## Integral of Tangential Component of a Force Along a Curve

Given a curve
$\mathbf{r}=\mathbf{r}(x,y)$
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
$\mathbf{r}=\mathbf{r}(x(s),y(s))$
then
Where
$s$
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{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}
.