11 - Line Integrals

Scalar Line Integral

Let $\mathbf{F}(r)$ be a vector field

Integral of a vector field $\mathbf{F}(\mathbf{r})$ along a path $C$ defined by $\mathbf{r}=\mathbf{r}(t)$
from $\mathbf{r}(t_0)$ to $\mathbf{r}(t_1)$ is

$${\int }_C\mathbf{F}(\mathbf{r})\cdot d\mathbf{r}={\int }_{t_0}^{t_1}\mathbf{F}(t)\frac{d\mathbf{r}}{dt}dt$$

Length of a Curve

Let $\mathbf{F}(\mathbf{r})$ be a vector field then
Integral of a vector field $\mathbf{F}(\mathbf{r})$ along a path $C$ defined by $\mathbf{r}=\mathbf{r}(t)$
from $\mathbf{r}(t_0)$ to $\mathbf{r}(t_1)$ is

$${\int }_C\mathbf{F}(\mathbf{r})\cdot d\mathbf{r}={\int }_{t_0}^{t_1}|\dot{\mathbf{r}}(t)|dt={\int }_{t_0}^{t_1}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$

Proof

Let $t$ represent time then
$r(t)$ represents the point on curve $C$ at time $t$ hence ${\mathbf{r}}^{\prime }(t)$ represents the velocity of the point as it moves along curve $C$

Suppose

$$\mathbf{F}(t)=\frac{\dot{\mathbf{r}}(t)}{|\dot{\mathbf{r}}(t)|}$$

Then $\mathbf{F}(r)$ is a unit vector in the direction of vector $\dot{\mathbf{r}}(t)$

Hence

$$\mathbf{F}(t)\cdot \frac{d\mathbf{r}}{dt}=\frac{\dot{\mathbf{r}}(t)}{|\dot{\mathbf{r}}(t)|}\cdot \dot{\mathbf{r}}(t)=|\dot{\mathbf{r}}(t)|$$

So the Scalar Line Integral becomes

$${\int }_C\mathbf{F}(\mathbf{r})\cdot d\mathbf{r}={\int }_{t_0}^{t_1}|\dot{\mathbf{r}}(t)|dt={\int }_{t_0}^{t_1}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$