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I have this question from "Div,Grad,Curl and all that" by Schey

I have: $$F =e_{\theta}/r$$ Find the line integral F t from the point $P_{1}(0,-1,0)$ to point $P_{2}(0,1,0)$ over two difference paths:$C_{R}$, the right hand side of the circle of radius 1 lying on xy plane centered at the origin, and $C_{L}$, the left hand side of the same circle.

The problem is, the question states that the result of line integration over the two different paths ($C_{R}$ and $C_{L}$) is not the same (path dependent line integral); but my calculations shows that it is path independent i.e. Line integration over path $C_{R}$ = line integration over path $C_{L}$...!? Am I wrong?!

Figure demonstrating C_{R} and C_{L} paths

user -1
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2 Answers2

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Very likely you did something like this:

On the right arc $P_1P_2$, $\mathrm{d}\vec\ell=\vec{e}_\theta\mathrm{d}\theta$, hence $$\vec F\cdot\mathrm{d}\vec\ell=1.$$ On this path, $\theta$ ranges from $-\pi/2$ to $\pi/2$, hence the circulation is $\pi$.

On the left arc $P_1P_2$, $\mathrm{d}\vec\ell=-\vec{e}_\theta\mathrm{d}\theta$, hence $$\vec F\cdot\mathrm{d}\vec\ell=-1.$$ On this path, $\theta$ ranges from $3\pi/2$ to $\pi/2$ (it's decreasing!), hence the circulation is $-1\times(\pi/2-3\pi/2)=-1\times(-\pi)=\pi$.

Let's recall the general method to compute the circulation of a vector field along a path: the first thing you need is a parametrization of this path, say $$[a,b]\longrightarrow\mathbb{R}^2:t\longmapsto\bigl(x(t),y(t)\bigr).$$ (with $a<b$, and that's important). Then, given a vector field $\vec F$, the circulation is given by $$\mathscr{C}=\int_a^b\vec F\bigl(x(t),y(t)\bigr)\cdot\bigl(x'(t),y'(t)\bigr)\,\mathrm{d}t.$$

Now in your case, the left arc $P_1P_2$ is parametrized by $$\begin{cases}x(t)=\cos(t)\\y(t)=\sin(t)\end{cases}\qquad t\in[-\pi/2,\pi/2].$$ Observe that $\bigl(x'(t),y'(t)\bigr)=\vec e_\theta$ hence $$\mathscr{C}=\int_{-\pi/2}^{\pi/2}\,\mathrm{d}\theta=\pi$$ (since on this path, $r=1$).

The right arc $P_1P_2$ is parametrized by $$\begin{cases}x(t)=-\cos(t)\\y(t)=\sin(t)\end{cases}\qquad t\in[-\pi/2,\pi/2].$$ Observe that $\bigl(x'(t),y'(t)\bigr)=-\vec e_\theta$ hence $$\mathscr{C}=-\int_{-\pi/2}^{\pi/2}\,\mathrm{d}\theta=-\pi$$ (since on this path, $r=1$).

Try to see where (and why) your reasoning was wrong.

gniourf_gniourf
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  • I did: $r(t)=\cos(\theta)i + r\sin(\theta)j$ I parameterized the circle as a whole. but even we assumed your assumption is right for the right hand side, in the left hand side where the parameterization is: $x(t)=-cos(t)$ and $y(t)=sin(t)$, but $\bigl(x'(t),y'(t)\bigr)$, does not equal to $-e_{\theta}$.....$e_{\theta}=-\sin(t)i+\cos(t)j$ so $-\vec e_\theta=\sin(t)i-\cos(t)j$....right?!?! – user -1 Jun 29 '14 at 18:48
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    @user3556914 Don't confuse the parameter $t$ of the parametrization of the curve, and $\theta$, the angular coordinate of a point in the plane. With $x(t)=-\cos t$ and $y(t)=\sin t$, the point $\bigl(x(t),y(t)\bigr)$ has angular coordinate $\theta=2\pi-t$ (draw a picture), and there,$$\vec e_\theta=-\sin(2\pi-t)\vec i+\cos(2\pi-t)\vec j=\sin(t)\vec i+\cos(t)\vec j=x'(t)\vec i+y'(t)\vec j.$$ – gniourf_gniourf Jun 29 '14 at 18:55
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    I was pondering more into this, probably I am digging too deep and confusing myself. I got that: $\vec e_\theta=-\sin(2\pi-t)\vec i+\cos(2\pi-t)\vec j=\sin(t)\vec i+\cos(t)\vec j=x'(t)\vec i+y'(t)\vec j$. but how does that equals to $-\vec e_\theta$ as you implied in your original answer where you said: "observe that $\bigl(x'(t),y'(t)\bigr)=-\vec e_\theta$". the negative sign is what bugging me, I tried to convince myself the negative sign came from the reverese of the rotation, but I can't find a Mathematical explanation to that!!!! Thanks in advance. – user -1 Jun 30 '14 at 01:24
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    @user3556914 I made a mistake in my previous comment: here's what I should have written (sorry about this): Don't confuse the parameter $t$ of the parametrization of the curve, and $\theta$, the angular coordinate of a point in the plane. With $x(t)=−\cos t$ and $y(t)=\sin t$, the point $\bigl(x(t),y(t)\bigr)$ has angular coordinate $\theta=\pi−t$ (symmetric wrt the $y$-axis of the point of coordinates $(\cos t,\sin t)$—draw a picture), and there, $$\vec e_\theta=−\sin(\pi−t)\vec i +\cos(\pi−t)\vec j=-\sin(t)\vec i-\cos(t)\vec j=x′(t)\vec i+y′(t)\vec j.$$ – gniourf_gniourf Jun 30 '14 at 09:21
  • @ gniourf_gniourf Thank You so much for the clarification. Indeed the reflection is by $\pi$ not by 2$\pi$. However I believe you made another small mistake $\vec e_\theta \neq x'(t) \vec i + y'(t) \vec i$. it should be like this for left side half $C_L: \int \vec F.t=\int \vec F(r(t)).r'(t) dt$ where $r'(t)=sin(t)\vec i+cos(t) \vec j$ and $\vec F(r(t))=\vec e_\theta=-sin(t)\vec i -cos(t)\vec j$ So: $\int \vec e_\theta d\theta$=$\int (-sin(t)\vec i -cos(t)\vec j).(sin(t)\vec i+cos(t) \vec j) dt$ = $-\int_{-\pi/2}^{\pi/2},\mathrm{d}t=-\pi$ – user -1 Jun 30 '14 at 16:23
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I think that the integral over the right is $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d \theta=\pi$$ and the left is

$$\int_{-\frac{\pi}{2}}^{-\frac{3\pi}{2}}d \theta=-\pi$$

Rene Schipperus
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