## Section9.2Path integrals

Note: 2–3 lectures

### Subsection9.2.1Piecewise smooth paths

Let $$\gamma \colon [a,b] \to \R^n$$ be a function and write $$\gamma = (\gamma_1,\gamma_2,\ldots,\gamma_n)\text{.}$$ Suppose $$\gamma$$ is continuously differentiable, that is, it is differentiable and the derivative is continuous. In other words, there exists a continuous function $$\gamma^{\:\prime} \colon [a,b] \to \R^n$$ such that for every $$t \in [a,b]\text{,}$$ we have $$\lim\limits_{h \to 0} \frac{\snorm{\gamma(t+h)-\gamma(t) - \gamma^{\:\prime}(t) \, h}}{\sabs{h}} = 0\text{.}$$ We treat $$\gamma^{\:\prime}(t)$$ either as a linear operator (an $$n \times 1$$ matrix) or a vector, $$\gamma^{\:\prime}(t) = \bigl( \gamma_1^{\:\prime}(t), \gamma_2^{\:\prime}(t), \ldots, \gamma_n^{\:\prime}(t) \bigr)\text{.}$$ Equivalently, $$\gamma_j$$ is a continuously differentiable function on $$[a,b]$$ for every $$j=1,2,\ldots,n\text{.}$$ By Exercise 8.2.6, the operator norm of the operator $$\gamma^{\:\prime}(t)$$ is equal to the euclidean norm of the corresponding vector, so there is no confusion when writing $$\snorm{\gamma^{\:\prime}(t)}\text{.}$$

#### Definition9.2.1.

A continuously differentiable function $$\gamma \colon [a,b] \to \R^n$$ is called a smooth path or a continuously differentiable path 1  if $$\gamma$$ is continuously differentiable and $$\gamma^{\:\prime}(t) \not= 0$$ for all $$t \in [a,b]\text{.}$$

The function $$\gamma \colon [a,b] \to \R^n$$ is called a piecewise smooth path or a piecewise continuously differentiable path if there exist finitely many points $$t_0 = a < t_1 < t_2 < \cdots < t_k = b$$ such that the restriction $$\gamma|_{[t_{j-1},t_j]}$$ is smooth path for every $$j=1,2,\ldots,k\text{.}$$

A path $$\gamma$$ is a closed path if $$\gamma(a) = \gamma(b)\text{,}$$ that is if the path starts and ends in the same point. A path $$\gamma$$ is a simple path if either 1) $$\gamma$$ is a one-to-one function, or 2) $$\gamma|_{[a,b)}$$ is one-to-one and $$\gamma(a)=\gamma(b)$$ ($$\gamma$$ is a simple closed path).

#### Example9.2.2.

Let $$\gamma \colon [0,4] \to \R^2$$ be defined by

\begin{equation*} \gamma(t) := \begin{cases} (t,0) & \text{if } t \in [0,1],\\ (1,t-1) & \text{if } t \in (1,2],\\ (3-t,1) & \text{if } t \in (2,3],\\ (0,4-t) & \text{if } t \in (3,4]. \end{cases} \end{equation*}

The path $$\gamma$$ is the unit square traversed counterclockwise. See Figure 9.2. It is a piecewise smooth path. For example, $$\gamma|_{[1,2]}(t) = (1,t-1)$$ and so $$(\gamma|_{[1,2]})'(t) = (0,1) \not= 0\text{.}$$ Similarly for the other 3 sides. Notice that $$(\gamma|_{[1,2]})'(1) = (0,1)\text{,}$$ $$(\gamma|_{[0,1]})'(1) = (1,0)\text{,}$$ but $$\gamma^{\:\prime}(1)$$ does not exist. At the corners $$\gamma$$ is not differentiable. The path $$\gamma$$ is a simple closed path, as $$\gamma|_{[0,4)}$$ is one-to-one and $$\gamma(0)=\gamma(4)\text{.}$$

The definition of a piecewise smooth path as we have given it implies continuity (exercise). For general functions, many authors also allow finitely many discontinuities, when they use the term piecewise smooth, and so one may say that we defined a piecewise smooth path to be a continuous piecewise smooth function. While one may get by with smooth paths, for computations, the simplest paths to write down are often piecewise smooth.

Generally, we are interested in the direct image $$\gamma\bigl([a,b]\bigr)\text{,}$$ rather than the specific parametrization, although that is also important to some degree. When we informally talk about a path or a curve, we often mean the set $$\gamma\bigl([a,b]\bigr)\text{,}$$ depending on context.

