5.1 Arclength and Line Integrals

5.1 Arclength and Line Integrals

Arclength and the Line Integral of a Real-Valued Function

The Line Integral of a Vector Field

The Fundamental Theorem of Line Integrals

Problems

\(\Leftarrow\) \(\Uparrow\) \(\Rightarrow\)

The integral allows us to compute lengths of curves, as well as the values of functions taken as we move along a curve. In particular, these functions can be vector-valued, so that we can compute the effect of vectors (for example, from a magnetic or gravitational field) on an object moving along a path.

Arclength and the Line Integral of a Real-Valued Function

Let \(\mathbf g:[a,b]\to \R^n\) be a function that is \(C^1\) on an open interval containing \([a,b]\), and which is injective on \((a,b)\) and \(\mathbf g'(t)\ne {\bf 0}\) for all \(t\in (a,b)\). Let \(C\subset \R^n\) be the image of \(\mathbf g\left([a,b]\right)\). Then the arclength of \(C\) is \[\begin{equation}\label{arclength.def} \int_a^b |\mathbf g'(t)|\, dt. \end{equation}\]

This is equivalent to the single-variable definition of arclength: if \(x=g_1(t), y=g_2(t)\), then \[ |\mathbf g'(t)|=\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}. \]

For the same \(\mathbf g\) and \(C\), if \(f:C\to \R\) is a continuous function, the line integral of \(f\) over \(C\) is \[\begin{equation}\label{lis.def} \int_C f\, ds \ = \int_a^b f(\mathbf g(t)) \ |\mathbf g'(t)|\, dt. \end{equation}\]

Note that the arclength of \(C\) equals \(\int_C 1\, ds\).

Example 1.

Let \(C\) be the curve parametrized by \(\mathbf x = \mathbf g(t) = \left(t^3 , \frac 32 t^4, \frac 6 5 t^5\right)\), for \(0\le t \le 2\). Compute the arclength of \(C\) and the integral \(\int_C xyz\, ds\)

Solution

We compute \[ \mathbf g'(t) = \left(3t^2, 6t^3, 6t^4\right) = 3t^2\left(1,2t, 2t^2\right) \] and hence \[ |\mathbf g'(t)| = 3t^2\sqrt{1+ 4t^2 + 4t^4} = 3t^2\left(1+2t^2\right). \] Thus \[ \text{arclength}(C) = \int_0^2 3t^2+ 6t^4\, dt = 2^3+ \frac 65 2^5 \] and

\[ \int_C xyz\, ds = \int_0^2 \left(t^3\right)\left(\frac 32 t^4\right)\left(\frac 65 t^5\right) \left(3t^2+ 6 t^4\right)\, dt \\ = \frac {27}5\int_0^2 t^{14} + 2t^{16}\, dt = \frac {27}5\left(\frac 1{15}2^{15} +\frac 2{17}\cdot 2^{17}\right) \]

An Invariant Property

We have said that the definition is the arclength of the curve \(C\), but computing it depends on having a parameterization for \(C\). We should make sure that the integral of a real-valued 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.

Suppose that \(\mathbf g:[a,b]\to \R^n\) satisfies the requirements in the definition of arclength, and that \(\mathbf h:[c,d]\to \R^n\) is another parameterization of the same set \(C\), which also satisfies the requirements.

Then for any continuous \(f:\R^n\to \R\), computing \(\int_Cf\, ds\) using either parametrization \(\mathbf g\) or \(\mathbf h\) yields the same answer: \[ \int_C f\ ds = \int_a^b f(\mathbf g(t))|\mathbf g'(t)|\, dt = \int_c^d f(\mathbf h(u))|\mathbf h'(u)|\, du. \] In particular, \[ \text{arclength}(C) = \int_a^b |\mathbf g'(t)|\, dt = \int_c^d|\mathbf h'(u)|\, du. \]

The proof uses Inverse Function Theorem, Chain Rule, and Change of Variables. First, we constuct a function \(\phi:[c,d]\to [a,b]\) such that \(\mathbf h=\mathbf g\circ\phi\). Since \(h\left([c,d]\right)= C=g\left([a,b]\right)\), we want to define \(\phi:(c,d)\to(a,b)\) by \(\phi(u)=t\), where \(g(t)=h(u)\). There is a unique such \(t\) if \(t \in(a,b)\), since both functions are injective on those sets. But we run into trouble if \(g(a)=g(b)=h(u^*)\). Even worse, \(\phi\) will not be continuous at this point, since the values less than \(u^*\) will be near one endpoint, while the values greater than \(u^*\) will be near the other. Fortunately, there can only be one such point, and we will be integrating, so it will not effect the result (the set of discontinuities has zero content). By Inverse Function Theorem, \(\phi\) is \(C^1\) on \((a,b)\), or on \((a,u^*)\cup(u^*,b)\). The values of \(\phi(c)\) and \(\phi(d)\) are determined by continuity, so we have a function which will change between any two parameterizations.

