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The basic definitions in this section are arclength in equation \eqref{arclength.def}, the line integral of a scalar function in \eqref{lis.def}, and the line integral of a vector field, in \eqref{livf.def}. It is important to commit these to memory.
Suppose that $C$ is a curve in $\R^n$ parametrized by a function $$ \bfx = \bfg(t), \quad a\le t\le b. $$ We will normally assume that $\bfg$ is $C^1$ on an open set interval that contains $[a,b]$, that $\bfg'(t)\ne {\bf 0}$ for all $t\in (a,b)$, and that $\bfg$ is one-to-one in $(a,b)$.
Also assume that $f$ is a continuous (scalar) function on $\R^n$ and that $\bfF$ is a continuous vector field on $\R^n$. We define \begin{equation}\label{arclength.def} \mbox{arclength of }C \ := \ \int_a^b |\bfg'(t)|\, dt\, , \end{equation} and \begin{equation}\label{lis.def} \mbox{integral of }f\mbox{ over }C := \int_C f\, ds \ := \ \int_a^b f(\bfg(t)) \ |\bfg'(t)|\, dt \, , \end{equation}
We also sometimes say or write line integral
instead of integral
when referring to integrals as in \eqref{lis.def}.
Note that the arclength of $C$ equals $\int_C 1\, ds$.
The integral of a function $f$ over a curve $C$ is interpreted as: $$ \mbox{ integral of }f\mbox{ over }C \ = \ (\mbox{arclength of }C)\times(\mbox{average of }f\mbox{ over }C). $$ The rigorous meaning of the above formula is that it is the definition of the average of $f$ over $C$ (that is, as integral/arclength). But since we have an intuitive sense of what both terms on the right-hand side should mean, its non-rigorous meaning is that it tells us how to think about the line integral of a function.
Example 1. Let $C$ be the curve paramaetrized by $\bfx = \bfg(t) = (t^3 , \frac 32 t^4, \frac 6 5 t^5)$, for $0\le t \le 2$. Compute the arclength of $C$ and the integral $\int_C xyz\, ds$
Solution. We compute $$ \bfg'(t) = (3t^2, 6t^3, 6t^4) = 3t^2(1,2t, 2t^2) $$ and hence $$ |\bfg'(t)| = 3t^2\sqrt{1+ 4t^2 + 4t^4} = 3t^2(1+2t^2). $$ Thus $$ \mbox{arclength}(C) = \int_0^2 3t^2+ 6t^4\, dt = 2^3+ \frac 65 2^5 $$ and
$$ \int_C xyz\, ds = \int_0^2 (t^3)(\frac 32 t^4)(\frac 65 t^5) (3t^2+ 6 t^4)\, dt = \frac {27}5\int_0^2 t^{14} + 2t^{16}\, dt = \frac {27}5(\frac 1{15}2^{15} +\frac 2{17}\cdot 2^{17}) $$
It follows from a short argument that the integral of a scalar function $f$ over a curve $C$ is independent of the parametrization. This implies in particular that the arclength of a curve $C$ is independent of the parametrization. (See Theorem 1 below for precise statements.)
This is nice to know, because if our definition of arclength depended on the parametrization, then it would not be a good definition; arclength should be an intrinsic property of a curve, independent of how we choose to represent the curve.
It also means that it makes sense to describe a curve in a way that does not specifically involve a parametrization, and then to ask you to compute either its arclength or an integral $\int_C f\, ds$. To carry out the computation, you would need to find a parametrization, but the invariance property guarantees that you will get the same result, no matter which parametrization you choose.
Example 2. Let $C$ be the ellipse $(x/a)^2+(y/b)^2 = 1$. Write down an integral formula for the arclength of $C$, and evaluate it if possible.
Solution.
The fact that arclength is independent of parametrization
means that we can choose any paramtrization
for $C$, and we can use it to write down
an integral for the arclength. For example, we can choose
$$
\bfx = \bfg(t) := (a\cos t, b\sin t), \qquad 0\le t\le 2\pi.
$$
Then
$$
\bfg'(t) = (-a\sin t, b\cos t), \qquad
\mbox{ so }|\bfg'(t)| = \sqrt{a^2 \sin^2 t + b^2\cos^2 t}.
$$
It follows that
$$
\mbox{arclength}(C) = \int_0^{2\pi}\sqrt{a^2 \sin^2 t + b^2\cos^2 t}\, dt.
$$
Unfortunately, this integral is impossible to evaluate
using elementary functions. (In fact, a particular kind of
function, called an elliptic integral
, was invented
for this purpose.)