#### Example9.2.3.

The condition $$\gamma^{\:\prime}(t) \not= 0$$ means that the image $$\gamma\bigl([a,b]\bigr)$$ has no “corners” where $$\gamma$$ is smooth. Consider

\begin{equation*} \gamma(t) := \begin{cases} (t^2,0) & \text{if } t < 0,\\ (0,t^2) & \text{if } t \geq 0. \end{cases} \end{equation*}

See Figure 9.3. It is left for the reader to check that $$\gamma$$ is continuously differentiable, yet the image $$\gamma(\R) = \bigl\{ (x,y) \in \R^2 : (x,y) = (s,0) \text{ or } (x,y) = (0,s) \text{ for some } s \geq 0 \bigr\}$$ has a “corner” at the origin. And that is because $$\gamma^{\:\prime}(0) = (0,0)\text{.}$$ More complicated examples with, say, infinitely many corners exist, see the exercises.

The condition $$\gamma^{\:\prime}(t) \not= 0$$ even at the endpoints guarantees not only no corners, but also that the path ends nicely, that is, it can extend a little bit past the endpoints. Again, see the exercises.

#### Example9.2.4.

A graph of a continuously differentiable function $$f \colon [a,b] \to \R$$ is a smooth path. Define $$\gamma \colon [a,b] \to \R^2$$ by

\begin{equation*} \gamma(t) := \bigl(t,f(t)\bigr) . \end{equation*}

Then $$\gamma^{\:\prime}(t) = \bigl( 1 , f'(t) \bigr)\text{,}$$ which is never zero, and $$\gamma\bigl([a,b]\bigr)$$ is the graph of $$f\text{.}$$

There are other ways of parametrizing the path. That is, there are different paths with the same image. The function $$t \mapsto (1-t)a+tb\text{,}$$ takes the interval $$[0,1]$$ to $$[a,b]\text{.}$$ Define $$\alpha \colon [0,1] \to \R^2$$ by

\begin{equation*} \alpha(t) := \bigl((1-t)a+tb,f((1-t)a+tb)\bigr) . \end{equation*}

Then $$\alpha'(t) = \bigl( b-a ,~ (b-a)f'((1-t)a+tb) \bigr)\text{,}$$ which is never zero. As sets, $$\alpha\bigl([0,1]\bigr) = \gamma\bigl([a,b]\bigr) = \bigl\{ (x,y) \in \R^2 : x \in [a,b] \text{ and } f(x) = y \bigr\}\text{,}$$ which is just the graph of $$f\text{.}$$

The last example leads us to a definition.

#### Definition9.2.5.

Let $$\gamma \colon [a,b] \to \R^n$$ be a smooth path and $$h \colon [c,d] \to [a,b]$$ a continuously differentiable bijective function such that $$h'(t) \not= 0$$ for all $$t \in [c,d]\text{.}$$ Then the composition $$\gamma \circ h$$ is called a smooth reparametrization of $$\gamma\text{.}$$

Let $$\gamma$$ be a piecewise smooth path, and $$h$$ a piecewise smooth bijective function with nonzero one-sided limits of $$h'\text{.}$$ The composition $$\gamma \circ h$$ is called a piecewise smooth reparametrization of $$\gamma\text{.}$$

If $$h$$ is strictly increasing, then $$h$$ is said to preserve orientation. If $$h$$ does not preserve orientation, then $$h$$ is said to reverse orientation.

A reparametrization is another path for the same set. That is, $$(\gamma \circ h)\bigl([c,d]\bigr) = \gamma \bigl([a,b]\bigr)\text{.}$$

The conditions on the piecewise smooth $$h$$ mean that there is some partition $$t_0 = c < t_1 < t_2 < \cdots < t_k = d\text{,}$$ such that $$h|_{[t_{j-1},t_j]}$$ is continuously differentiable and $$(h|_{[t_{j-1},t_j]})'(t) \not= 0$$ for all $$t \in [t_{j-1},t_j]\text{.}$$ Since $$h$$ is bijective, it is either strictly increasing or strictly decreasing. So either $$(h|_{[t_{j-1},t_j]})'(t) > 0$$ for all $$t$$ or $$(h|_{[t_{j-1},t_j]})'(t) < 0$$ for all $$t\text{.}$$

#### Proof.

Assume that $$h$$ preserves orientation, that is, $$h$$ is strictly increasing. If $$h \colon [c,d] \to [a,b]$$ gives a piecewise smooth reparametrization, then for some partition $$r_0 = c < r_1 < r_2 < \cdots < r_\ell = d\text{,}$$ the restriction $$h|_{[r_{j-1},r_j]}$$ is continuously differentiable with a positive derivative.