Writing \(\mathbf h\) as a function of a variable \(u\in [c,d]\), the Chain Rule implies that \[ \mathbf h'(u) = \mathbf g'(\phi(u)) \phi'(u), \] so \[ \int_c^d f(\mathbf h(u))|\mathbf h'(u)|\, du = \int_c^d f(\mathbf g(\phi(u)))\, |\mathbf g'(\phi(u))| \, |\phi'(u)|\, du. \] We define the new variable \(t = \phi(u)\). Since \(\phi:[c,d]\to [a,b]\) is a bijection, the Change of Variables Theorem implies that \[ \int_c^d f(\mathbf g(\phi(u)))\, |\mathbf g'(\phi(u))| \, |\phi'(u)|\, du. = \int_a^b f(\mathbf g(t)) |\mathbf g'(t)|\, dt, \] and this implies the conclusion of the theorem.

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 to compute either its arclength or a line integral \(\int_C f\, ds\). To carry out the computation, we can use any parametrization we like.

Example 2.

Let \(C\) be the ellipse \((x/a)^2+(y/b)^2 = 1\). Write down an integral formula for the arclength of \(C\).

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 \[ \mathbf x = \mathbf g(t) = (a\cos t, b\sin t), \qquad 0\le t\le 2\pi. \] Then \[ \mathbf g'(t) = (-a\sin t, b\cos t), \qquad \text{ so }|\mathbf g'(t)| = \sqrt{a^2 \sin^2 t + b^2\cos^2 t}. \] It follows that \[ \text{arclength}(C) = \int_0^{2\pi}\sqrt{a^2 \sin^2 t + b^2\cos^2 t}\, dt. \] Unfortunately, this integral is impossible to evaluate using elementary functions. The elliptic integral (see the last Problem) was invented for this purpose by Giulio Fagnano (Italy, 1682-1786), also known for showing \(\pi=2i\log\left(\frac{1-i}{1+i}\right)\), and for solving an integration problem posed by Brook Taylor to “non-English mathematicians” as an attempt to discredit European mathematics.

The fact that arclength is independent of the parametrization ensures that if we had chosen another parametrization, say \(\mathbf x = (a\sin t, -b\cos t)\) for \(-\pi \le t\le \pi\), or \(\mathbf x = (a\sin(e^t), b\cos(e^t))\) 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.)

Another Approach to Arclength

Let \(C\) be a curve in \(\R^n\) parametrized by \(\mathbf x = \mathbf g(t)\) 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 \(\mathbf g(t_{j-1})\) to \(\mathbf g(t_j)\) for \(j=1,\ldots, J\), which we will denote \(\mathcal L_P(C)\), so that \[ \mathcal L_P(C) = \sum_{j=1}^J |\mathbf g(t_{j}) - \mathbf g(t_{j-1})|. \]

A curve \(C\) as above is rectifiable if \[ \left\{ \mathcal L_P(C) : P\text{ is a partition of }[a,b]\right\} \] is bounded. In this case, we define the arclength of \(C\) to be \[ \mathcal L(C) = \sup\left\{ \mathcal L_P(C) : P\text{ is a partition of }[a,b]\right\}. \]

The following theorem guarantees that this definition of arclength is consistant with the one we gave above.

If \(\mathbf g:[a,b]\to \R^n\) is \(C^1\), then \[ \mathcal L(C) = \int_a^b |\mathbf g'(t)|\, dt. \]

The proof follows by applying the definition of the derivative of \(\mathbf g\) and showing that \(\forall \varepsilon>0,\) there is a partition \(P\) with \(U_P(|\mathbf g'|)-\varepsilon \leq \mathcal L(C) \leq L_P(|\mathbf g'|)+\varepsilon\).

