1.6: Integrating Along a Curve

Suppose that we have a curve \(\mathcal{C}\) that is parametrized as \(\vecs{r} (t)\) with \(a\le t\le b\text{.}\) Suppose further that \(\mathcal{C}\) is actually a piece of wire and that the density (i.e. mass per unit length) of the wire at the point \(\vecs{r} \) is \(\rho(\vecs{r} )\text{.}\) How do we figure out the mass of \(\mathcal{C}\text{?}\) Of course we use the standard Calculus divide and conquer strategy. We select a natural number \(n\) and

• divide the interval \(a\le t\le b\) into \(n\) equal subintervals, each of length \(\Delta t=\frac{b-a}{n}\text{.}\) We denote by \(t_\ell = a + \ell\Delta t\) the right hand end of interval number \(\ell\text{.}\)
• Then we approximate the length of the part of the curve between \(\vecs{r} \big(t_{\ell-1}\big)\) and \(\vecs{r} \big(t_\ell\big)\) by \(\big|\vecs{r} \big(t_\ell\big)-\vecs{r} \big(t_{\ell-1}\big)\big|\) and the mass of the part of the curve between \(\vecs{r} \big(t_{\ell-1}\big)\) and \(\vecs{r} \big(t_\ell\big)\) by \(\rho\big(\vecs{r} (t_\ell)\big) \big|\vecs{r} \big(t_\ell\big)-\vecs{r} \big(t_{\ell-1}\big)\big|\text{.}\)

  

• This gives us, as an approximate mass for \(\mathcal{C}\) of

  \[ \sum_{\ell=1}^n \rho\big(\vecs{r} (t_\ell)\big) \big|\vecs{r} \big(t_\ell\big)-\vecs{r} \big(t_{\ell-1}\big)\big| =\sum_{\ell=1}^n \rho\big(\vecs{r} (t_\ell)\big) \bigg|\frac{\vecs{r} \big(t_\ell\big)-\vecs{r} \big(t_{\ell-1}\big)} {t_\ell-t_{\ell-1}}\bigg|\Delta t \nonumber \]

Then we take the limit as \(n\rightarrow\infty\text{.}\) Assuming¹ that \(\vecs{r} (t)\) is continuously differentiable and that \(\rho(\vecs{r} )\) is continuous we get

\[ \text{Mass of } \mathcal{C} = \int_a^b \rho\big(\vecs{r} (t)\big) \left|\dfrac{d\vecs{r} }{dt}(t)\right|\,\text{d}t \nonumber \]

which we take to be a definition.

Definition 1.6.1

1. For a parametrized curve \(\big(x(t),y(t), z(t)\big)\text{,}\) \(a\le t\le b\text{,}\) in \(\mathbb{R}^3\) that we call \(\mathcal{C}\text{,}\) and for a function \(f(x,y,z)\text{,}\) we define

   \[ \int_\mathcal{C} f(x,y,z)\,\text{d}s =\int_a^b f\big(x(t), y(t) , z(t) \big)\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\ \text{d}t \nonumber \]

   In this notation the subscript \(\mathcal{C}\) specifies the curve, and \(\text{d}s\) signifies arc length.
2. For a curve \(y=f(x)\text{,}\) \(a\le x\le b\text{,}\) in \(\mathbb{R}^2\) that we call \(C\text{,}\) and for a function \(g(x,y)\text{,}\) we define

   \[ \int_C g(x,y)\,\text{d}s =\int_a^b g\big(x, f(x) \big)\sqrt{1+f'(x)^2}\ \text{d}x \nonumber \]

