Why "A line integral of a scalar field is thus a line integral of a vector field where the vectors are always tangential to the line"?

Question

I was reading about line integrals of a scalar field and line integrals of a vector field on Wikipedia. You can find the links here: line integral for a scalar field and line integral for a vector field.

In the second link the sentence: "A line integral of a scalar field is thus a line integral of a vector field where the vectors are always tangential to the line."

However I don't really understand why this is true. Also, because the scalar field outputs scalars whereas the vector field outputs vectors.

How do we explain the sentence above and more in general, what is the relationship between a line integral of a vector field and the line integral of a scalar field?

Answer 1

A line integral of a scalar field $g $ is $\displaystyle\int g\left(\overrightarrow{r}(t)\right)\left|\overrightarrow{r}'(t)\right|\,\mathrm dt$ along the curve. A line integral of a vector field $\overrightarrow{F}$ is $\displaystyle\int \overrightarrow{F}\left(\overrightarrow{r}(t)\right)\bullet\overrightarrow{r}'(t)\,\mathrm dt$. By a property of dot products, this equals $\displaystyle\int \left|\overrightarrow{F}\left(\overrightarrow{r}(t)\right)\right|\left|\overrightarrow{r}'(t)\right|\cos\left(\theta (t)\right)\,\mathrm dt$, where $\theta (t) $ is the angle between $ \overrightarrow{F}\left(\overrightarrow{r}(t)\right)$ (the direction of the vector field) and $\overrightarrow{r}'(t)$ (the direction of the curve).

If we choose $$\overrightarrow{F}\left(\overrightarrow{r}(t)\right)=g\left(\overrightarrow{r}(t)\right)*\frac{\overrightarrow{r}'(t)}{\left|\overrightarrow{r}'(t)\right|}\text{,}$$ then $\overrightarrow{F}$ has magnitude $|g|$ and direction: either the same as $\overrightarrow{r}'(t)$ or directly opposite (precisely when $g <0$). These directions make $\cos\left(\theta (t)\right) = \pm 1$ according to the sign of $g $. This way, $\left|\overrightarrow{F}\left(\overrightarrow{r}(t)\right)\right|\left|\overrightarrow{r}'(t)\right|\cos\left(\theta (t)\right)=g\left(\overrightarrow{r}(t)\right)\left|\overrightarrow{r}'(t)\right|$, so the vector field integral becomes the scalar field integral.

In summary, by choosing $\overrightarrow{F}$ so that it goes in the same direction of the curve (up to a minus sign), we can make a vector field integral that equals the scalar field integral.

Answer 2

If $\vec{F}= \pmatrix{P\\Q}$ is your vector field, then: $$ \underbrace{\int_C \vec{F} \cdot d\vec{r}}_{\text{vector field line integral}} = \int_C \pmatrix{P\\Q}\cdot \pmatrix{dx\\dy} = \int_C P\; dx + Q\; dy = \underbrace{\int_C P\; dx}_{\text{scalar field line integral}} +\underbrace{\int_C Q\; dy}_{\text{scalar field line integral}} $$

So in other words, your initial vector field line integral is the sum of two scalar field line integrals, in the direction of the $x$ and $y$ axes, respectively.

Note that if $\vec{r}(t)$ is a parametrization of $C$, you can define the line integral of a vector field as follows: $$ \int_C \vec{F} \cdot d\vec{r} = \int_C \vec{F} \cdot\vec{T}\; ds = \int_t \vec{F}(\vec{r}(t)) \cdot\frac{\vec{r}'(t)}{\mid \mid \vec{r}'(t) \mid \mid}\;\mid \mid \vec{r}'(t) \mid \mid dt = \int_t \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t)\; dt $$ The last integral is also a scalar field integral.