## 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{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}
.
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
$\mathbf{r(t)}=\mathbf{i}+t(2 \mathbf{j}+ 3 \mathbf{k})=\mathbf{i} + 2t \mathbf{j}+3t \mathbf{k}, 0<=t<=1$
, 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{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}