Let $$t_0 = a < t_1 < t_2 < \cdots < t_k = b$$ be the partition from the definition of piecewise smooth for $$\gamma$$ together with the points $$\{ h(r_0), h(r_1), h(r_2), \ldots, h(r_\ell) \}\text{.}$$ Let $$s_j := h^{-1}(t_j)\text{.}$$ Then $$s_0 = c < s_1 < s_2 < \cdots < s_k = d$$ is a partition that includes (is a refinement of) the $$\{ r_0,r_1,\ldots,r_\ell \}\text{.}$$ If $$\tau \in [s_{j-1},s_j]\text{,}$$ then $$h(\tau) \in [t_{j-1},t_j]$$ since $$h(s_{j-1}) = t_{j-1}\text{,}$$ $$h(s_{j}) = t_j\text{,}$$ and $$h$$ is strictly increasing. Also $$h|_{[s_{j-1},s_j]}$$ is continuously differentiable, and $$\gamma|_{[t_{j-1},t_j]}$$ is also continuously differentiable. Then

\begin{equation*} (\gamma \circ h)|_{[s_{j-1},s_{j}]} (\tau) = \gamma|_{[t_{j-1},t_{j}]} \bigl( h|_{[s_{j-1},s_j]}(\tau) \bigr) . \end{equation*}

The function $$(\gamma \circ h)|_{[s_{j-1},s_{j}]}$$ is therefore continuously differentiable and by the chain rule

\begin{equation*} \bigl( (\gamma \circ h)|_{[s_{j-1},s_{j}]} \bigr) ' (\tau) = \bigl( \gamma|_{[t_{j-1},t_{j}]} \bigr)' \bigl( h(\tau) \bigr) (h|_{[s_{j-1},s_j]})'(\tau) \not= 0 . \end{equation*}

Consequently, $$\gamma \circ h$$ is a piecewise smooth path. Orientation reversing $$h$$ is left as an exercise.

If two paths are simple and their images are the same, it is left as an exercise that there exists a reparametrization. Here is where our assumption that $$\gamma'$$ is never zero is important.

### Subsection9.2.2Path integral of a one-form

#### Definition9.2.7.

Let $$(x_1,x_2,\ldots,x_n) \in \R^n$$ be our coordinates. Given $$n$$ real-valued continuous functions $$\omega_1,\omega_2,\ldots,\omega_n$$ defined on a set $$S \subset \R^n\text{,}$$ we define a one-form to be an object of the form

\begin{equation*} \omega = \omega_1 \,dx_1 + \omega_2 \,dx_2 + \cdots + \omega_n \,dx_n . \end{equation*}

We could represent $$\omega$$ as a continuous function from $$S$$ to $$\R^n\text{,}$$ although it is better to think of it as a different object.

#### Example9.2.8.

\begin{equation*} \omega(x,y) := \frac{-y}{x^2+y^2} \,dx + \frac{x}{x^2+y^2} \,dy \end{equation*}

is a one-form defined on $$\R^2 \setminus \{ (0,0) \}\text{.}$$

#### Definition9.2.9.

Let $$\gamma \colon [a,b] \to \R^n$$ be a smooth path and let

\begin{equation*} \omega = \omega_1 \,dx_1 + \omega_2 \,dx_2 + \cdots + \omega_n \,dx_n , \end{equation*}

be a one-form defined on the direct image $$\gamma\bigl([a,b]\bigr)\text{.}$$ Write $$\gamma = (\gamma_1,\gamma_2,\ldots,\gamma_n)\text{.}$$ Define:

\begin{equation*} \begin{split} \int_{\gamma} \omega & := \int_a^b \Bigl( \omega_1\bigl(\gamma(t)\bigr) \gamma_1^{\:\prime}(t) + \omega_2\bigl(\gamma(t)\bigr) \gamma_2^{\:\prime}(t) + \cdots + \omega_n\bigl(\gamma(t)\bigr) \gamma_n^{\:\prime}(t) \Bigr) dt \\ &\phantom{:}= \int_a^b \left( \sum_{j=1}^n \omega_j\bigl(\gamma(t)\bigr) \gamma_j^{\:\prime}(t) \right) dt . \end{split} \end{equation*}

To remember the definition note that $$x_j$$ is $$\gamma_j(t)\text{,}$$ so $$dx_j$$ becomes $$\gamma_j^{\:\prime}(t) \, dt\text{.}$$

If $$\gamma$$ is piecewise smooth, take the corresponding partition $$t_0 = a < t_1 < t_2 < \ldots < t_k = b\text{,}$$ and assume the partition is minimal in the sense that $$\gamma$$ is not differentiable at $$t_1,t_2,\ldots,t_{k-1}\text{.}$$ As each $$\gamma|_{[t_{j-1},t_j]}$$ is a smooth path, define

\begin{equation*} \int_{\gamma} \omega := \int_{\gamma|_{[t_0,t_1]}} \omega \, + \, \int_{\gamma|_{[t_1,t_2]}} \omega \, + \, \cdots \, + \, \int_{\gamma|_{[t_{k-1},t_k]}} \omega . \end{equation*}

The notation makes sense from the formula you remember from calculus, let us state it somewhat informally: If $$x_j(t) = \gamma_j(t)\text{,}$$ then $$dx_j = \gamma_j^{\:\prime}(t) \, dt\text{.}$$

Paths can be cut up or concatenated. The proof is a direct application of the additivity of the Riemann integral, and is left as an exercise. The proposition justifies why we defined the integral over a piecewise smooth path in the way we did, and it justifies that we may as well have taken any partition not just the minimal one in the definition.

