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Evaluate the line integral

$$ \int_C (\ln y) e^{-x} \,dx - \dfrac{e^{-x}}{y}\,dy + z\,dz $$

where C is the curve parametrized by $r(t)=(t-1)i+e^{t^4}j+(t^2+1)k$ for $0\leq t\leq 1$

I know that the potential function is $$f(x, y, z) = -e^{-x}\ln y + \frac{z^2}{2} + C$$ but exactly how would I use that to evaluate the line integral?


marked as duplicate by Simon S, Winther, Aditya Hase, ml0105, hardmath Nov 28 '14 at 3:35

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  • $\begingroup$ What course is this from? The identical integral was asked earlier today. I'll find the link. $\endgroup$ – Simon S Nov 28 '14 at 1:55
  • $\begingroup$ Would 1/2 be the correct answer? $\endgroup$ – Jon Nov 28 '14 at 3:02

Pretty much like an ordinary real integral. The potential function takes the place of the antiderivative; if the the path goes from $A$ to $B$ then the integral is $$f(B)-f(A)\ .$$ In this case $A$ is $r(0)$ and $B$ is $r(1)$. Can you do the calculations?

  • $\begingroup$ So I just do f(1,1,1)-f(0,0,0)? $\endgroup$ – Jon Nov 28 '14 at 1:51
  • $\begingroup$ I don't think $r(0)$ is $(0,0,0)$ is it? Look at the formula for $r(t)$. $\endgroup$ – David Nov 28 '14 at 1:51
  • $\begingroup$ Whoops I got r(0)=-i+j+k and r(1)=ej+2k. So exactly what would I do with that? $\endgroup$ – Jon Nov 28 '14 at 1:58
  • $\begingroup$ Now you have to calculate $f(B)-f(A)$. You know $f$, you know $B$, you know $A$, so you should not have any difficulty with this. $\endgroup$ – David Nov 28 '14 at 2:08
  • $\begingroup$ Just to check, so f(0,e,2)-f(-1,1,1) will give the answer? $\endgroup$ – Jon Nov 28 '14 at 2:20

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