Example 1.6.2

Suppose that we have a helical wire²

\[ \vecs{r} (t) = \big(x(t)\,,\,y(t)\,,\,z(t)\big) =\big(a\cos t\,,\,a\sin t\,,\, bt\big)\qquad 0\le t\le 2\pi \nonumber \]

and that this wire has constant mass density \(\rho\text{.}\) Let's find the centre of mass of the wire. Recall that the centre of mass is \(\big(\bar x,\bar y,\bar z)\) with, for example, \(\bar x\) being the weighted average

\[ \bar x = \frac{\int x\rho \text{d}s}{\int \rho\text{d}s} = \frac{\int x \text{d}s}{\int \text{d}s} \qquad\text{(since $\rho$ is constant)} \nonumber \]

of \(x\) over the wire. Similarly \(\bar y = \frac{\int y \text{d}s}{\int \text{d}s}\) and \(\bar z = \frac{\int z \text{d}s}{\int \text{d}s} \text{.}\) For the given curve

\[\begin{align*} \big(x(t)\,,\,y(t)\,,\,z(t)\big) &=\big(a\cos t\,,\,a\sin t\,,\, bt\big)\\ \big(x'(t)\,,\,y'(t)\,,\,z'(t)\big) &=\big(-a\sin t\,,\,a\cos t\,,\, b\big)\\ \dfrac{ds}{dt}(t) &=\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\\ &=\sqrt{a^2\sin^2t+a^2\cos^2t+b^2}\\ &=\sqrt{a^2+b^2} \end{align*}\]

so that

\[\begin{align*} \bar x &= \frac{\int x \text{d}s}{\int \text{d}s} = \frac{\int_0^{2\pi} x(t) \sqrt{a^2+b^2}\,\text{d}t} {\int_0^{2\pi} \sqrt{a^2+b^2}\,\text{d}t} = \frac{\int_0^{2\pi} a\cos(t) \,\text{d}t}{2\pi} =0\\ \bar y &= \frac{\int y \text{d}s}{\int \text{d}s} = \frac{\int_0^{2\pi} y(t) \sqrt{a^2+b^2}\,\text{d}t} {\int_0^{2\pi} \sqrt{a^2+b^2}\,\text{d}t} = \frac{\int_0^{2\pi} a\sin(t) \,\text{d}t}{2\pi} =0\\ \bar z &= \frac{\int z \text{d}s}{\int \text{d}s} = \frac{\int_0^{2\pi} z(t) \sqrt{a^2+b^2}\,\text{d}t} {\int_0^{2\pi} \sqrt{a^2+b^2}\,\text{d}t} = \frac{\int_0^{2\pi} bt \,\text{d}t}{2\pi} =\frac{b}{2\pi}\Big[\frac{t^2}{2}\Big]_0^{2\pi} =b\pi \end{align*}\]

So the centre of mass is right on the axis of the helix, half way up, which makes perfect sense.

Exercises

Stage 1

1

Give an equation for arclength of a curve \(C\) as a line integral.

2

1. Show that the integral \(\int_\mathcal{C} f(x,y)\,ds\) along the curve \(\mathcal{C}\) given in polar coordinates by \(r=r(\theta)\text{,}\) \(\theta_1\le \theta\le\theta_2\text{,}\) is

   \[ \int_{\theta_1}^{\theta_2}f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2+\left(\dfrac{dr}{d\theta}(\theta)\right)^2}\, \text{d}\theta \nonumber \]

2. Compute the arc length of \(r=1+\cos\theta,\ 0\le \theta\le 2\pi\text{.}\) You may use the formula

   \[ 1+\cos\theta=2\cos^2\frac{\theta}{2} \nonumber \]

   to simplify the computation.

Stage 2

3

Calculate \(\int_C \left(\frac{xy}{z}\right)\text{d}s\text{,}\) where \(C\) is the curve \(\left( \frac23t^3 , \sqrt{3}t^2 , 3t \right)\) from \(t=1\) to \(t=2\text{.}\)

4

A hoop of radius \(r\) traces out the curve \(x^2+y^2=1\text{,}\) where \(x\) and \(y\) are measured in metres. At a point \((x,y)\text{,}\) its density is \(x^2\) kg per metre. What is the mass of the hoop?