#### Example9.2.11.

Let the one-form $$\omega$$ and the path $$\gamma \colon [0,2\pi] \to \R^2$$ be defined by

\begin{equation*} \omega(x,y) := \frac{-y}{x^2+y^2} \,dx + \frac{x}{x^2+y^2} \,dy, \qquad \gamma(t) := \bigl(\cos(t),\sin(t)\bigr) . \end{equation*}

Then

\begin{equation*} \begin{split} \int_{\gamma} \omega & = \int_0^{2\pi} \Biggl( \frac{-\sin(t)}{{\bigl(\cos(t)\bigr)}^2+{\bigl(\sin(t)\bigr)}^2} \bigl(-\sin(t)\bigr) + \frac{\cos(t)}{{\bigl(\cos(t)\bigr)}^2+{\bigl(\sin(t)\bigr)}^2} \bigl(\cos(t)\bigr) \Biggr) \, dt \\ & = \int_0^{2\pi} 1 \, dt = 2\pi . \end{split} \end{equation*}

Next, let us parametrize the same curve as $$\alpha \colon [0,1] \to \R^2$$ defined by $$\alpha(t) := \bigl(\cos(2\pi t),\sin(2 \pi t)\bigr)\text{,}$$ that is $$\alpha$$ is a smooth reparametrization of $$\gamma\text{.}$$ Then

\begin{equation*} \begin{split} \int_{\alpha} \omega & = \int_0^{1} \Biggl( \frac{-\sin(2\pi t)}{{\bigl(\cos(2\pi t)\bigr)}^2+{\bigl(\sin(2\pi t)\bigr)}^2} \bigl(-2\pi \sin(2\pi t)\bigr) \\ & \phantom{=\int_0^1\Biggl(~} + \frac{\cos(2 \pi t)}{{\bigl(\cos(2 \pi t)\bigr)}^2+{\bigl(\sin(2 \pi t)\bigr)}^2} \bigl(2 \pi \cos(2 \pi t)\bigr) \Biggr) \, dt \\ & = \int_0^{1} 2\pi \, dt = 2\pi . \end{split} \end{equation*}

Now let us reparametrize with $$\beta \colon [0,2\pi] \to \R^2$$ as $$\beta(t) := \bigl(\cos(-t),\sin(-t)\bigr)\text{.}$$ Then

\begin{equation*} \begin{split} \int_{\beta} \omega & = \int_0^{2\pi} \Biggl( \frac{-\sin(-t)}{{\bigl(\cos(-t)\bigr)}^2+{\bigl(\sin(-t)\bigr)}^2} \bigl(\sin(-t)\bigr) + \frac{\cos(-t)}{{\bigl(\cos(-t)\bigr)}^2+{\bigl(\sin(-t)\bigr)}^2} \bigl(-\cos(-t)\bigr) \Biggr) \, dt \\ & = \int_0^{2\pi} (-1) \, dt = -2\pi . \end{split} \end{equation*}

The path $$\alpha$$ is an orientation preserving reparametrization of $$\gamma\text{,}$$ and the integrals are the same. The path $$\beta$$ is an orientation reversing reparametrization of $$\gamma$$ and the integral is minus the original. See Figure 9.4.

The previous example is not a fluke. The path integral does not depend on the parametrization of the curve, the only thing that matters is the direction in which the curve is traversed.

#### Proof.

Assume first that $$\gamma$$ and $$h$$ are both smooth. Write $$\omega = \omega_1 \, dx_1 + \omega_2 \, dx_2 + \cdots + \omega_n \, dx_n\text{.}$$ Suppose that $$h$$ is orientation preserving. Use the change of variables formula for the Riemann integral:

\begin{equation*} \begin{split} \int_{\gamma} \omega & = \int_a^b \left( \sum_{j=1}^n \omega_j\bigl(\gamma(t)\bigr) \gamma_j^{\:\prime}(t) \right) dt \\ & = \int_c^d \left( \sum_{j=1}^n \omega_j\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) \gamma_j^{\:\prime}\bigl(h(\tau)\bigr) \right) h'(\tau) \, d\tau \\ & = \int_c^d \left( \sum_{j=1}^n \omega_j\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) (\gamma_j \circ h)'(\tau) \right) d\tau = \int_{\gamma \circ h} \omega . \end{split} \end{equation*}

If $$h$$ is orientation reversing, it swaps the order of the limits on the integral and introduces a minus sign. The details, along with finishing the proof for piecewise smooth paths, is left as Exercise 9.2.4.

