Line integral over a vector field using a potential function. Let $X(t)=\left ( \pi\cos t, t, e^{\sin t} \right ), 0\leq t\leq \pi $.
And let $\textbf{F}(x,y,z)=\left ( 2xy+\frac{-y}{x^2+y^2}, x^2+ze^{yz}+\frac x{x^2+y^2}, ye^{yz} \right )$ be a vector field.
I want to calculate $\int _X\textbf{F}\cdot d\textbf{s}$.
First, I found that $\varphi (x,y,z)=x^2y+e^{yz}+\arctan \frac yx$ is a potential function of $\textbf{F}$.
So I had $\int _X\textbf{F}\cdot d\textbf{s}=\int _Xgrad\varphi \cdot d\textbf{s}=\int_{0}^{\pi }grad\varphi (X(t))\cdot X'(t)dt=\int_{0}^{\pi}\frac d{dt}\left ( \varphi (X(t)) \right )dt=\varphi (X(\pi ))-\varphi (X(0))$.
But I figured out that $\varphi$ is not a potential function of $\textbf{F}$ since $\varphi$ is not differentiable at which $x=0$. I know that if $X$ does not pass through $x=0$, then there is no problem with above calculation. However, the given $X$ passes through $(0,\frac{\pi}2,e)$. Calculation by definition of line intgral is too hard for me. What can I do?
 A: Your path is $\textbf{r}=(\pi \cos t,t,e^{\sin t})\space(t\in[0,\pi])$. Hence $\textbf{dr}=(-\pi \sin t,1,e^{\sin t}\cos t)dt$. Along the path $\textbf{F}$ is 
$$\left(2\pi t\cos t-{t\over \pi^2\cos ^2t+t^2},\pi^2\cos ^2t+e^{\sin t+te^{\sin t}}+{\pi \cos t\over \pi^2\cos ^2t+t^2},te^{te^{\sin t}}\right)$$
And F$\cdot$dr/dt is $$-\pi^2 t\sin 2t+{\pi t\sin t\over \pi^2\cos ^2t+t^2}+\pi^2\cos ^2t+e^{\sin t+te^{\sin t}}+{\pi \cos t\over \pi^2\cos ^2t+t^2}+\cos t.te^{te^{\sin t}+\sin t}$$
Over $[0,\pi]$ integrals of both $1$st and $3$rd terms evaluate to ${\pi^3\over 2}$. Now notice that $${de^{te^{\sin t}}\over dt}=e^{te^{\sin t}+\sin t}{\left(1+t\cos t\right)}=\text{integrand for the 4th term+last term }$$  As a result contribution of those terms in the integral is $e^{\pi}-1$.
Anti-derivative of the sum of the $2$nd and $5$-th term is
$$\pi \int{t\sin t+\cos t\over \pi^2\cos ^2t+t^2}dt=\pi \int{t\sec t.\tan  t+\sec t\over \pi^2+t^2\sec ^2t}dt=\pi \int{{d(t\sec t)\over dt}\over  \pi^2+t^2\sec ^2t}dt=\pi \int{1\over\pi^2+u^2}du=\tan  ^{-1}{u\over \pi}+C=\tan  ^{-1}{t\sec t\over \pi}+C$$
So contribution of those terms evaluates to $\tan  ^{-1}(\sec \pi)={3\pi \over 4}.$
So the final line integral
is$${\pi^3}+{3\pi \over 4}+e^{\pi}-1$$
Same as what your potential function gives.....
A: There is no problem with differentiability unless both $x$ and $y$ are $0$, which cannot happen for your curve where $x(t)=\pi\cos t$ and $y(t)=t$. If $y$ is $0$, then $t=0$, in which case $x=x(0)\ne 0$. So I think everything's okay.