The fact that arclength is independent of the parametrization ensures that if we had chosen another parametrization, say $\bfx = (a\sin t, -b\cos t)$ for $-\pi \le t\le \pi$, or $\bfx = (a\sin(e^t), b\cos(e^t))$ for $\ln \pi \le t \le \ln 3\pi$, we would have gotten the same answer (that is, we would have come up with an equivalent impossible-to-evaluate integral.)
A precise statement of the invariance property is given by the following.
Theorem 1.
Assume that $\bfg:[a,b]\to \R^n$ is
parametrization of a curve $C$ that is $C^1$ on an open
interval containing $C$, such that $\bfg'(t)\ne \bf 0$ for
$t\in [a,b]$. Assume also that $\bfg$ is one-to-one on $(a,b)$.
Also, let $\phi:[c,d]\to [a,b]$ be a function that is one-to-one and
onto, and assume that $\phi$ is $C^1$, with $\phi'\ne 0$, on an interval
that contains $[c,d]$. Finally, let $\bfh = \bfg\circ\phi$.
Then for any continuous $f:\R^n\to \R$, computing $\int_Cf\, ds$ using either parametrization $\bfg$ or $\bfh$ yields the same answer: $$ \int_C f\ ds = \int_a^b f(\bfg(t))|\bfg'(t)|\, dt = \int_c^d f(\bfh(u))|\bfh'(u)|\, du. $$ In particular, $$ \mbox{arclength}(C) = \int_a^b |\bfg'(t)|\, dt = \int_c^d|\bfh'(u)|\, du. $$
The proof is a straightforward change of variables. Writing $\bfh$ as a function of a variable $u\in [c,d]$, the Chain Rule implies that $$ \bfh'(u) = \bfg'(\phi(u)) \phi'(u), $$ so $$ \int_c^d f(\bfh(u))|\bfh'(u)|\, du = \int_c^d f(\bfg(\phi(u)))\, |\bfg'(\phi(u))| \, |\phi'(u)|\, du. $$ We define the new variable $t = \phi(u)$. Since $\phi:[c,d]\to [a,b]$ is one-to-one and onto, the Change of Variables Theorem implies that $$ \int_c^d f(\bfg(\phi(u)))\, |\bfg'(\phi(u))| \, |\phi'(u)|\, du. = \int_a^b f(\bfg(t)) |\bfg'(t)|\, dt, $$ and this implies the conclusion of the theorem. $\quad\Box$
Let $C$ be a curve in $\R^n$ parametrized by $\bfx = \bfg(t)$ for $a\le t\le b$.
If $P =\{ t_0\ldots, t_J\}$ is any partition of $[a,b]$, so that $a=t_0 < t_1 < \ldots < t_J=b$, then the length of $C$ may be approximated by the sum of the lengths of the line segments connecting $\bfg(t_{j-1})$ to $\bfg(t_j)$ for $j=1,\ldots, J$, which we will denote $$ L_P(C) := \sum_{j=1}^J |\bfg(t_{j}) - \bfg(t_{j-1})|. $$ We say that $C$ is rectifiable if $$ \{ L_P(C) : P\mbox{ is a partition of }[a,b]\} \mbox{ is bounded}, $$ and in this case, we define $$ \mbox{arclength of }C := L(C) := \sup\{ L_P(C) : P\mbox{ is a partition of }[a,b]\}. $$
The following theorem guarantees that this definition of arclength is consistant with the one we gave above.
Theorem 2. If $\bfg:[a,b]\to \R^n$ is $C^1$, then $$ L(C) = \int_a^b |\bfg'(t)|\, dt. $$
The proof may be added to these notes later.
We continue to assume that $C$ is a curve in $\R^n$ parametrized by a function $$ \bfx = \bfg(t), \quad a\le t\le b, $$ with the same assumptions about $\bfg$ as above, e.g. $\bfg'$ exists and is nonzero everywhere in an open interval containing $[a,b]$.
A vector field in $\R^n$ is a function $\bfF:S\to \R^n$,
where $S$ is a subset of $\R^n$.
When we use the term vector field
(rather
than, say, transformation
) we have in mind that
for any $\bfx\in S$, $\bfF(\bfx)$ is a vector in the strict sense,
that is, a quantity with a direction and magnitude, such as
a force or a velocity (rather than, say, a position). Vector fields arise very often
in physics, precisely because there are many physical
situations where one is interested in a vectorial
quantity (direction + magnitude) that varies from point
to point in physical space.