The Line Integral of a Vector Field

We continue to suppose that \(C\) is a curve in \(\R^n\) parametrized by a function \[ \mathbf x = \mathbf g(t), \quad a\le t\le b, \] with the same assumptions about \(\mathbf g\) as above, e.g. \(\mathbf g'\) exists and is nonzero everywhere in an open interval containing \([a,b]\).

A vector field in \(\R^n\) is a function \(\mathbf F:S\to \R^n\), 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 \(\mathbf x\in S\), \(\mathbf F(\mathbf x)\) is a vector in the strict sense, that is, a quantity with a direction and magnitude, such as a force or a velocity. We want to think of \(\mathbf F\) as assigning an \(n-\)dimensional vector to every point in \(S\). Vector fields arise often in physics, precisely because there are many physical situations where vectors vary from point to point (for example, the force of gravity on an object rotating around the Earth).

Given a vector field \(\mathbf F:S\to \R^n\), and a curve \(C\subset S \subseteq \R^n\), parameterized by \(\mathbf g\), we define the line integral of \(\mathbf F\) over \(C\) to be \[\begin{equation}\label{livf.def} \int_C \mathbf F\cdot d\mathbf x = \int_a^b \mathbf F(\mathbf g(t)) \cdot \mathbf g'(t)\, dt. \end{equation}\]

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 \(\mathbf x = \mathbf g(t)\), \(a\le t\le b\), and if there is a force (due for example to gravity, or electromagnetic forces) \(\mathbf F(\mathbf x)\) acting on the particle at position \(\mathbf x\), then \(\int_C \mathbf F\cdot d\mathbf x\) is the work done by the force field on the particle.

In the next example, we will show that the line integral of a vector field is not independent of the parametrization.

Example 3.

Let \(S = \R^2\setminus \left\{ {\bf 0}\right\}\), and let \(\mathbf F\) be the vector field on \(S\) defined by \[ \mathbf F(x,y) = ( -y, x) . \] Let \(C\) be the unit circle \(C = \left\{(x,y)\in \R^2 : x^2+y^2=1\right\}\). Compute the line integral of \(\mathbf F\) using two different parametrizations of \(C\), one given by \[ \mathbf x = \mathbf g(t) = (\cos t, \sin t), \quad 0\le t \le 2\pi \] and the other given by \[ \mathbf x = \mathbf h(t) = (\cos t, -\sin t), \quad 0\le t \le 2\pi. \]

We have \(\mathbf g'(t) = (-\sin t, \cos t)\), and \(\mathbf F\circ \mathbf g(t) = (-\sin t, \cos t)\), so

\[ \int_0^{2\pi} \mathbf F(\mathbf g(t))\cdot \mathbf g'(t)\, dt = \int_0^{2\pi}(\sin^2t+\cos^2t)dt = 2\pi. \]

Similar computations show that

\[ \int_0^{2\pi} \mathbf F(\mathbf h(t))\cdot \mathbf h'(t)\, dt = -\int_0^{2\pi}(\sin^2t+\cos^2t)dt = -2\pi. \]

Thus, in this case, we get different answers from the two parametrizations.

Line integrals of vector fields do have a certain invariant property - notice that the two integrals above differ only by a negative sign - and it depends on one aspect of the parameterization.

Orientation

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 \(\mathbf x = \mathbf g(t), a\le t\le b\), then the orientation depends on whether

the curve starts at \(\bf p\) and ends at \(\bf q\) (that is, \(\mathbf g(a)=\bf p\) and \(\mathbf g(b)= \bf q\)), or
the curve starts at \(\bf q\) and ends at \(\bf p\) (that is, \(\mathbf g(a)=\bf q\) and \(\mathbf g(b)= \bf p\)).

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 \(\mathbf g(t)\) 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 \(\mathbf g(t)\) 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 \(\mathbf F\) over a curve \(C\) depends on the orientation of \(C\) but is otherwise independent of the parametrization.

In fact, changing the orientation of \(C\) can only change the sign of the line integral \(\int_C \mathbf F\cdot d\mathbf x\).

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 \(\int_C \mathbf F\cdot d\mathbf x\), without having to specify a particular parametrization.

Suppose that \(C\) is a curve with two parameterizations, \(\mathbf g:[a,b]\to \R^n\) and \(h:[c,d]\to \R^n\), that both satisfy all the requirements in the definition of arclength, and such that there is a \(C^1\) function \(\phi:[c,d]\to[a,b]\) which is a bijection.

