Line Integrals - Work Done

As a particle subject to a force
\[\mathbf{F}\]
moves from a point A to a point B, it does work. The work done is equal to
\[W=\int^B_A \mathbf{F}\cdot d \mathbf{r}\]
. Any curve
\[\mathbf{r}= \mathbf{r(x,y,z)}\]
can be written in parametric form
\[\mathbf{r}=\mathbf{r(x(t),y(t),z(t))}\]
. For example, suppose a particle subject to a force
\[\mathbf{F}=xy\mathbf{i}+x^2y\mathbf{j}+yz\mathbf{k}\]
moves from a point
\[A(1,0,0)\]
to a point
\[B(1,2,3)\]
In general the work done will depend on the path taken. Only in the specific case where
\[\nabla \times \mathbf{F}=0\]
is the integral independent of the path. In this case
\[ \begin{equation} \begin{aligned}\nabla \times \mathbf{F} &=(\frac{\partial}{\partial x}\mathbf{i}+\frac{\partial}{\partial y}\mathbf{j}+\frac{\partial}{\partial z}\mathbf{i})\times (xy\mathbf{i}+x^2y\mathbf{j}+yz\mathbf{k}) {} \\ &=(\frac{\partial (yz)}{\partial y}-\frac{\partial (x^2 y)}{\partial z})\mathbf{i}+(\frac{\partial (xy)}{\partial z}-\frac{\partial (xy)}{\partial z})\mathbf{j}+(\frac{\partial (x^2 y)}{\partial x}-\frac{\partial (xy)}{\partial y})\mathbf{k}\\{} \\ &=z \mathbf{i}+(2xy-x) \mathbf{k}\neq 0 \end{aligned} \end{equation}\]
.
The integral does depend on the curve followed between A and B. If we dollow a straight line between A and B, we can use the curve
\[\mathbf{r(t)}=\mathbf{i}+t(2 \mathbf{j}+ 3 \mathbf{k})=\mathbf{i} + 2t \mathbf{j}+3t \mathbf{k}, 0, then
\[d \mathbf{r} =2 \mathbf{j} +3 \mathbf{k}\]

In terms of t,
\[\mathbf{F}=xy\mathbf{i}+x^2y\mathbf{j}+yz\mathbf{k}=2t \mathbf{i} +2t \mathbf{i}+6t^2 \mathbf{k}\]

The line integral is
\[ \begin{equation} \begin{aligned} W &=\int^B_A \mathbf{F} \cdot d \mathbf{r} \\ &=\int^1_0 (2t \mathbf{i} +2t \mathbf{j}+6t^2 \mathbf{k}) \cdot (2 \mathbf{j} +3 \mathbf{k}) d \mathbf{r} \\ &=\int^1_0 4t+18t^2 dt \\ &=[2t^2+6t^3]^1_0 \\ &=(2+6)-(0+0)=8. \end{aligned} \end{equation} \]