Given a vector field $\bfF:\R^n\to \R^n$, we define \begin{equation}\label{livf.def} \mbox{the (line) integral of }\bfF\mbox{ over }C := \int_C \bfF\cdot d\bfx \ := \ \int_a^b \bfF(\bfg(t)) \cdot \bfg'(t)\, dt\, . \end{equation}
The line integral of a vector field arises naturally in a variety of applications. For example, if a particle traverses a curve $C$ parametrized by $\bfx = \bfg(t)$, $a\le t\le b$, and if there is a force (due for example to gravity, or electromagnetic forces) $\bfF(\bfx)$ acting on the particle at position $\bfx$, then $\int_C \bfF\cdot d\bfx$ is the work done by the force field on the particle.
In the next example, we will examine whether the line integral of a vector field is independent of the parametrization.
Example 3. Let $S := \R^2\setminus \{ {\bf 0}\}$, and let $\bfF$ be the vector field on $S$ defined by $$ \bfF(x,y) = ( -y, x) . $$ Let $C$ be the unit circle $C := \{(x,y)\in \R^2 : x^2+y^2=1\}$, and let us consider two different parametrizations of $C$, one given by $$ \bfx = \bfg_1(t) := (\cos t, \sin t), \quad 0\le t \le 2\pi $$ and the other given by $$ \bfx = \bfg_2(t) := (\cos t, -\sin t), \quad 0\le t \le 2\pi. $$
Then $\bfg_1'(t) = (-\sin t, \cos t)$, and $\bfF\circ \bfg_1(t) = (-\sin t, \cos t)$, so
$$ \int_0^{2\pi} \bfF(\bfg_1(t))\cdot \bfg_1'(t)\, dt = \int_0^{2\pi}(\sin^2t+\cos^2t)dt = 2\pi. $$
Similar computations show that
$$ \int_0^{2\pi} \bfF(\bfg_2(t))\cdot \bfg_2'(t)\, dt = -\int_0^{2\pi}(\sin^2t+\cos^2t)dt = -2\pi. $$
Thus, in this case, we get different answers from the two parametrizations.
In turns out that, despite this example, line integrals of vector fields in fact do have a certain invariance property (reflected in the fact that the two integrals above differ only by a minus sign.)
For a parametrized curve, we often speak of the orientation. For example, if $C$ is a curve with endpoints $\bf p$ and $\bf q$ such that $\bf p \ne q$, and if $C$ is parametrized by $\bfx = \bfg(t), a\le t\le b$, then the orientation depends on whether
Similarly, if $C$ is a circle in the $xy$ plane (or an ellipse, or the boundary of a connected set that does not have any holes
), then the orientation depends on
whether $\bfg(t)$ traverses $C$ in a clockwise or counterclockwise direction, as $t$ increases from $a$ to $b$.
In either case, the curve has only two possible orientations.
In drawing pictures, the orientation is typically represented by an arrow which indicates the direction that $\bfg(t)$ travels along the curve $C$ as $t$ increases.
The invariance property of line integral of vector fields is that the line integral of a vector field $\bfF$ over a curve $C$ depends on the orientation of $C$ but is otherwise independent of the parametrization.
In fact, changing the orientation of $C$ changes (only) the sign of the line integral $\int_C \bfF\cdot d\bfx$.
This means that if we want to compute the line integral of a vector field, we need to specify the orientation of the curve. Once we have specified the orientation, we can speak unambiguously about $\int_C \bfF\cdot d\bfx$, without having to specify a particular parametrization.
Theorem 3. Assume that $\bfg:[a,b]\to \R^n$ is parametrization of a curve $C$ that is $C^1$ on an open interval containing $[a,b]$, such that $\bfg'(t)\ne \bf 0$ for $t\in [a,b]$. Assume also that $\bfg$ is one-to-one on $(a,b)$.
Also, let $\phi:[c,d]\to [a,b]$ be a function that is one-to-one and onto, and assume that $\phi$ is $C^1$, with $\phi'\ne 0$, on an interval that contains $[c,d]$. Finally, let $\bfh = \bfg\circ\phi$.
