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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?

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    $\begingroup$ On your specific question about what to do with the curve $y = x, z = \ln(1+x)$, the parametrization $(t, t, \ln(1+t))$ is representing this curve. You parametrized looking at the start and end points which is not the right way. You should look at the equation of the surfaces to parametrize the intersection curve. The start and end points then help in finding the limits of the variable(s), you are using for parametrization. $\endgroup$
    – Math Lover
    Nov 28 '21 at 16:24
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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$

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  • $\begingroup$ I'm not sure i understand the how $F$ can be non conservative while $F_1$ is conservative. Is it that $F_1=2xsin(\pi y)-e^z$ so it's only the first part that is conservative. Also how did you get $F_2=(0,-3,0)$ $\endgroup$
    – Zert44
    Nov 28 '21 at 16:42
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    $\begingroup$ I separated $ - 3 e^z$ from the y-component of $F$. Also $F_2$ is not $(0, -3, 0)$. It is $(0, -3e^z, 0)$. Rest is $F_1$ and is conservative. $\endgroup$
    – Math Lover
    Nov 28 '21 at 16:45
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    $\begingroup$ I have clearly written what is $F_1$ and what is $F_2$ and what is $f(x, y, z)$. When a vector field is conservative, what do you do? You find the scalar function whose gradient is the given vector field right? $f(x, y, z)$ is that scalar function as $ \nabla f(x, y, z) = \vec F_1$ $\endgroup$
    – Math Lover
    Nov 28 '21 at 16:48
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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 $t=1$, $r(1)=(1,1,\ln 2)$.

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  • $\begingroup$ So both $z=ln(1+t)$ and $y=t$ are given in $r(t)$ and I would not need to use them later? $\endgroup$
    – Zert44
    Nov 28 '21 at 15:44
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    $\begingroup$ Correct. All you need to do is to calculate $d r$ from the equation of the curve, use the dot product, and integrate from $t=0$ to $t=1$. $\endgroup$
    – Andrei
    Nov 28 '21 at 16:07

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