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A line integral (with respect to arc length) can be interpreted geometrically as the area under $f(x,y)$ along $C$ as in the picture. You sum up the areas of all the infinitesimally small 'rectangles' formed by $f(x,y)$ and $ds$.

enter image description here

What I'm wondering is how do I interpret line integrals with respect to $x$ or $y$ geometrically?

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  • $\begingroup$ You rather integrate with respect to the arclength, that's why you use $dS$. In turn, this will depend on $dx$ and $dy$ for cartesian cordinates or on $d\theta$ and $dr$ for polar coordinates, i.e. (and don't take this strictly) $$ds^2 = dx^2+dy^2$$ $$ds^2 = dr^2+r^2d\theta^2$$ $\endgroup$
    – Pedro
    Commented Mar 4, 2012 at 19:44
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    $\begingroup$ You get the "algebraic sum" of the projected areas. E.g., if $C$ is a circle in the $(x,y)$-plane you get zero. $\endgroup$ Commented Mar 4, 2012 at 21:07

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Let's compare the definitions of these three related, but distinct concepts. Let $C$ be a parametrized curve with respect to the parameter $t\in[a,b]$. Then

\begin{equation}\tag{1} \int_C f(x,y)\,ds := \int_a^b f(x(t),y(t))\,\color{blue}{\sqrt{[x'(t)]^2+[y'(t)]^2}}\,dt \end{equation} whereas \begin{align} \int_C f(x,y)\,dx &:= \int_a^b f(x(t),y(t))\,\color{red}{x'(t)}\,dt,\tag{2}\\ \int_C f(x,y)\,dy &:= \int_a^b f(x(t),y(t))\,\color{green}{y'(t)}\,dt.\tag{3} \end{align}

You seem to understand the geometric interpretation of (1): it is the area of the "fence" built along the curve $C$ whose height along any point $(x,y)$ on $C$ is given by $f(x,y)$. Alternatively, focus on the multiplier in blue in (1): we are weighting the integrand $f(x(t),y(t))$ by the length of the velocity vector along $C$.

On the other hand, in (2), we are weighting the integrand by only the $x$ component of the velocity vector.

In (3), we are weighting the integrand by only the $y$ component of the velocity vector.

As a simple example, consider $f(x,y)=1$.

\begin{align} \int_C 1\,ds&=\int_a^b \sqrt{[x'(t)]^2+[y'(t)]^2}\,dt =\text{length of }C\\ \int_C 1\,dx&=\int_a^b x'(t)\,dt =x(b)-x(a)=\text{net displacement in $x$ direction as $C$ is traversed}\\ \int_C 1\,dy&=\int_a^b y'(t)\,dt =y(b)-y(a)=\text{net displacement in $y$ direction as $C$ is traversed}. \end{align}

Draw a simple example of something like an $S$ shaped curve for $C$ and look at the three quantities above in that setting.

Edit: Here is an admittedly crude graphical interpretation of what (2) and (3) mean in the particular case when $f(x,y)=1$ (and I realize that in the picture $f(x,y)\not= 1$).

enter image description here

$\int_C 1\,dx$ corresponds to the dark red line on the $x$ axis while $\int_C 1\,dy$ corresponds to the dark blue line on the $y$ axis.

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    $\begingroup$ This diagram came from James Stewart's Calculus. One example from the text was $\int_C 2x\,ds$, where $C$ goes from $(0, 0)$ to $(1,1)$ along $y = x^2$ and from $(1, 1)$ to $(1, 2)$ along the vertical line segment. With the parametrization $x = x$, $y = x^2$ in the parabola and $x = 1$, $y = y$ in the line segment, you will get $\int_C f(x,y)\,ds = \int_{0}^{1} 2x\sqrt{1+4x^2}\,dx + \int_{1}^{2} 2\,dy$. I'm not seeing this as displacements in the x- or y- direction. Rather, as Mr. Treble indicated below, the parametrization just "decomposed" as such. Please kindly comment, TY! $\endgroup$
    – Andy Tam
    Commented Nov 11, 2014 at 4:01
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    $\begingroup$ I said that was the interpretation when $f(x,y)=1$. $\endgroup$
    – JohnD
    Commented Nov 11, 2014 at 4:07
  • $\begingroup$ What happens when we project $C$ onto the $(y,z)$ plane? The curve is going back and forth along the same $y$ values. So what would this mean? $\endgroup$
    – user5826
    Commented Jun 17, 2018 at 18:50
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The geometric interpretation of line integral with respect to $x$ is as the following figure. I can't give a geometrical interpretation of the line integral with respect to $y$ in this case because the direction of $y$ back and forth when $t$ increase. enter image description here

