Line integral of non conservative vector field I'm having trouble finding the line integral of this problem. I have been given a vector field
$F=(2x\sin(\pi y)-e^z,\pi x^2\cos(\pi y)-3e^z,-xe^z)$
Where the curve $C$ intercepts between $z=\ln(1+x)$ and $y=x$ from $(0,0,0)$ to $(1,1,\ln(2)$.
So I try to do this by solving with: $\int_{C}^{} F \cdot dr$
I started of by determining conservativity and it is non conservative,
$\frac{\partial f_1}{\partial y} = \frac{\partial f_2}{\partial x}=2x\pi \cos(\pi y)$
$\frac{\partial f_1}{\partial z} = \frac{\partial f_3}{\partial x}=-e^z$
$-3e^z=\frac{\partial f_2}{\partial z} \ne \frac{\partial f_3}{\partial y}=0$
Next I found the vector function by using the two coordinats $(0,0,0)$ & $(1,1,\ln(2)$,
$r(t)=(t)\hat{i}+(t)\hat{j}+(\ln(2)t)\hat{k}$
From this vector function we could say that,
$x=t$, $y=t$ & $z=\ln(2)t$
Now usually I believe I should substitute $(x,y,z)$ into $F$ and then take the dot product between $F$ and $r'(t)$. But I'm not sure this correct, because I also have to take $z=\ln(1+x)$ and $y=x$ into consideration.
So now I'm stuck on how to proceed. What am I supposed to do with  $z=\ln(1+x)$ and $y=x$? And is it even the correct approach to this problem?
 A: The given vector field is $ \vec F=(2x\sin(\pi y)-e^z,\pi x^2\cos(\pi y)-3e^z,-xe^z)$.
$r(t)= (t, t, \ln(1+t)), 0 \leq t \leq 1$. As called out in the other answer you have a mistake in the z-component.
You are correct that the vector field is not conservative but what may help notice is that vector field $\vec F_1 = $
$(2x\sin(\pi y)-e^z,\pi x^2\cos(\pi y),-xe^z)$ is conservative. Its curl is zero and the potential function is,
$f(x, y, z) = x^2 \sin (\pi y) - x e^z $.
So, $ \displaystyle \int_C \vec F_1 \cdot dr = \int_C (\nabla f(x, y, z) \cdot dr = f(r(1)) - f(r(0))$
So all you are left with is to find line integral of $ \vec F_2 = (0, - 3e^z, 0)$ over the given curve which is straightforward.
So the line integral of $\vec F$ over the given curve is,
$ \displaystyle f(r(1)) - f(r(0)) + \int_0^1 \vec F_2 (r(t)) \cdot r'(t) ~ dt$
A: Instead of
$$r(t)=(t)\hat{i}+(t)\hat{j}+(\ln(2)t)\hat{k}$$
you have $$r(t)=(t)\hat{i}+(t)\hat{j}+\ln(1+t)\hat{k}$$
Notice that when $t=1$, $r(1)=(1,1,\ln 2)$.
