Solving a line integral A vector field $\overrightarrow{E}=7x^2\hat{e}_x +3y\hat{e}_y-2xz\hat{e}_z$
Evaluate the line integral I$=\int_{c} \overrightarrow{E}\bullet d\overrightarrow{I}$ where the contour C is the straight line from the point (0,0,0) to (1,2,0)
I was wondering if we are allowed to split the intergral up, so for example
$I= \int_{0}^{1} 7x^2 dx + \int_{0}^{2} 3yzdx$
So do the line integral from (0,0,0) to (1,0,0) in the first integral then (1,0,0) to (1,2,0).
or is there an easier method.
 A: You could choose the path $(0,0,0)$ to $(1,0,0)$ to $(1,2,0)$. Please note that at each step you have only one of $x, y, z$ change.
$\vec{E}=7x^2 \, \hat{x} +3y \, \hat{y}-2xz \, \hat{z}$
So, $ \displaystyle \int_{c} \vec{E}\cdot d\vec{I} = \int_0^1 7x^2 \, dx + \int_0^2 3y \, dy$
Or you can use the parameterized equation of line between $(0,0,0)$ to $(1,2,0)$.
$ \displaystyle \frac{x-0}{1-0} = \frac{y-0}{2-0} = \frac{z-0}{0-0} = t \implies x = t, y = 2t, z =0$.
$dx = dt, dy = 2dt$.
When $(0,0,0), t = 0$; when $(1,2,0), t = 1$.
So, $ \displaystyle \int_{c} \vec{E}\cdot d\vec{I} = \int_0^1 7t^2 \, dt +  3(2t)(2dt) = \int_0^1 (7t^2 +  12t) dt = [\frac{7t^3}{3}+6t^2]_0^1 = \frac{25}{3}$
EDIT: I earlier missed to call out that if the vector field is not conservative, we would have to choose the path that is given as line integral would not be path independent. However in this specific case, choosing the path $C_1$ from $(0, 0, 0)$ to $(1, 0, 0)$ to $(1, 2, 0)$ works. The direction vector from $(0, 0, 0)$ to $(1, 2, 0)$ is in $XY$ plane (no $z$ component) and if we take only $XY$ component of our vector field ($7x^2, 3y$), it is conservative ( $\nabla (\frac{7}{3}x^3, \frac{3}{2}y^2)$). Our chosen path $C_1$ also ensures that we are only moving in $XY$ plane.
