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I need to calculate $\int_\Gamma F(x) \, dx$ from $(0,1)$ to $(-1,0)$ of the unit circle.

$$F(x) = (x,y)$$

Now the answer is: enter image description here

But I don't understand what they did. Why $\Gamma(t) = (\cos t, \sin t)$ & $t: \pi/2 \mapsto \pi$? And why the calculate I like this?

I want to understand this kind of integration and I'd like to get some help. Thanks for any kind of help.

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    $\begingroup$ Is it possible that you read $F(x)\cdot dx$, with the dot indicating the dot product of vectors, and you wrote it as $F(x)\,dx$ without the dot? ${}\qquad{}$ $\endgroup$ Commented Aug 1, 2014 at 16:57
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    $\begingroup$ They parametrized the circle. $\endgroup$ Commented Aug 1, 2014 at 16:57
  • $\begingroup$ @MichaelHardy - yes, x is a vector. My bad. $\endgroup$
    – momd
    Commented Aug 1, 2014 at 16:58
  • $\begingroup$ @MhenniBenghorbal - for any circle they'll parametrized like this? for (cost, sint)? $\endgroup$
    – momd
    Commented Aug 1, 2014 at 16:58
  • $\begingroup$ @momd: See the answer. $\endgroup$ Commented Aug 1, 2014 at 17:01

3 Answers 3

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It looks like you've copied a few things down wrong.

The unit circle is parametrised by $\Gamma(t) = (\cos t, \sin t)$, where $0 \le t \le 2\pi$. This parametrisation starts at $(1,0)$ and runs, anti-clockwise, all the way around and back to $(1,0)$. You can substitute in some $t$-values and you'll see that $\Gamma(\frac{\pi}{2})=(0,1)$ and $\Gamma(\pi) = (-1,0)$. So, taking $\frac{\pi}{2} \le t \le \pi$ runs around the circle, anti-clockwise, from $(0,1)$ to $(-1,0)$.

For the line integral of a force you find the integral $$\int {\bf F}({\bf x}) \cdot \mathrm{d}{\bf x}$$

If we make the substitution ${\bf x} = \Gamma(t)=(\cos t, \sin t)$ then ${\bf F}({\bf x}) = (x,y) = (\cos t, \sin t)$. The differential $\mathrm{d}{\bf x}$ changes as well. We have $$\mathrm{d}{\bf x} = \mathrm{d}\left[\Gamma(t)\right] = \frac{\mathrm{d}\Gamma}{\mathrm{d}t}\mathrm{d}t$$ Since $\Gamma(t) = (\cos t, \sin t)$ we have $\mathrm{d}\Gamma/\mathrm{d}t = (-\sin t, \cos t)$. That gives $$\int_{\pi/2}^{\pi} (\cos t, \sin t) \cdot (-\sin t, \cos t)~\mathrm{d}t$$

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  • $\begingroup$ @momd No problem. Don't forget to up-vote any posts you found helpful (click the up-arrows next to the posts), and to select the most useful answer (click the tick next to the post). $\endgroup$ Commented Aug 1, 2014 at 17:43
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Your circle has radius $r=1$. For any circle with radius $a$ and centered at the origin we have the parametrization

$$ (a\cos(t),a\sin(t)). $$

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Recall from trigonometry that the point $(\cos t,\sin t)$ moves around the circle of unit radius centered at $(0,0)$, and has the following values: $$ \begin{array}{c|c} t & (\cos t,\sin t) \\ \hline 0 & (1,0) \\ \pi/2 & (0,1) \\ \pi & (-1,0) \\ 3\pi/2 & (0,-1) \\ 2\pi & (1,0) \\ \hline \end{array} $$ Then it has gone all the way around the circle and starts over. This explains why $t$ goes from $\pi/2$ to $\pi$.

Then, differentiating, we get \begin{align} x & = (\cos t,\sin t) \\[8pt] dx & = (-\sin t,\cos t)\,dt \end{align} and $F(x)$ is the identity function, so we're integrating $x\cdot dx$.

So $F(x)\cdot dx = (\cos t, \sin t)\cdot(-\sin t,\cos t)\,dt$.

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