Due to this proposition (and the exercises), if $$\Gamma \subset \R^n$$ is the image of a simple piecewise smooth path $$\gamma\bigl([a,b]\bigr)\text{,}$$ then as long as we somehow indicate the orientation, that is, the direction in which we traverse the curve, we can write

\begin{equation*} \int_{\Gamma} \omega , \end{equation*}

without mentioning the specific $$\gamma\text{.}$$ Furthermore, for a simple closed path, it does not even matter where we start the parametrization. See the exercises.

Recall that simple means that $$\gamma$$ is one-to-one except perhaps at the endpoints, in particular it is one-to-one when restricted to $$[a,b)\text{.}$$ We may relax the condition that the path is simple a little bit. For example, it is enough to suppose that $$\gamma \colon [a,b] \to \R^n$$ is one-to-one except at finitely many points. See Exercise 9.2.14. But we cannot remove the condition completely as is illustrated by the following example.

#### Example9.2.13.

Suppose $$\gamma \colon [0,2\pi] \to \R^2$$ is given by $$\gamma(t) := \bigl(\cos(t),\sin(t)\bigr)\text{,}$$ and $$\beta \colon [0,2\pi] \to \R^2$$ is given by $$\beta(t) := \bigl(\cos(2t),\sin(2t)\bigr)\text{.}$$ Notice that $$\gamma\bigl([0,2\pi]\bigr) = \beta\bigl([0,2\pi]\bigr)\text{,}$$ and we travel around the same curve, the unit circle. But $$\gamma$$ goes around the unit circle once in the counter clockwise direction, and $$\beta$$ goes around the unit circle twice (in the same direction). See Figure 9.5.

Compute

\begin{equation*} \begin{aligned} & \int_{\gamma} -y\, dx + x\,dy = \int_0^{2\pi} \Bigl( \bigl(-\sin(t) \bigr) \bigl(-\sin(t) \bigr) + \cos(t) \cos(t) \Bigr) dt = 2 \pi,\\ & \int_{\beta} -y\, dx + x\,dy = \int_0^{2\pi} \Bigl( \bigl(-\sin(2t) \bigr) \bigl(-2\sin(2t) \bigr) + \cos(t) \bigl(2\cos(t)\bigr) \Bigr) dt = 4 \pi. \end{aligned} \end{equation*}

It is sometimes convenient to define a path integral over $$\gamma \colon [a,b] \to \R^n$$ that is not a path. Define

\begin{equation*} \int_{\gamma} \omega := \int_a^b \left( \sum_{j=1}^n \omega_j\bigl(\gamma(t)\bigr) \gamma_j^{\:\prime}(t) \right) dt \end{equation*}

for every continuously differentiable $$\gamma\text{.}$$ A case that comes up naturally is when $$\gamma$$ is constant. Then $$\gamma^{\:\prime}(t) = 0$$ for all $$t\text{,}$$ and $$\gamma\bigl([a,b]\bigr)$$ is a single point, which we regard as a “curve” of length zero. Then, $$\int_{\gamma} \omega = 0$$ for every $$\omega\text{.}$$

### Subsection9.2.3Path integral of a function

Next we integrate a function against the so-called arc-length measure $$ds\text{.}$$ The geometric picture we have in mind is the area under the graph of the function over a path. Imagine a fence erected over $$\gamma$$ with height given by the function and the integral is the area of the fence. See Figure 9.6.

#### Definition9.2.14.

Suppose $$\gamma \colon [a,b] \to \R^n$$ is a smooth path, and $$f$$ is a continuous function defined on the image $$\gamma\bigl([a,b]\bigr)\text{.}$$ Then define

\begin{equation*} \int_{\gamma} f \,ds := \int_a^b f\bigl( \gamma(t) \bigr) \snorm{\gamma^{\:\prime}(t)} \, dt . \end{equation*}

To emphasize the variables we may use

\begin{equation*} \int_{\gamma} f(x) \,ds(x) := \int_{\gamma} f \,ds . \end{equation*}

The definition for a piecewise smooth path is similar as before and is left to the reader.

The path integral of a function is also independent of the parametrization, and in this case, the orientation does not matter.

