## Integral of Normal Component of a Vector Field Along a Curve

Give a vector field
$\mathbf{F}$
and a curve
$C$
the integral of the component of
$\mathbf{F}$
normal to
$C$
along
$C$
from
$A$
to
$B$
is given by
$\int^B_A \mathbf{F_n} \cdot d \mathbf{r}$
where
$\mathbf{F_n}$
is the component of
$\mathbf{F}$
normal to
$C$
.
In two the
$xy$
plane we can take
$\mathbf{n} = \mathbf{T} \times \mathbf{k}=(\frac{dx}{ds} \mathbf{i} + \frac{dy}{ds}\mathbf{j}) \times \mathbf{k}=\frac{dy}{ds}\mathbf{i} -\frac{dx}{ds}\mathbf{j}$
.
With
$\mathbf{F}=F_1 \mathbf{i} + F_2 \mathbf{j}$
we have

$\int^B_A \mathbf{F_n} d \mathbf{r} =\int^B_A \mathbf{F} \cdot \mathbf{n} ds= \int^B_A -F_2 dx + F_1 dy$

Take
$A=(0,0),B=(1,4)$
, take
$\mathbf{F}=e^x \mathbf{i} + e^{3x} \mathbf{j}$
and take the curve
$C$
to be
$y=x^2 +3x \rightarrow dy=2xdx+3dx=(2x+3)dx$
.
Then
\begin{aligned} \int^B_A \mathbf{F_n} \cdot d \mathbf{r} &= \int^1_0 -e^{3x} dx + int^1_0 e^x (2x+3) dx \\ &= [- \frac{1}{3}e^{3x}]^1_0 +[(2x+3)e^x - 2e^x]^1_0 \\ &= - \frac{1}{3}(e^3 -1)+ 3e-1 \\ &=3e-\frac{1}{3}e^3 - \frac{2}{3} \end{aligned}