Then for any vector field \(\mathbf F\) that is continuous everywhere on \(C\), \[ \int_a^b \mathbf F(\mathbf g(t))\cdot \mathbf g'(t)\, dt =\begin{cases} \int_c^d\mathbf F(\mathbf h(u))\cdot \mathbf h'(u)\, du&\text{ if }\mathbf g\text{ and }\mathbf h\text{ have the same orientation}\\ -\int_c^d\mathbf F(\mathbf h(u))\cdot \mathbf h'(u)\, du&\text{ if }\mathbf g\text{ and }\mathbf h\text{ have the opposite orientation}. \end{cases} \]

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 \[\begin{align} \phi\text{ is increasing } &\ \iff \ \phi(c)= a\text{ and }\phi(d) = b \nonumber \\ &\ \iff \ t=\phi(u)\text{ increases from }a\text{ to }b\text{ as }u\text{ increases from }c\text{ to }d. \nonumber \\ &\ \iff \ \text{ parametrizations }\mathbf g\text{ and }\mathbf h\text{ share the same orientation}. \nonumber \end{align}\] We omit the other details.

Remark 1.

Since \(\mathbf g'(t)\ne 0\), it makes sense to define \(\mathbf T(t) = \frac{\mathbf g'(t)}{|\mathbf g'(t)|}\), which is the unit tangent vector to the curve \(C\) at \(\mathbf g(t)\), pointing in the direction given by the orientation. We can then define \(F_{tgt}(\mathbf g(t)) = \mathbf F(\mathbf g(t)) \cdot \mathbf T(t)\) to be the tangential component of \(\mathbf F\) at \(\mathbf g(t)\). Then \[ \int_{C}\mathbf F\cdot d\mathbf x = \int_a^b \mathbf F(\mathbf g(t)) \cdot (\frac{\mathbf g'(t)}{|\mathbf g'(t)|})|\mathbf g'(t)| dt = \int_C F_{tgt} \, ds. \] Thus we can interpret \(\int_C\mathbf F\cdot d\mathbf x\) as the scalar integral of the tangential component of \(\mathbf F\) over the curve \(C\). A reparametrization leaves \(F_{tgt}\) the same if it preserves the orientation, and only changes the sign of \(F_{tgt}\) 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 real-valued functions.

Other Notation

It is often useful to think of \(d\mathbf x\) as a vector \(d\mathbf x = (dx_1,\ldots,dx_n)\). Then given an oriented curve \(C\) and a vector field \(\mathbf F = (F_1,\ldots, F_n)\), we can write \[ \int_C \mathbf F\cdot d\mathbf x = \int_C F_1 \, dx_1+\ldots+ F_n dx_n. \] Here, for each \(j\), we define \[ \int_C F_j\, dx_j = \int_a^b F_j(\mathbf g(t)) \, g_j'(t)\, dt, \] where \(\mathbf x = \mathbf g(t)\), \(a\le t\le b\) is a parametrization of \(C\) with the right orientation. If we write \(\mathbf x(t)\) to indicate that \(\mathbf x = (x_1(t),\ldots, x_n(t))\) is a function of \(t\), then this looks nicer: \[ \int_C F_j\, dx_j = \int_a^b F_j(\mathbf x(t)) \, x_j'(t)\, dt. \]

We also sometimes write \(d\mathbf x = (dx, dy)\) in \(2\) dimensions, or \(d\mathbf x = (dx, dy, dz)\) in \(3\) dimensions.

Sometimes, as in the statement of Green’s Theorem (see the next section), it is traditional to write a vector field \(\mathbf F:\R^2\to \R^2\) with components \((P,Q)\), rather than \((F_1, F_2)\). Then the integrand \(\mathbf F\cdot d\mathbf x\) is rewritten as \(P\,dx+Q\,dy\), and \[ \int_C \mathbf F\cdot d\mathbf x \ = \ \int_C P\, dx + Q\, dy = \int_a^b \left[ P(\mathbf x(t))x'(t) + Q(\mathbf x(t))y'(t) \right] dt \] using the casual but suggestive notation \(\mathbf x(t) = (x(t),y(t))\), \(a\le t\le b\) 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. Compute \[ \int_C -y\,dx+x\,dy. \] This is the same as \(\int_C \mathbf F\cdot d\mathbf x\), where \(\mathbf F(x,y) = (-y,x)\), just with different notation.

