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There is a line integral over a straight line as follows (the problem here: https://tutorial.math.lamar.edu/Solutions/CalcIII/LineIntegralsPtI/Prob1.aspx):

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Line Integral: $$\int 3x^2 - 2y\ ds$$ Equation of the line: $$2y=7x-9$$

In the solution, The author suggests that due to the reverse direction of line, parametrization is necessary but I don't know why. In fact, I can't figure out why my solution (below) leads to a wrong result (negative the correct one), despite the fact that I'm defining the integration interval($x$) from 3 to 1 so it is supposed to be consistent with the direction. $$\int 3x^2 - 2y\ ds \rightarrow \int_{3}^{1} 3x^2 - 2y \sqrt{1+\left (\frac{7}{2} \right )^2} dx $$

Edit: once I simplified the main function, but now it is as in the problem.

Hooman
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  • Since the function being integrated along this line, $1$, is constant, should the value of this line integral change if we rotate the path around its midpoint? – Eric Towers Dec 02 '20 at 23:10
  • in fact in the main problem the function is $ 3x^2 - 2y $. But I thought changing the main function wont change the problem – Hooman Dec 02 '20 at 23:13
  • A detailed answer for a similar problem: https://math.stackexchange.com/a/3326676/700004 – Hooman Dec 03 '20 at 12:36

1 Answers1

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You are traversing the line in the opposite direction than the "natural one" (usually $x$ increases), this is why the result should be negative.

If you want your answer to be correct, you need to do $\int_3^1$, not $\int_1^3$, since you should start from the point where $x$ is 3, and finish where $x$ is 1.

Edit: The line segment is the set of points $(x,\frac{7x-9}2)$ from $x=3$ to $x=1$. Thus $ds=-\sqrt{1+\big(\frac72\big)^2}\,dx$, and we have $$\int_\mathcal C 3x^2-2y\,ds = \int_3^1\Big(3x^2-2\Big(\frac{7x-9}{2}\Big)\Big)\Big(-\sqrt{1+\Big(\frac72\Big)^2}\,dx\Big) = 8\sqrt{53}.$$ Notice that $ds$ has a minus sign because the orientation is reversed. If this is confusing, you can just pick a parametrisation where the directions are aligned with the integral. The strangeness here occurs because we are going backwards. (Intuitively, it should make sense that $\Delta s \approx - c\,\Delta x$.)

Luke Collins
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  • Thank you for your answer. Isn't the start point $a$ and the end point $b$ in this integral?: $\int_{a}^{b}$ – Hooman Dec 02 '20 at 23:07
  • Yes, $a$ is the starting point and $b$ is the end point of $x$ (since you are integrating w.r.t. $x$). If you had to follow the arrows on the line in your diagram, you should start from 3, not from 1. – Luke Collins Dec 02 '20 at 23:09
  • well i'm using the same integral which gives me the wrong result: $\int_{3}^{1} 3x^2 - 2y \sqrt{1+\left (\frac{7}{2} \right )^2} dx$ – Hooman Dec 02 '20 at 23:19
  • Let me edit my answer to explain more – Luke Collins Dec 02 '20 at 23:24
  • Thank you, For clarification I'm Integrating the function, then evaluating it at $x=1$ ($F(1)$) and $x=3$($F(3)$), then $F(1)-F(3)$. – Hooman Dec 02 '20 at 23:31
  • Is there any convention in which direction should ds be positive and in which it should be considered negative? – Hooman Dec 03 '20 at 00:12
  • $ds$ is positive if a tiny change in $x$ (i.e., going from $x$ to $x+dx$) would result in an increase in the arclength $s$ we obtain ($dx$ denotes a positive change in $x$ always). If you go from $x$ to $x+dx$ according to how we are travelling (from right to left), we would actually decrease the amount of arclength we've covered; so that's why the change $ds$ is negative.

    e.g. if we are at $x=2$, then we've travelled half way along the path. If we go to $2+0.1$, we decrease how much we've travelled. This is why $ds$ is negative.

     – Luke Collins Dec 03 '20 at 00:14
  • then we can build the path from $(3,6)$. each $dx$ to left adds a ds to the path from $(3,6)$. Also, for the reverse direction of the problem, $\int_{1}^{3}$ we also have to consider $ds$ negative. This one is building the path form (1, -1). – Hooman Dec 03 '20 at 00:27
  • $s$ is the "length we've travelled so far". Think of $ds$ as the change in that. If $x$ becomes $x+dx$, then $s$ becomes $s+ds$, and clearly $ds$ and $dx$ have to have opposite signs.

    Look at this picture. If we increase $x$ by $dx$, obtaining $x+dx$, then $s$ becomes $s+ds$. Clearly $ds$ has to be negative.

     – Luke Collins Dec 03 '20 at 00:36
  • I think $ds$ is an unsigned value and can not be negative https://math.stackexchange.com/a/3326676/700004 – Hooman Dec 03 '20 at 12:52
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    @Hooman It technically depends how you formalise things, $ds$ doesn't have to be unsigned, he replaces $dx$ with $-dx$ instead there, which I choose to have as the unsigned value in my formalisation. There is no fixed convention of how to reverse orientation; the important thing is that $dx$ and $ds$ must have opposite signs; you choose which does what. His reasoning is a Riemann–Stieltjes integral with respect to $g(x)=-x$.

    Whatever interpretation helps you understand the most is the best one :)

     – Luke Collins Dec 03 '20 at 12:59