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$\int_{\gamma}(2xz)\,dx+(\sin(z))\,dy+(x^2+y*\cos(z)+2\sin^2(z))\,dz$

If $\gamma$ is an oriented curve of class $C^1$ in $R^3$ from $M(1,0,0)$ as a start point and $N(0,1,\frac{\pi}{2})$ as an end point.

I know that I need to find a function $F$ and in the end calculate $F(N)-F(M)$, but I don't know how to find it in $R^3$. What derivatives do I need to compute to show that the function is indeed path independent and how to find $F$.

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The theorem you should use the fundamental theorem for line integrals

$$ \int_C \nabla \cdot f ~dr = f(b) -f(a) $$

where $C$ is the differentiable curve with starting point $a$ and endpoint $b$. Thus by taking the integral of each part and fixing the integration constants (which are functions) you can find the solution for which the fundamental theorem for line integrals applies. Thus for

$$ \int 2xz ~dx = x^2z + j(z,y),$$

$$ \int sin(z) ~dy = y\sin(z) + k(x,z),$$

$$ \int x^2+y\cos(z)+2\sin^2(z) ~dz = x^2z+y\sin(z)+ \frac{2z-\sin(2z)}{2} + l(x,y),$$

there can be found functions $j(z,y)$, $k(x,z)$ and $l(x,y)$ so that the three integrals are equals, and thus equal to $f$ in the fundamental theorem for line integrals.

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    And how can i find those functions? – 2017-02-11
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    Set the integrals equal to each other, if you do this for the first and second then you see that $j(z,y)$ will contain a term $y\sin(z)$ and $k(x,z)$ a term $x^2z$, setting the integral equal to the third will give rise to another term in the functions and this will be your $f$ – 2017-02-12