# Work done by a Force Field( Green's Theorem)

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 $• Your idea is good but I think your limits when doing polar coordinates with the closed path are wrong, as for example the radius goes from two to one as the path you have is a semicircle of radius one around$\;(1,0)\;$, not around the origin$\;(0,0)\;$....You may want either to correct this or to do a traslation to the origin. – DonAntonio Jul 24 '16 at 19:48 • Thank you Don! I can see where is my mistake. – JJWho Jul 24 '16 at 20:09 ## 1 Answer 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$C = C_1 + C_2 + \cdots + C_N$. $$\\$$$\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;

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

• 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... – JJWho Jul 24 '16 at 19:43
• I made an edit. Your coordinate-change should help you get this now. – Faraad Armwood Jul 24 '16 at 19:57
• Thanks Faraad! I will edit and put the correct resolution. – JJWho Jul 24 '16 at 20:09
• Sorry, Faraad I didn't know that I had to do it. – JJWho Jul 25 '16 at 19:54
• It's just so the question won't stay open. – Faraad Armwood Jul 25 '16 at 19:55