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Question: Compute the work done by a force Field $ F(x,y)=(2xe^y-x^2y-\frac{y^3}{3},x^2e^y+sin(y)) $ when a particle moves moves around the path describe by $ r(t)=(1+cos(t),sin(t)),0 \leq t \leq \pi $.

My try: I think to use the Green's Theorem but the curve is not closed then I choose a way to close it and use the theorem because all the other conditions would be satisfy, but I'm doing something wrong that I can't see,because the answer is $3\frac{\pi}{4}-4. Here is what I did,

Let $P=2xe^y-x^2y-\frac{y^3}{3}$, $Q=x^2e^y+sin(y)$ and $\partial D= \phi_1 \cup \phi_2 $, where $\phi_1(t) = r(t)$ and $\phi_2(t) = (2t,0),0 \leq t \leq \ 1 $. In that way, D is the region in the plane limited by $\partial D$. Then

$\oint_{\partial D} (P\, dx+Q\, dy) = \iint_D \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dx\,dy$

Now I will define $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y}= 2xe^y- x^2 - y^2 $ ; $\frac{\partial Q}{\partial x}= 2xe^y $

Hence,

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}= x^2+y^2 $

Then already using polar coordinates I calculated that integral:

$\iint_D \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dx\,dy= \int_{0}^{\pi}\int_{0}^{1} r^3 drdt = \frac{\pi}{4} $

So,

$ \oint_{\partial D} (P\, dx+Q\, dy) = \frac{\pi}{4} $

Using the property of Line Integrals I have,

$\oint_{\partial D} (P\, dx+Q\, dy)= \int_{\phi_1} (P\, dx+Q\, dy)+ \int_{\phi_2} (P\, dx+Q\, dy) $

Follows that,

$ \int_{\phi_1} (P\, dx+Q\, dy)=\oint_{\partial D} (P\, dx+Q\, dy)- \int_{\phi_2} (P\, dx+Q\, dy) $

Where,

$ \int_{\phi_2} (P\, dx+Q\, dy) = \int_{0}^{1} 2(2*(2t)*e^0-2t*0) dt = 4 $

I omitted the second part of the above integral because we multiply it by zero from the second component of $(\phi_2)'(t)$

Finally,

$ \int_{\phi_1} (P\, dx+Q\, dy)=\oint_{\partial D} (P\, dx+Q\, dy)- \int_{\phi_2} (P\, dx+Q\, dy) = \frac{\pi}{4} - 4 $

What is my mistake? thanks for the help community ;)

Note: I'm learning to use the TeX,I'm sorry for my mistakes...

EDIT: Credits to Don and Faraad:

Here I need to consider that the circle is not at de origin so that:

$\iint_D \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dx\,dy= \int_{0}^{\frac{\pi}{2}}\int_{0}^{2cos\theta} r^3 drd\theta = \frac{3\pi}{4} $

Then,

$ \int_{\phi_1} (P\, dx+Q\, dy)=\oint_{\partial D} (P\, dx+Q\, dy)- \int_{\phi_2} (P\, dx+Q\, dy) = \frac{3\pi}{4} - 4 $

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  • $\begingroup$ Your idea is good but I think your limits when doing polar coordinates with the closed path are wrong, as for example the radius goes from two to one as the path you have is a semicircle of radius one around $\;(1,0)\;$ , not around the origin $\;(0,0)\;$ ....You may want either to correct this or to do a traslation to the origin. $\endgroup$ – DonAntonio Jul 24 '16 at 19:48
  • $\begingroup$ Thank you Don! I can see where is my mistake. $\endgroup$ – JJWho Jul 24 '16 at 20:09
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Green's Theorem is for closed curves! Use the vector-line integral computation namely;

$$\textrm{Work} = \int_{c} \textbf{F} \cdot d \textbf{s} = \int_{t=a}^b \textbf{F}(c(t)) \cdot c'(t) \ dt$$

For a handwavy reason why you use this computation is because $\textbf{F}(c(t))$ is interpreted as the force acting on a particle at position $c(t)$. Then if your segment is small enough, the displacement is approximately $c'(t_i^*) \Delta t$ for $t_i^* \in [t_{i-1},t_i]$ and so we have;

$$\textrm{Work}=\lim_{\Delta t \to 0} \sum_{i=1}^N \ \textbf{F}(c(t_i^*)) \ c'(t_i^*)\ \Delta t =\int_{c} \textbf{F} \cdot d \textbf{s} = \int_{t=a}^b \textbf{F}(c(t)) \cdot c'(t) \ dt$$

In the above, $\textbf{F}(c(t_i^*)) c'(t_i^*) \delta t$ represents the work along a curve $C_i$ where we divided the original curve $C$ into $N$ parts i.e $C = C_1 + C_2 + \cdots + C_N$.

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$\textbf{Edit}$ (Using Green's Theorem): If we increase the interval to $[0,2\pi]$, define $C_1$ to be the portion of the curve on $[\pi, 2 \pi]$ then we get a closed curve and we have;

$$\textrm{Work} = \oint_D \textbf{F} \cdot d \textbf{s} = \int_C \textbf{F} \cdot d\textbf{s} + \int_{C_1} \textbf{F} \cdot d \textbf{s}$$

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  • $\begingroup$ Hi Faraand, I tried to close it, apply the theorem and after that take out the work generate from the segment that I use to close the curve. I've learned it with my teacher and tried to apply it. And if you see the force field is complicated. Solve that line integral without a computer is insane, i think... $\endgroup$ – JJWho Jul 24 '16 at 19:43
  • $\begingroup$ I made an edit. Your coordinate-change should help you get this now. $\endgroup$ – Faraad Armwood Jul 24 '16 at 19:57
  • $\begingroup$ Thanks Faraad! I will edit and put the correct resolution. $\endgroup$ – JJWho Jul 24 '16 at 20:09
  • $\begingroup$ Sorry, Faraad I didn't know that I had to do it. $\endgroup$ – JJWho Jul 25 '16 at 19:54
  • $\begingroup$ It's just so the question won't stay open. $\endgroup$ – Faraad Armwood Jul 25 '16 at 19:55

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