#### Proof.

Suppose first that $$h$$ is orientation preserving and that $$\gamma$$ and $$h$$ are both smooth. Then

\begin{equation*} \begin{split} \int_{\gamma} f \, ds & = \int_a^b f\bigl(\gamma(t)\bigr) \snorm{\gamma^{\:\prime}(t)} \, dt \\ & = \int_c^d f\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) \snorm{\gamma^{\:\prime}\bigl(h(\tau)\bigr)} h'(\tau) \, d\tau \\ & = \int_c^d f\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) \snorm{\gamma^{\:\prime}\bigl(h(\tau)\bigr) h'(\tau)} \, d\tau \\ & = \int_c^d f\bigl((\gamma \circ h)(\tau)\bigr) \snorm{(\gamma \circ h)'(\tau)} \, d\tau \\ & = \int_{\gamma \circ h} f \, ds . \end{split} \end{equation*}

If $$h$$ is orientation reversing it swaps the order of the limits on the integral, but you also have to introduce a minus sign in order to take $$h'$$ inside the norm. The details, along with finishing the proof for piecewise smooth paths is left to the reader as Exercise 9.2.5.

As before, due to this proposition (and the exercises), if $$\gamma$$ is simple, it does not matter which parametrization we use. Therefore, if $$\Gamma = \gamma\bigl( [a,b] \bigr)\text{,}$$ we can simply write

\begin{equation*} \int_\Gamma f\, ds . \end{equation*}

In this case we also do not need to worry about orientation, either way we get the same integral.

#### Example9.2.16.

Let $$f(x,y) := x\text{.}$$ Let $$C \subset \R^2$$ be half of the unit circle for $$x \geq 0\text{.}$$ We wish to compute

\begin{equation*} \int_C f \, ds . \end{equation*}

Parametrize the curve $$C$$ via $$\gamma \colon [\nicefrac{-\pi}{2},\nicefrac{\pi}{2}] \to \R^2$$ defined as $$\gamma(t) := \bigl(\cos(t),\sin(t)\bigr)\text{.}$$ Then $$\gamma^{\:\prime}(t) = \bigl(-\sin(t),\cos(t)\bigr)\text{,}$$ and

\begin{equation*} \int_C f \, ds = \int_\gamma f \, ds = \int_{-\pi/2}^{\pi/2} \cos(t) \sqrt{ {\bigl(-\sin(t)\bigr)}^2 + {\bigl(\cos(t)\bigr)}^2 } \, dt = \int_{-\pi/2}^{\pi/2} \cos(t) \, dt = 2. \end{equation*}

#### Definition9.2.17.

Suppose $$\Gamma \subset \R^n$$ is parametrized by a simple piecewise smooth path $$\gamma \colon [a,b] \to \R^n\text{,}$$ that is $$\gamma\bigl( [a,b] \bigr) = \Gamma\text{.}$$ We define the length by

\begin{equation*} \ell(\Gamma) := \int_{\Gamma} ds = \int_{\gamma} ds . \end{equation*}

If $$\gamma$$ is smooth,

\begin{equation*} \ell(\Gamma) = \int_a^b \snorm{\gamma^{\:\prime}(t)}\, dt . \end{equation*}

This may be a good time to mention that it is common to write $$\int_a^b \snorm{\gamma^{\:\prime}(t)}\, dt$$ even if the path is only piecewise smooth. That is because $$\snorm{\gamma^{\:\prime}(t)}$$ is defined and continuous at all but finitely many points and is bounded, and so the integral exists.

#### Example9.2.18.

Let $$x,y \in \R^n$$ be two points and write $$[x,y]$$ as the straight line segment between the two points $$x$$ and $$y\text{.}$$ Parametrize $$[x,y]$$ by $$\gamma(t) := (1-t)x + ty$$ for $$t$$ running between $$0$$ and $$1\text{.}$$ See Figure 9.7. Then $$\gamma^{\:\prime}(t) = y-x\text{,}$$ and therefore

\begin{equation*} \ell\bigl([x,y]\bigr) = \int_{[x,y]} ds = \int_0^1 \snorm{y-x} \, dt = \snorm{y-x} . \end{equation*}

So the length of $$[x,y]$$ is the standard euclidean distance between $$x$$ and $$y\text{,}$$ justifying the name.

A simple piecewise smooth path $$\gamma \colon [0,r] \to \R^n$$ is said to be an arc-length parametrization if for all $$t \in [0,r]\text{,}$$ we have

\begin{equation*} \ell\bigl( \gamma\bigl([0,t]\bigr) \bigr) = t . \end{equation*}

If $$\gamma$$ is smooth, then

\begin{equation*} \int_0^t d\tau = t = \ell\bigl( \gamma\bigl([0,t]\bigr) \bigr) = \int_0^t \snorm{\gamma^{\:\prime}(\tau)} \, d\tau \end{equation*}

for all $$t\text{,}$$ which means that $$\snorm{\gamma^{\:\prime}(t)} = 1$$ for all $$t\text{.}$$ Similarly for piecewise smooth $$\gamma\text{,}$$ we get $$\snorm{\gamma^{\:\prime}(t)} = 1$$ for all $$t$$ where the derivative exists. So you can think of such a parametrization as moving around your curve at speed 1. If $$\gamma \colon [0,r] \to \R^n$$ is an arclength parametrization, it is common to use $$s$$ as the variable as $$\int_\gamma f \,ds = \int_0^r f\bigl(\gamma(s)\bigr) \,ds\text{.}$$

### Subsection9.2.4Exercises

#### Exercise9.2.1.

Show that if $$\varphi \colon [a,b] \to \R^n$$ is a piecewise smooth path as we defined it, then $$\varphi$$ is a continuous function.