Then for any vector field $\bfF$ that is continuous everywhere on $C$, $$ \int_a^b \bfF(\bfg(t))\cdot \bfg'(t)\, dt =\begin{cases} \int_c^d\bfF(\bfh(u))\cdot \bfh'(u)\, du&\mbox{ if }\bfg\mbox{ and }\bfh\mbox{ have the same orientation}\\ -\int_c^d\bfF(\bfh(u))\cdot \bfh'(u)\, du&\mbox{ if }\bfg\mbox{ and }\bfh\mbox{ have the opposite orientation}. \end{cases} $$
The proof relies on a change of variables, like the proof of Theorem 1. The new ingredient here is that we have to pay attention to orientations. The main point is that \begin{align} \phi\mbox{ is increasing } &\ \iff \ \phi(c)= a\mbox{ and }\phi(d) = b \nonumber \\ &\ \iff \ t=\phi(u)\mbox{ increases from }a\mbox{ to }b\mbox{ as }u\mbox{ increases from }c\mbox{ to }d. \nonumber \\ &\ \iff \ \mbox{ parametrizations }\bfg\mbox{ and }\bfh\mbox{ share the same orientation}. \nonumber \end{align} We omit the other details. $\quad \Box$
Remark 1. Note that since we have assumed that $\bfg'(t)\ne 0$, it makes sense to define $\bfT(t) := \frac{\bfg'(t)}{|\bfg'(t)|}$, which is the unit tangent vector to the curve $C$ at $\bfg(t)$, pointing in the direction given by the orientation. We can then define $F_{tang}(\bfg(t)) := \bfF(\bfg(t)) \cdot \bfT(t)$ to be the tangential component of $\bfF$ at $\bfg(t)$. Then $$ \int_{C}\bfF\cdot d\bfx = \int_a^b \bfF(\bfg(t)) \cdot (\frac{\bfg'(t)}{|\bfg'(t)|})|\bfg'(t)| dt = \int_C F_{tang} \, ds. $$ Thus we can interpret $\int_C\bfF\cdot d\bfx$ as the scalar integral of the tangential component of $\bfF$ over the curve $C$. A reparametrization leaves $F_{tang}$ the same if it preserves the orientation, and changes (only) the sign of $F_{tang}$ if it reverses the orientation. This gives a way of understanding the relation between the different invariance properties for line itegrals of vector fields and scalar functions.
It is often useful to think of $d\bfx$ as a vector $d\bfx = (dx_1,\ldots,dx_n)$. Then given an oriented curve $C$ and a vector field $\bfF = (F_1,\ldots, F_n)$, we can write $$ \int_C \bfF\cdot d\bfx = \int_C F_1 \, dx_1+\ldots+ F_n dx_n. $$ Here, for each $j$, we define $$ \int_C F_j\, dx_j = \int_a^b F_j(\bfg(t)) \, g_j'(t)\, dt, $$ where $\bfx = \bfg(t)$, $a\le t\le b$ is a parametrization of $C$ with the right orientation. If we are casual about notation and just write $\bfx(t)$ to indicate that $\bfx = (x_1(t),\ldots, x_n(t))$ is a function of $t$, then this looks nicer: $$ \int_C F_j\, dx_j = \int_a^b F_j(\bfx(t)) \, x_j'(t)\, dt. $$
We also sometimes write $d\bfx = (dx, dy)$ in $2$ dimensions, or $d\bfx = (dx, dy, dz)$ in $3$ dimensions.
Sometimes, as in the statement of Green's Theorem (see the next section), it is traditional to write a vector field $\bfF:\R^2\to \R^2$ with components $(P,Q)$, rather than $(F_1, F_2)$. Then the integrand $\bfF\cdot d\bfx$ is rewritten as $P\,dx+Q\,dy$, and $$ \int_C \bfF\cdot d\bfx \ = \ \int_C P\, dx + Q\, dy = \int_a^b \left[ P(\bfx(t))x'(t) + Q(\bfx(t))y'(t) \right] dt $$ using the casual but suggestive notation $\bfx(t) = (x(t),y(t))$, $a\le t\le b$ for a parametrization of the curve with the right orientation.
Please note that with this notation, $\int_C f \, ds$ and $\int_C f\, dx$ mean very different things.
Example 3 revisited. Let $C$ be the unit circle in the $xy$-plane, oriented counterclockwise, and compute $$ \int_C -y\,dx+x\,dy. $$ (This is exactly the same as $\int_C \bfF\cdot d\bfx$, where $\bfF(x,y) = (-y,x)$, just with different notation.)
