-y)$ $\endgroup$ – simon Jan 14 at 6:03

$\int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy$

$= \int_\gamma \sin(y-x)\ dx +\left(\sin(x-y)\right) \ dy \ + \ \int_\gamma 2xy \ dy$

But please note that $\big(\sin(y-x), \sin(x-y)\big) = \nabla f(x,y) \ $ where $\ f(x,y) = \cos(y-x)$ and $f(1,1) - f(0,0) = 1 -1 = 0$.

So the line integral for part of the vector field is zero and all you are left with is to find -

$\displaystyle \int_{\gamma} 2xy \ dy = \int_0^1 t \ dt = \frac{1}{2}$

  • $\begingroup$ Q is $2xy+\sin(x-y)$ so $F$ is not conservative vector field, sorry i was unclear $\endgroup$ – simon Jan 14 at 6:02
  • $\begingroup$ OK so contribution of the part of the vector field in my answer is zero. So all you need to work on is $\int_{\gamma} \ 2xy \ dy$ $\endgroup$ – Math Lover Jan 14 at 6:06
  • $\begingroup$ Thank you Math Lover , your answer is correct but since this is the first time that i see a vector field being partially conservative, i wonder what theorem have you used for this and what is called such a vector filed? I thought a vector filed is called conservative if the requirement $Q'_x=P'_y$ would be met for the whole vector field but now the requirement is met only partially? $\endgroup$ – simon Jan 14 at 6:30
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    $\begingroup$ Thank you i see it now :) $\endgroup$ – simon Jan 14 at 6:33
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    $\begingroup$ @simon when you see vector fields having plus or minus components, you should check if splitting it helps. It is possible that part of the vector field is path independent and may be even zero... and hopefully the one that requires more work to integrate :) $\endgroup$ – Math Lover Jan 14 at 6:35


$$\int \sin(t-\sqrt{t})(\frac{1}{2\sqrt{t}}-1)dt=$$ $$\cos(t-\sqrt{t})$$


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