# Changing a differential expression into form $f ' (x, y, z)$

I need to evaluate line integral of $$(2xy-z) dx + (yz) dy + (x) dz$$ over any path from $(1,0,0)$ to $(2,1,4)$.

I thought of integrating once from $(1,0,0)$ to $(2,0,0)$ then to $(2,1,0)$ and lastly to $(2,1,4)$.

...but that wouldn't necessarily be valid for any path. I think I need to convert the expression in the form $d(f(x, y, z))$ but I'm being unable to do so.

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That is not a conservative field. Are you sure the problem is asking for a path-independent solution, or could it be asking you to integrate over any path you so choose? – apnorton Jan 31 '13 at 14:42
@anorton yes. F = (2xy−z)i+(yz)j+(x)k Evaluate ∫F.dr along any path from (1,0,0) to (2,1,4) – Shashwat Black Jan 31 '13 at 15:09

Let $C$ be the path of integration, $\mathbf{F}(x,y,z)=(2xy−z,yz,x)$ the vector field, and $d\mathbf{r}=(dx,dy,dz)$ an infinitesimal displacement along $C$.
Choose $C$ to be the line segment from $(1,0,0)$ to $(2,1,4)$. A parametrization for the line is given by $\mathbf{r}(t)=(1,0,0)+t(1,1,4)=(t+1,t,4t)$ for $t \in [0,1]$.
Note that $\mathbf{F}(\mathbf{r}(t))=(2t^2-2t,4t^2,t+1)$ and $\mathbf{r}'(t)=(1,1,4)$. Now we can evaluate the line integral:
$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{t=0}^{1} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \,dt = \int_{t=0}^{1} 6t^2+2t+4 \,dt = \left[2t^3+t^2+4t\right]_{t=0}^{1} = 7$$
However, it is to be noted that $\boldsymbol{\nabla}\times\mathbf{F}=(-y,-2,-2x)\ne\mathbf{0}$, so the integral depends on $C$.