To evaluate the integral, we first find a parametrization that traverses $C$ in a counterclockwise direction, say $\bfg(t) = (\cos t, \sin t)$ for $0\le t\le 2\pi$. Then it makes sense to write $x(t)=\cos t$ and $y(t)=\sin t$. So convert $\int_C -y\,dx+x\,dy$ into a concrete integral, we just
With $x(t)=\cos t$ and $y(t)=\sin t$, this leads to $$ \int_C -y\,dx+x\,dy = \int_0^{2\pi}(-\sin t) (-\sin t\, dt) +(\cos t)(\cos t\, dt) = \int_0^{2\pi}1\, dt = 2\pi. $$
Example 4. Let $C$ be the ellipse $x^2+4y^2 = 16$, oriented counterclockwise, and compute $$ \int_C xy\,dx - x\,dy $$
Solution. We can parametrize $C$ by $$ \bfx = (4\cos t, 2\sin t), \qquad 0\le t\le 2\pi. $$ We then simply make the replacements
Thus the integral becomes $$ \int_0^{2\pi}-32 \cos t \sin^2 t \, dt - \int_0^{2\pi} 8\cos^2 t\, dt = 0 - 8\pi = -8\pi. $$
Example 5. Assume that $\psi:[a,b]\to \R$ is a $C^1$ function, and compute $\int_C y\, dx$, where $C$ is the set $\{(x, \psi(x)) : a\le x \le b\}$, oriented from $(b,\psi(b))$ to $(a,\psi(a))$.
Solution. The natural way to parametrize the curve is by $\bfg(t) = (t, \psi(t))$ for $a\le t\le b$. Unfortunately, this has the wrong orientation. To fix this, we have two options:
We could for example choose something like $\bfg(t) = (a -t, \psi(a-t))$ for $a-b \le t \le 0$. This parametrizes $C$ and has the right orientation, but it seems unnatural and requires some thought to be sure we are doing the right thing.
It is simpler to use the natural parametrization $\bfg(t) = (t,\psi(t))$ for $a\le t\le b$ with the wrong orientation, and then mutliply the result by $-1$ to correct for the orientation. Theorem 3 guarantees that this will give the right answer. This leads to $$ \int_C y\, dx = -\int_a^b \psi(t)\, dt. $$ (You can check that if you had opted for the more parametrization with the correct oriemtation, you would have gotten the same result.)
Assume that $f:\R^n\to \R$ is a function of class $C^1$, and that $C$ is a curve oriented by $\bfx = \bfg(t), a\le t\le b$. Then we can compute the line integral of the vector field $\bfF = \nabla f$: \begin{align} \int_C \nabla f\cdot d\bfx &= \int_a^b \nabla f(\bfg(t)) \cdot \bfg'(t)\, dt & \nonumber \\ &= \int_a^b\frac d{dt} f(\bfg(t)) \, dt &\mbox{ by the chain rule} \nonumber \\ &= f(\bfg(b)) - f(\bfg(a)) &\mbox{by the Fundamental Theorem of Calculus}. \end{align} This is sometimes called the Fundamental Theorem of Calculus for line integrals of vector fields, or more briefly, the Fundamental Theorem of Line Integrals.
You must know the definitions of the arclength of a curve $C$, and the line integral of a scalar function or a vector field and you must be able to use them.
Compute the arclength of the following curves. Do enough examples (and no more) to fix in your memory the definition of the arclength of a curve.
Evaluate the following integrals. (Do enough examples to fix in your memory the definition of $\int_C f\, ds$.)
Compute the following integrals. (Again, do enough examples to make sure that you remember the definition of $\int_C \bfF\cdot d\bfx$.)
In the integrals below, suppose that $f$ is an extremely complicated function, for example $f(x,y) =(2+3x) e^{x^2\cos(xy-3y^3)}$, or a similarly complicated function of more variables.
For each of the following curves, compute $\int_C ds, \int_C dx$ and $\int_C dy$. If possible, do them in your head. (This will get easier after you have done several examples.) Keep on doing examples until you can do them more or less instantly.
Let $C$ be a curve in $\R^n$, and assume that $f:\R^n\to \R$ is a continuous function such that $f(\bfx)=0$ for every $\bfx\in C$.
Let $C$ be an oriented curve in $\R^n$, and assume that $\bfF:\R^n\to \R^n$ is a continuous vector field such that $\bfF(\bfx)=\bf 0$ for every $\bfx\in C$.
The elliptic integral of the second kind is the function $E(k)$, defined for $0 < k\le 1$, defined by $$ E(k) = \int_0^{\pi/2} \sqrt{1 - k^2\sin^2 t}\, dt. $$ Assume that $C$ is the ellipse $(x/a)^2+(y/b)^2=1$. Also, assume for concreteness that $a>b$. Express the arclength of $C$ as a multiple of $E(k)$, for some $k$ that will depend on $a$ and $b$.