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Given a vector field $\vec{F} = zy\hat{x} + zx\hat{y} + xy\hat{z}$ evaluate the integral $\vec{F} \cdot d\vec{l}$ from $(0,0,0)$ to $(1,2,3)$. Note please do not parameterize the path.

Attempt: Let $d\vec{l} = \Big( (1-0)\hat{x} + (2-0)\hat{y} + (3-0)\hat{z} \Big)dz$
Therefore $\int\vec{F} \cdot d\vec{l} = \int_{0}^{3} (zy\hat{x} + zx\hat{y} + xy\hat{z}) \cdot ( \hat{x} + 2\hat{y} + 3\hat{z})dz = \frac{9y}{2} + 9x + 9xy$.

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One thing I need to mention is that: $$\vec{dl}=dx \hat x+dy \hat y+dz \hat z$$

So the integral now becomes $$\begin{align} & \int_{(0,0,0)}^{(1,2,3)} (zydx+zxdy+xydz) \\ & =\int_{(0,0,0)}^{(1,2,3)} d(xyz) \\ & =1\cdot 2\cdot 3 -0 \\ & =6\end{align}$$

EDIT: In case you want to parametrize the path, proceed as follows-

Now let $x=t$, $y=2t$ and $z=3t$.

Then the integral becomes $$\int_{0}^{1} (6t^2\cdot dt+3t^2\cdot 2dt+2t^2\cdot 3dt)$$

Hope this helps you.

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    I forgot to mention that I wanted to avoid parameterizing the path. I found the answer this way but it was not used in class so I want to avoid it. The method my professor used in class can be found here https://www.docdroid.net/umdgVal/microsoft-word-fwlec3-9-22-04.pdf.html and it's not very helpful with determining the $d\vec{l}$ – 2017-02-24
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    @Drew See if this method suits you. Updated the answer. – 2017-02-24
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    @Drew Note that this method works here only because it is a perfect differential. – 2017-02-24