Solution

To evaluate the integral, we first find a parametrization that traverses \(C\) in a counterclockwise direction, say \(\mathbf g(t) = (\cos t, \sin t)\) 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

replace \(x\) by \(x(t)\) and \(y\) by \(y(t)\); and
replace \(dx\) by \(x'(t)\,dt\), and replace \(dy\) by \(y'(t)\,dt\)
then integrate \(\int\cdots dt\) from \(t=0\) to \(t=2\pi\).

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. Compute \[ \int_C xy\,dx - x\,dy \]

Solution

We can parametrize \(C\) by \[ \mathbf x = (4\cos t, 2\sin t), \qquad 0\le t\le 2\pi. \] We then make the replacements

replace \(x\) by \(x(t) = 4\cos t\) and replace \(y\) by \(y(t) = 2\sin t\).
replace \(dx\) by \(x'(t)dt = -4\sin t\ dt\) and replace \(dy\) by \(y'(t)dt = 2\cos t\, dt\).

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.

Suppose that \(\psi:[a,b]\to \R\) is a \(C^1\) function. Compute \(\int_C y\, dx\), where \(C\) is the set \(\left\{(x, \psi(x)) : a\le x \le b\right\}\), oriented from \((b,\psi(b))\) to \((a,\psi(a))\).

Solution

The natural way to parametrize the curve is by \(\mathbf g(t) = (t, \psi(t))\) for \(a\le t\le b\). Unfortunately, this has the wrong orientation. To fix this, we have two options:

We could choose something like \(\mathbf g(t) = (a -t, \psi(a-t))\) for \(a-b \le t \le 0\). This parametrizes \(C\) and has the right orientation, but it seems unnatural and requires careful substitutions.
We could use the natural parametrization \(\mathbf g(t) = (t,\psi(t))\) 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 orientation, you would have gotten the same result.)

The Fundamental Theorem of Line Integrals

Suppose that \(f:\R^n\to \R\) is a function of class \(C^1\), and that \(C\) is a curve oriented by \(\mathbf x = \mathbf g(t), a\le t\le b\). Then we can compute the line integral of the vector field \(\mathbf F = \nabla f\): \[\begin{align} \int_C \nabla f\cdot d\mathbf x &= \int_a^b \nabla f(\mathbf g(t)) \cdot \mathbf g'(t)\, dt & \nonumber \\ &= \int_a^b\frac d{dt} f(\mathbf g(t)) \, dt &\text{ by the chain rule} \nonumber \\ &= f(\mathbf g(b)) - f(\mathbf g(a)) &\text{by the Fundamental Theorem of Calculus}. \end{align}\]

This is sometimes called the Fundamental Theorem of Calculus for Line Integrals of Vector Fields, but we prefer the Fundamental Theorem of Line Integrals.

Problems

Basic

Compute the arclength of the following curves.
- \(C\) parametrized by \(\mathbf x = \mathbf g(t) = (t , t+ \frac 12 \sin 2t, 2\sin t)\) for $0 t $.
  
  Hint
  Rewrite \(1+\cos 2t\) using a trigonometric identity.
- \(C\) parametrized by \((\ln t, 2t, t^2)\) for \(1\le t \le 2\).
- The arc of the parabola \(y=\frac 12 x^2\) that stretches between the origin and the point \((2,2)\).
  
