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