5

Compute \(\int_C (xy+z) \text{d}s\) where \(C\) is the straight line from \((1,2,3)\) to \((2,4,5)\text{.}\)

6

Evaluate the path integral \(\int_\mathcal{C} f(x,y,z)\,\text{d}s\) for

1. \(f(x,y,z)=x\cos z\text{,}\) \(\mathcal{C}:\vecs{r} (t)=t\hat{\pmb{\imath}}+t^2\hat{\pmb{\jmath}}\text{,}\) \(0\le t\le 1\text{.}\)
2. \(f(x,y,z)=\frac{x+y}{y+z}\text{,}\) \(\mathcal{C}:\vecs{r} (t)= \big(t,\frac{2}{3}t^{3/2},t\big)\text{,}\) \(1\le t\le 2\text{.}\)

7

Evaluate \(\int_C \sin x\,\text{d}s\text{,}\) where \(C\) is the curve \((\textrm{arcsec}(t), \ln t)\text{,}\) \(1 \le t \le \sqrt{2}\text{.}\)

8. ✳

A particle of mass \(m = 1\) has position \(\vecs{r} (0) = \hat{\pmb{\jmath}}\) and velocity \(\vecs{v} _0 = \hat{\pmb{\imath}} + \hat{\mathbf{k}}\) at time \(t = 0\text{.}\) The particle moves under a force

\[ \vecs{F} (t) = \hat{\pmb{\jmath}} - \sin t\,\hat{\mathbf{k}} \nonumber \]

where \(t\) denotes time.

1. Find the position \(\vecs{r} (t)\) of the particle as a function of \(t\text{.}\)
2. Find the position \(\vecs{r} (t_1)\) of the particle when it crosses the plane \(x = \pi/2\) for the first time at \(t_1\text{.}\)
3. Determine the work done by \(\vecs{F} \) in moving the particle from \(\vecs{r} (0)\) to \(\vecs{r} (t_1)\text{.}\)

Stage 3

9. ✳

Evaluate the line integral \(\int_C \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}s\) where \(\vecs{F} (x,y) = xy^2 \,\hat{\pmb{\imath}} + ye^x \,\hat{\pmb{\jmath}}\), \(C\) is the boundary of the rectangle \(R\text{:}\) \(0 \le x \le 3\text{,}\) \(-1 \le y \le 1\text{,}\) and \(\hat{\textbf{n}}\) is the unit vector, normal to \(C\text{,}\) pointing to the outside of the rectangle.

10. ✳

Let \(\mathcal{C}\) be the curve given by

\[ \vecs{r} (t)=t\cos t\,\hat{\pmb{\imath}}+t\sin t\,\hat{\pmb{\jmath}}+t^2\,\hat{\mathbf{k}},\qquad 0\le t\le \pi \nonumber \]

1. Find the unit tangent \(\hat{\textbf{T}}\) to \(\mathcal{C}\) at the point \((-\pi,0,\pi^2)\text{.}\)
2. Calculate the line integral

   \[ \int_\mathcal{C} \sqrt{x^2+y^2}\ \text{d}s \nonumber \]

3. Find the equation of a smooth surface in \(3\)-space containing the curve \(\mathcal{C}\text{.}\)
4. Sketch the curve \(\mathcal{C}\text{.}\)

11

A wire traces out a path \(C\) described by the curve \((t+\frac12t^2 , t-\frac12t^2 , \frac{4}{3}\,t^{3/2})\text{,}\) \(0 \leq t \leq 4\text{.}\) Its density at the point \((x,y,z)\) is \(\rho(x,y,z)={\left( \frac{x+y}{2}\right)}\text{.}\) Find its centre of mass.

1. We could relax these conditions somewhat by instead assuming that \(\vecs{r} '(t)\) and \(\rho(t)\) are bounded and are continuous except at a finite number of points. (\(\vecs{r} '(t)\) need not exist at all at those points.)
2. For example, your favourite solenoid or spring or slinky.