Work done by force field in moving particle along given path Path: $y=x^2$, from ($-\pi$, $\pi^2$) to ($\pi$,$\pi^2$)
Field: $F(x,y) = e^y\sin(x)$ i$-(e^y\cos(x)-\sqrt {1+y})$ j
I began this problem by parametrizing the path by $x=t$, $y=t^2$.
Next, setting $r(t)=t$ i$+t^2$ j, found $r'(t)$.
I then used the formula for work, $W=\int_a^bF\bullet r'(t)dt$, which forces me to take the integral of $e^{t^2}\sin(t)$. This is impossible as far as I know. 
I tried a different parametrization, $x=\sqrt{\ln(t)}, y=\ln(t)$ but also ran into an undeterminable integral.
Where am I going wrong?
 A: I got
$$\int_{-\pi}^\pi(e^{t^2}\sin t-2te^{t^2}\cos t+2t\sqrt{1+t^2})\,dt.$$
I then thought about the derivative of $e^{t^2}\cos t$.
A: This can actually be solved rather nicely using Green's Theorem in the plane by connecting your parameterized arc to the arc $r(t) = (-t,\pi^2)$ as $t$ ranges from $-\pi$ to $\pi$.  This closes a region in the plane, and if we acknowledge that the work done by the force along this contour is zero then we can immediately apply Green's Theorem.  Note that 
\begin{eqnarray*}
F(r(t))\cdot r'(t) & = & \left(-e^{\pi^2}\sin(t), \; -e^{\pi^2}\cos(t)+ \sqrt{1+\pi^2} \right ) \cdot (-1,0) \\
& = & e^{\pi^2}\sin(t).
\end{eqnarray*}
Integrating this between $\pi$ and $-\pi$ yields zero.  Hence if $C_1$ was the contour described in your problem, and $C_2$ is the contour I just described here, we have that 
$$
\oint_C F(r(t))\cdot r'(t)dt \;\; =\;\; \int_{C_1}F(r(t))\cdot r'(t)dt + \underbrace{\int_{C_2} F(r(t))\cdot r'(t)dt}_{=0} \;\; =\;\; \iint \left (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right ) dA.
$$
What we notice from Green's theorem though is that 
$$
\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \;\; =\;\; e^y\sin(x) -e^y\sin(x) \;\; =\;\; 0. 
$$
This proves that work done by your force along the curve $(t,t^2)$ is zero.