#### Exercise9.2.2.

Finish the proof of Proposition 9.2.6 for orientation reversing reparametrizations.

#### Exercise9.2.4.

Finish the proof of Proposition 9.2.12 for

1. orientation reversing reparametrizations, and

2. piecewise smooth paths and reparametrizations.

#### Exercise9.2.5.

Finish the proof of Proposition 9.2.15 for

1. orientation reversing reparametrizations, and

2. piecewise smooth paths and reparametrizations.

#### Exercise9.2.6.

Suppose $$\gamma \colon [a,b] \to \R^n$$ is a piecewise smooth path, and $$f$$ is a continuous function defined on the image $$\gamma\bigl([a,b]\bigr)\text{.}$$ Provide a definition of $$\int_{\gamma} f \,ds\text{.}$$

#### Exercise9.2.7.

Directly using the definitions compute:

1. The arc-length of the unit square from Example 9.2.2 using the given parametrization.

2. The arc-length of the unit circle using the parametrization $$\gamma \colon [0,1] \to \R^2\text{,}$$ $$\gamma(t) := \bigl(\cos(2\pi t),\sin(2\pi t)\bigr)\text{.}$$

3. The arc-length of the unit circle using the parametrization $$\beta \colon [0,2\pi] \to \R^2\text{,}$$ $$\beta(t) := \bigl(\cos(t),\sin(t)\bigr)\text{.}$$

Note: Feel free to use what you know about sine and cosine from calculus.

#### Exercise9.2.8.

Suppose $$\gamma \colon [0,1] \to \R^n$$ is a smooth path, and $$\omega$$ is a one-form defined on the image $$\gamma\bigl([a,b]\bigr)\text{.}$$ For $$r \in [0,1]\text{,}$$ let $$\gamma_r \colon [0,r] \to \R^n$$ be defined as simply the restriction of $$\gamma$$ to $$[0,r]\text{.}$$ Show that the function $$h(r) := \int_{\gamma_r} \omega$$ is a continuously differentiable function on $$[0,1]\text{.}$$

#### Exercise9.2.9.

Suppose $$\gamma \colon [a,b] \to \R^n$$ is a smooth path. Show that there exists an $$\epsilon > 0$$ and a smooth function $$\widetilde{\gamma} \colon (a-\epsilon,b+\epsilon) \to \R^n$$ with $$\widetilde{\gamma}(t) = \gamma(t)$$ for all $$t \in [a,b]$$ and $$\widetilde{\gamma}^{\:\prime}(t) \not= 0$$ for all $$t \in (a-\epsilon,b+\epsilon)\text{.}$$ That is, prove that a smooth path extends some small distance past the end points.

#### Exercise9.2.10.

Suppose $$\alpha \colon [a,b] \to \R^n$$ and $$\beta \colon [c,d] \to \R^n$$ are piecewise smooth paths such that $$\Gamma := \alpha\bigl([a,b]\bigr) = \beta\bigl([c,d]\bigr)\text{.}$$ Show that there exist finitely many points $$\{ p_1,p_2,\ldots,p_k\} \in \Gamma\text{,}$$ such that the sets $$\alpha^{-1}\bigl( \{ p_1,p_2,\ldots,p_k\} \bigr)$$ and $$\beta^{-1}\bigl( \{ p_1,p_2,\ldots,p_k\} \bigr)$$ are partitions of $$[a,b]$$ and $$[c,d]$$ such that on every subinterval the paths are smooth (that is, they are partitions as in the definition of piecewise smooth path).

#### Exercise9.2.11.

1. Suppose $$\gamma \colon [a,b] \to \R^n$$ and $$\alpha \colon [c,d] \to \R^n$$ are two smooth paths that are one-to-one and $$\gamma\bigl([a,b]\bigr) = \alpha\bigl([c,d]\bigr)\text{.}$$ Then there exists a smooth reparametrization $$h \colon [a,b] \to [c,d]$$ such that $$\gamma = \alpha \circ h\text{.}$$
Hint 1: It is not hard to show $$h$$ exists. The trick is to prove it is continuously differentiable with a nonzero derivative. Apply the implicit function theorem though it may at first seem the dimensions are wrong.
Hint 2: Worry about derivative of $$h$$ in $$(a,b)$$ first.