  Hint
  It may be a good idea to look for a substitution involving the hyperbolic trigonometric functions \[ \cosh t = \frac 12 (e^t+e^{-t}), \qquad \sinh t = \frac 12 (e^t-e^{-t}), \] Among other properties, these satisfy \[ \cosh' t= \sinh t, \qquad \sinh' t = \cosh t, \qquad \cosh^2 t = 1 + \sinh^2 t. \]
- The curve \(C\) parametrized by \((t\cos t, t\sin t)\) for \(0\le t\le 2\pi\).
- Let \(C\) be a curve parametrized by \(\mathbf x = (t^{-1}, t, at^3)\) for \(3\le t \le 4\). Find a value of \(a\) that makes it possible for you to compute the arclength of \(C\), and then compute the arclength with that choice of \(a\).
Evaluate the following integrals.
- \(\int_C x \, ds\), where \(C\) is the curve parametrized by \((t,t+ \frac 12 \sin 2t,2\sin t)\) for $0 t $.
- \(\int_C y\,ds\), where \(C\) is the curve parametrized by \((\ln t, 2t, t^2)\) for \(1\le t \le 2\).
- \(\int_C x \, ds\), where \(C\) is the portion of the parabola \(y=x^2\) between \((1,1)\) and \((3,9)\).
- \(\int_C x y\, ds\), where \(C\) is the part of the ellipse \((x/2)^2+y^2=1\) in the first quadrant.
- \(\int_C \mathbf x \cdot d\mathbf x\), where \(C\) is the part of the ellipse \(x^2+4y^2=4\) in the first quadrant, oriented counterclockwise.
- \(\int_C z\,dx + x\,dy\), where \(C\) is the set of points satisfying both equations \[ x^2+y^2+z^2 = 4, \qquad 2z+x=0, \] and \(C\) is oriented counterclockwise (as seen from above).
- \(\int_C y \cos z \, dx\) where \(C\) is the curve parametrized by \((t,t^2, t^3)\) for \(0\le t\le 1\).
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.
- Compute \(\int_C f(x,y)\, dy\), where \(C\) is the line segment that starts at \((23,1)\) and ends at \((2,1)\).
- Compute \(\int_C f(x,y) \,dx\), where \(C\) is the line segment starting at \((1,-1)\) and ending at \((1, 7)\).
- Compute \(\int_C f(x,y)(x \,dx + y\, dy)\), where \(C\) is the unit circle, oriented clockwise.
- Compute \(\int_C f(x,y,z)\, dz\), where \(C\) is the line segment from \((1,2,3)\) to \((5,4,3)\)
For each of the following curves, compute \(\int_C ds, \int_C dx\) and \(\int_C dy\).
- \(C\) is the straight line in \(\R^2\) starting at the origin and ending at the point \((1,1)\).
- \(C\) is the straight line in \(\R^2\) starting at the origin and ending at the point \((3,4)\).
- \(C\) is the straight line in \(\R^2\) starting at the point \((3,4)\) and ending at the origin.
- \(C\) is the path in \(\R^2\) consisting of the horizontal and vertical sides of the right triangle with vertices at \((0,0), (3,0)\) and \((3,4)\), starting at the origin and ending at \((3,4)\).
- \(C\) is the unit circle in \(\R^2\), oriented counterclockwise.
- \(C\) is the unit circle in \(\R^2\), oriented clockwise.
- \(C\) is the straight line that starts at \((5,0)\) and ends at \((0,12)\).
Let \(C\) be a curve in \(\R^n\), and suppose that \(f:\R^n\to \R\) is a continuous function such that \(f(\mathbf x)=0\) for every \(\mathbf x\in C\).
- Prove that under the above hypotheses, \(\int_C f\, ds = 0\).
- Suppose that \(g:\R^2\to\R\) is an incredibly complicated continuous function, and let \(C\) denote the unit circle in the \(xy\)-plane. Compute \(\int_C (1-x^2-y^2)g(x,y) \, ds\) in your head.
Let \(C\) be an oriented curve in \(\R^n\), and suppose that \(\mathbf F:\R^n\to \R^n\) is a continuous vector field such that \(\mathbf F(\mathbf x)=\bf 0\) for every \(\mathbf x\in C\).
- Prove that under the above hypotheses, \(\int_C \mathbf F\cdot d\mathbf x = 0\).
- Let \(\mathbf G:\R^3\to \R^3\) be an unbelievably complicated continuous vector field, and let \(C\) be the curve parametrized by \((t, \cos t, 2\sin t)\) for \(0\le t \le 200\). Compute \(\int_C (4y^2+z^2-4)\mathbf G(x,y,z)\cdot d\mathbf x\) in your head.

Advanced

The elliptic integral of the second kind is the function \(E(k)\), defined for \(0 < k\le 1\), defined by \[ E(k) = \int_0^{\pi/2} \sqrt{1 - k^2\sin^2 t}\, dt. \] Suppose that \(C\) is the ellipse \((x/a)^2+(y/b)^2=1\). Also, suppose for concreteness that \(a>b\). Express the arclength of \(C\) as a multiple of \(E(k)\), for some \(k\) that depends on \(a\) and \(b\).

\(\Leftarrow\) \(\Uparrow\) \(\Rightarrow\)

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