I have an exercise that involves the line integral $\int_C \mathbf{r} \cdot d\mathbf{r}$ where $\mathbf{r}(t)$ is a vector function but I'm confused as to what it means:

If $C$ is a smooth curve given by a vector function $\mathbf{r}(t), a \leq t \leq b$, show that $$ \int_C \mathbf{r} \cdot d\mathbf{r} = \frac{1}{2} \left[ |\mathbf{r}(b)|^2 - |\mathbf{r}(a)|^2 \right] $$

It looks similar to $\int_C \mathbf{F} \cdot d\mathbf{r}$ which my textbook defines as

Let $\mathbf{F}$ be a continuous vector field defined on a smooth curve $C$ given by a vector function $\mathbf{r}(t), a \leq t \leq b$. Then the line integral of $\mathbf{F}$ along $C$ is $$ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt $$

But $\mathbf{r}(t)$ clearly isn't a vector field, but a vector function. If I were to apply the definition here, then I would end up with $$ \int_a^b \mathbf{r}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt $$

And this makes no sense since you can't plug a vector into a scalar function. I really don't know what this line integral even represents. Can anyone help me with this?

Also, my highest knowledge of math is just vector calculus (only up to line integrals because that's what I'm learning at the moment), so please don't explain the answer in advanced terms. (There are always those people who SOMEHOW find a way to explain it in terms of topology)

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    $\begingroup$ It is simply $\int_a^b \mathbf{r}(t) \cdot \mathbf{r}'(t) \, dt$. Indeed an abuse of notation, but a small one. $\endgroup$
    – John B
    Jul 1, 2023 at 18:37
  • 1
    $\begingroup$ Alternatively, you can consider this to be the case where $F$ is the identity vector field, i.e., $F(r) \mapsto r$, $\endgroup$
    – whpowell96
    Jul 1, 2023 at 18:44

1 Answer 1


This is simply a slight overloading of notation, one that should be quite common and familiar in the context of $1$D functions. If we have a $1$D function given by $f(x) = x^2-1$, in many contexts we use the expression $x^2-1$ as a stand-in for the function $f$, e.g. we will write $$ \int f(x) dx = \int (x^2 -1 )dx.$$ This is so familiar that the latter integral is perfectly meaningful even if we had not previously referred to $f$ at all.

This is precisely what is happening here: $\int_C \mathbf r \cdot d \mathbf r$ is just the line integral $\int_C \mathbf F(\mathbf r) \cdot d \mathbf r$ of the vector field $\mathbf F(\mathbf r) = \mathbf r$, i.e. the vector field that associates to the point at the position $\mathbf r$ the position vector $\mathbf r$ itself (often referred to as the identity map).

Hence the definition of the line integral of $\mathbf F(\mathbf r) = \mathbf r$ yields: $$\int_C \mathbf r \cdot d \mathbf r = \int_a^b \mathbf r(t) \cdot \mathbf r'(t) dt.$$ It now shouldn't take much more work to complete the exercise.


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