\[\mathbf{r}=\mathbf{r}(x,y)\]
and a vector field \[\mathbf{F}(x,y)=F_1(x,y) \mathbf{i} + F_2 (x,y) \mathbf{j}\]
, to evaluate the integral \[\int_C F_T d r\]
where \[F_T\]
is the component of \[\mathbf{F}\]
along the curve, write \[\int_C F_T d r = \int_C \mathbf{F} \cdot d \mathbf{r}=\int_C (F_1 \mathbf{i} + F_2 \mathbf{j}) \cdot (dx \mathbf{i} +dy \mathbf{j})=\int_C F_1 dx + F_2 dy\]
We can parametrize the curve
\[\mathbf{r}=\mathbf{r}(x(s),y(s))\]
thenWhere
\[s\]
is the distance along the curve.
\[\int_C F_T d r=\int_C F_1 dx + F_2 dy =\int_C (F_1 (s)\frac{dx}{ds} + F_2(s) \frac{dy}{ds}) ds\]
.Example: Evaluate
\[\int_C F_T d r\]
along thureve \[y=x\]
from \[(0,0)\]
to \[(1,1)\]
for the vector field \[\mathbf{F}=(x^2 -y^2 ) \mathbf{i} + 2xy \mathbf{j}\]
For the line
\[y=x\]
use the parametrization \[x=\frac{s}{\sqrt{2}} , y=\frac{s}{\sqrt{2}}\]
When
\[(x,y)=(0,0),(1,1)\]
\[s=0, \sqrt{2}\]
respectively.The integral becomes
\[\begin{equation} \begin{aligned} \int_C F_T d r &= \int_C (F_1 (s)\frac{dx}{ds} + F_2(s) ) ds \\ &=
\int_0^{\sqrt{2}} (((\frac{s}{\sqrt{2}})^2 -(\frac{s}{\sqrt{2}})^2) +2 \frac{s}{\sqrt{2}} \frac{s}{\sqrt{2}}) \frac{dy}{ds}) \frac{1}{\sqrt{2}}ds \\ &=
\int_0^{\sqrt{2}} s^2 \frac{1}{\sqrt{2}}ds \\ &= \frac{1}{\sqrt{2}} [\frac{s^3}{3}]_0^{\sqrt{2}} \\ &= \frac{1}{\sqrt{2}} \frac{(\sqrt{2})^3}{3}-0 \\ &= \frac{2 }{3} \end{aligned} \end{equation}\]
.