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  • $\begingroup$ How did you edit the image? $\endgroup$
    – Sensebe
    Commented Apr 18, 2017 at 7:04
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    $\begingroup$ @ImmortalPlayer By the software Paint built in Micro\$oft Windows. This image is from Stewart's Calculus. (sec. Line Integral.) $\endgroup$
    – bfhaha
    Commented Apr 18, 2017 at 9:43
  • $\begingroup$ What happens when we project $C$ onto the $(y,z)$ plane? The curve is going back and forth along the same $y$ values. So what would this mean? $\endgroup$
    – user5826
    Commented Jun 17, 2018 at 18:52
  • $\begingroup$ @AlJebr I believe that it means that some of the area is counted many times (i.e. every time there is overlap) $\endgroup$
    – D.R.
    Commented Apr 26, 2020 at 23:41
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    $\begingroup$ @D.R. I don't think that's quite right. Integrating with respect to $y$ will weight the integrand by the rate of change of $y$ with respect to $t$, not the absolute value of the rate of change. This means when the curve is going back on itself in the projection onto the $yz$-plane, since the rate of change of $y$ is negative, we are removing the portion of the area that is then being projected. $\endgroup$ Commented Jun 16, 2020 at 14:45
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Given a function $(x,y)\mapsto z=f(x,y)$ and a curve $$\gamma:\quad s\mapsto{\bf z}(s)=\bigl(x(s),y(s)\bigr)\qquad(a\leq s\leq b)$$ parametrized with respect to arc length, the integral $$\int_\gamma f(x,y)\>ds:=\int_a^b f\bigl(x(s),y(s)\bigr)\>ds\tag{0}$$ can be interpreted in various ways. You have chosen to interpret it as surfacr area of a "Christo curtain" $S$ displayed along $\gamma$ and having height $f(x,y)$ at the point $(x,y)\in\gamma$.

Now you want an interpretation of the integral

$$\int_\gamma f(x,y)\>dx:=\int_a^b f\bigl(x(s),y(s)\bigr)\>\dot x(s)\>ds\tag{1}$$ in a similar vein.

Since $\gamma$ is parametrized with respect to arc length one has $$\dot {\bf z}(s)=\bigl(\cos\theta(s),\sin\theta(s)\bigr)\ ,$$ where $\theta(s)$ is the angle between the positive $x$-axis and the tangent vector $\dot{\bf z}(s)$. So we can replace $(1)$ by $$\int_\gamma f(x,y)\>dx=\int_a^b f\bigl(x(s),y(s)\bigr)\>\cos\theta(s)\>ds\tag{2}$$ In $(2)$ an "infinitesimal curtain element" no longer weighs in with its area $dS=f\bigl(x(s),y(s)\bigr)\>ds$ as in $(0)$ but with the area $dS'$ of the shadow (or projection) of this element onto the $(x,z)$-plane. Therefore one is tempted to say that $(1)$ represents the total area of the projected curtain.

But there is more to it: Note that $\cos\theta(s)$ has a sign. When $\cos\theta(s)$ is positive then the curtain has its good side towards the $x$-axis, and when $\cos\theta(s)$ is negative its backside. The latter parts of the shadow are counted negative. Similarly, if the same part of the $(x,z)$-plane is "shadowed" several times by successive pleats of the curtain, all these "coverings" are summed up with their proper sign in $(1)$, resp. $(2)$.

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This is not a really detailed answer, however:

I think that pulling back the integral to the $x$ or $y$ axis is geometrically unnatural (and you have to decide how you want to do this, since generically $C$ will not be a graph over either axis), so I wouldn't expect a really good geometric interpretation. But you could do something like the following. Break up $C$ into pieces $C_i$ so that each $C_i$ is a graph over either the $x$ or $y$ axis, and then you can use the chain/substitution rule to literally write the line integral as an integral w.r.t. $x$ or $y$ for each $C_i$, and this makes the relationship explicit. This construction gives you (if $C_i$ is a graph of $r(x)$, for instance) something like the area under the curve $f(x, r(x))\sqrt{1 + r'(x)^2}$, so it doesn't seem very natural.

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    $\begingroup$ So if there is no geometric interpretation of line integrals wrt to x or y, what exactly am i calculating when i perform this operation...what is the purpose of line integrals wrt to x and y? $\endgroup$
    – Jim_CS
    Commented Mar 4, 2012 at 21:17
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    $\begingroup$ I guess I would turn it around and ask you, what do you mean by line integral w.r.t. x or w.r.t y? Whenever I do a line integral over a curve $C$, I usually want to think of $C$ as a parametrized curve, that is $C$ is the image of a function $$ s \to (f(s), g(s))$$ in which case it makes sense to use the line element $ds$ instead of trying to relate things back to the $x$ or $y$ axis. If I want $C$ to lie in the $x, y$ plane I can set $f(s) = x$ and $g(s) = y$ but under this setup it isn't clear that I should get anything meaningful if i try to integrate over the $x$ or $y$ axis. $\endgroup$
    – treble
    Commented Mar 4, 2012 at 21:22
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    $\begingroup$ By w.r.t. to x or y I mean using dx instead of ds. I am reading Chapter 16 of Calculus Early Transcendentals and it features these line integrals w.r.t. to x and y but it gives no information as to their meaning. $\endgroup$
    – Jim_CS
    Commented Mar 4, 2012 at 21:38
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    $\begingroup$ I took a calculus class from that book a long time ago, but I no longer have a copy and therefore I would need to see the page of text to which you are referring in order to figure out what is happening. But if I had to guess I would say they probably mean to consider $s = x$ so that $C$ is given by the parametric equations $(x, g(x))$. In this case you may use the definition of line integral to re-write the line integral as an ordinary single-variable calculus integral. $\endgroup$
    – treble
    Commented Mar 4, 2012 at 21:43
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Wouldn't a geometric representation of a line integral in respect to x be taking the surface whose area you would have calculated when taking the line integral in respect to arc length:

https://i.sstatic.net/8LVrc.png (the same image in another answer here)

and then projecting that surface onto the XZ plane. The area of that projection would essentially be the geometric representation of the line integral in respect to x. A line integral in respect to y would be the projection of that same surface onto the YZ plane. The projections would be like viewing that surface in the image above just through the XZ plane and then through the YZ plane. Is this a correct interpretation? Sorry, I am new here so I couldn't put the image in my post, so I just put the link.

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