2. Prove the same thing as part a, but now for simple closed paths with the further assumption that $$\gamma(a) = \gamma(b) = \alpha(c) = \alpha(d)\text{.}$$

3. Prove parts a) and b) but for piecewise smooth paths, obtaining piecewise smooth reparametrizations.
Hint: The trick is to find two partitions such that when restricted to a subinterval of the partition both paths have the same image and are smooth, see the exercise above.

#### Exercise9.2.12.

Suppose $$\alpha \colon [a,b] \to \R^n$$ and $$\beta \colon [b,c] \to \R^n$$ are piecewise smooth paths with $$\alpha(b)=\beta(b)\text{.}$$ Let $$\gamma \colon [a,c] \to \R^n$$ be defined by

\begin{equation*} \gamma(t) := \begin{cases} \alpha(t) & \text{if } t \in [a,b], \\ \beta(t) & \text{if } t \in (b,c]. \end{cases} \end{equation*}

Show that $$\gamma$$ is a piecewise smooth path, and that if $$\omega$$ is a one-form defined on the curve given by $$\gamma\text{,}$$ then

\begin{equation*} \int_{\gamma} \omega = \int_{\alpha} \omega + \int_{\beta} \omega . \end{equation*}

#### Exercise9.2.13.

Suppose $$\gamma \colon [a,b] \to \R^n$$ and $$\beta \colon [c,d] \to \R^n$$ are two simple closed piecewise smooth paths. That is, $$\gamma(a)=\gamma(b)$$ and $$\beta(c) = \beta(d)$$ and the restrictions $$\gamma|_{[a,b)}$$ and $$\beta|_{[c,d)}$$ are one-to-one. Suppose $$\Gamma = \gamma\bigl([a,b]\bigr) = \beta\bigl([c,d]\bigr)$$ and $$\omega$$ is a one-form defined on $$\Gamma \subset \R^n\text{.}$$ Show that either

\begin{equation*} \int_\gamma \omega = \int_\beta \omega, \qquad \text{or} \qquad \int_\gamma \omega = - \int_\beta \omega. \end{equation*}

In particular, the notation $$\int_{\Gamma} \omega$$ makes sense if we indicate the direction in which the integral is evaluated. Hint: See previous three exercises.

#### Exercise9.2.14.

Suppose $$\gamma \colon [a,b] \to \R^n$$ and $$\beta \colon [c,d] \to \R^n$$ are two piecewise smooth paths which are one-to-one except at finitely many points. That is, there exist finite sets $$S \subset [a,b]$$ and $$T \subset [c,d]$$ such that $$\gamma|_{[a,b]\setminus S}$$ and $$\beta|_{[c,d]\setminus T}$$ are one-to-one. Suppose $$\Gamma = \gamma\bigl([a,b]\bigr) = \beta\bigl([c,d]\bigr)$$ and $$\omega$$ is a one-form defined on $$\Gamma \subset \R^n\text{.}$$ Show that either

\begin{equation*} \int_\gamma \omega = \int_\beta \omega, \qquad \text{or} \qquad \int_\gamma \omega = - \int_\beta \omega. \end{equation*}

In particular, the notation $$\int_{\Gamma} \omega$$ makes sense if we indicate the direction in which the integral is evaluated.
Hint: Same hint as the last exercise.

#### Exercise9.2.15.

Define $$\gamma \colon [0,1] \to \R^2$$ by $$\gamma(t) := \Bigl( t^3 \sin(\nicefrac{1}{t}),\, t{\bigl(3t^2\sin(\nicefrac{1}{t})-t\cos(\nicefrac{1}{t})\bigr)}^2 \Bigr)$$ for $$t \not= 0$$ and $$\gamma(0) = (0,0)\text{.}$$ Show that

1. $$\gamma$$ is continuously differentiable on $$[0,1]\text{.}$$

2. Show that there exists an infinite sequence $$\{ t_n \}$$ in $$[0,1]$$ converging to 0, such that $$\gamma^{\:\prime}(t_n) = (0,0)\text{.}$$

3. Show that the points $$\gamma(t_n)$$ lie on the line $$y=0$$ and such that the $$x$$-coordinate of $$\gamma(t_n)$$ alternates between positive and negative (if they do not alternate you only found a subsequence, you need to find them all).

4. Show that there is no piecewise smooth $$\alpha$$ whose image equals $$\gamma\bigl([0,1]\bigr)\text{.}$$ Hint: Look at part c) and show that $$\alpha'$$ must be zero where it reaches the origin.

5. (Computer) If you know a plotting software that allows you to plot parametric curves, make a plot of the curve, but only for $$t$$ in the range $$[0,0.1]$$ otherwise you will not see the behavior. In particular, you should notice that $$\gamma\bigl([0,1]\bigr)$$ has infinitely many “corners” near the origin.

Note: Feel free to use what you know about sine and cosine from calculus.

The word “smooth” can sometimes mean “infinitely differentiable” in the literature.
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