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Scalar Line Integrals · Tutorial 734 of 999

Scalar Line Integrals on Helices and Space Curves

Use a space curve’s coordinate functions and speed to turn a scalar line integral into an ordinary definite integral.

Intermediate 10 min read

What You'll Learn

  • Write the scalar line integral along a parametrized space curve.
  • Find a curve’s speed from all components of its derivative.
  • Substitute the curve’s coordinates into a scalar field.
  • Evaluate scalar line integrals along standard and reparametrized helices.
  • Compute a scalar line integral along a non-helical space curve.
  • Recognize and correct a common error in calculating speed.

From a Space Curve to an Ordinary Integral

A scalar field assigns a number to each point in space. A curve traces out points where that field can be evaluated, and a scalar line integral accumulates those field values along the curve. To calculate one, we need two pieces of information: the field value at each point of the curve, and the rate at which the parametrization travels along it.

For a helix or another space curve, the calculation follows a consistent pattern. Write the curve as a vector-valued function, differentiate to find its speed, substitute its coordinate functions into the scalar field, and integrate the resulting expression over the parameter interval.

Definition: Let \(C\) be parametrized by \(\mathbf{r}(t)=(x(t),y(t),z(t))\), for \(a\le t\le b\), and let \(f\) be a scalar field defined along \(C\). If \(\mathbf{r}\) is continuously differentiable and \(f\) is continuous along the curve, then $$ \int_C f\,ds = \int_a^b f\bigl(x(t),y(t),z(t)\bigr)\,\lVert\mathbf{r}'(t)\rVert\,dt. $$ Here \(\lVert\mathbf{r}'(t)\rVert\) is the speed, and \(ds=\lVert\mathbf{r}'(t)\rVert\,dt\) is the arc-length element.

If a curve is piecewise continuously differentiable, the integral is evaluated on each smooth piece and the resulting integrals are added. The coordinate substitution and speed calculation are done using the parameter for the piece currently being considered.

Why Speed Supplies the Arc-Length Element

The parameter \(t\) is not generally a measure of distance along the curve. The speed corrects for that difference. For a small positive parameter change \(\Delta t\), differentiability gives

$$ \mathbf{r}(t+\Delta t)-\mathbf{r}(t) = \mathbf{r}'(t)\Delta t+o(\Delta t). $$

Taking lengths shows that the distance traveled over this small interval is approximately \(\lVert\mathbf{r}'(t)\rVert\Delta t\). The scalar field value over the same small piece is approximately \(f(\mathbf{r}(t))\). Thus a sum of field value times small arc length has terms approximately \(f(\mathbf{r}(t))\lVert\mathbf{r}'(t)\rVert\Delta t\). As the parameter intervals shrink, these sums give the definite integral in the definition.

$$ \int_C f\,ds = \int_a^b f\bigl(x(t),y(t),z(t)\bigr) \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}\,dt. $$

This component formula follows from the Euclidean length of the velocity vector. In three dimensions, omitting any one coordinate derivative can change the speed and therefore the integral.

Worked Example: A Standard Helix

Worked Example: A Standard Helix

Evaluate \(\int_C f\,ds\) for the helix \(\mathbf{r}(t)=(2\cos t,2\sin t,t)\), \(0\le t\le\pi\), with \(f(x,y,z)=x^2+y^2+z\).

Step 1: Differentiate the parametrization.

$$ \mathbf{r}'(t)=(-2\sin t,\,2\cos t,\,1). $$

Step 2: Calculate the speed using all three components.

$$ \begin{aligned} \lVert\mathbf{r}'(t)\rVert &=\sqrt{(-2\sin t)^2+(2\cos t)^2+1^2}\\ &=\sqrt{4\sin^2 t+4\cos^2 t+1}\\ &=\sqrt{4(\sin^2 t+\cos^2 t)+1}\\ &=\sqrt{5}. \end{aligned} $$

Step 3: Substitute the coordinates into the field. Along the curve, \(x=2\cos t\), \(y=2\sin t\), and \(z=t\), so

$$ f(\mathbf{r}(t)) =(2\cos t)^2+(2\sin t)^2+t =4(\cos^2 t+\sin^2 t)+t =4+t. $$

Step 4: Form and evaluate the definite integral.

$$ \begin{aligned} \int_C f\,ds &=\int_0^\pi (4+t)\sqrt{5}\,dt\\ &=\sqrt{5}\left[4t+\frac{t^2}{2}\right]_0^\pi\\ &=\sqrt{5}\left(4\pi+\frac{\pi^2}{2}\right). \end{aligned} $$

Check: Differentiating \(4t+\frac{t^2}{2}\) gives \(4+t\), the field value after substitution. The endpoints give \(4\pi+\frac{\pi^2}{2}\) at \(t=\pi\) and \(0\) at \(t=0\), as used above.

A Non-Helical Curve in Three Dimensions

The same method applies when the coordinates do not describe a helix. Each coordinate contributes its own derivative to the speed, while the field is evaluated at the point whose coordinates are given by the parametrization.

Worked Example: A Polynomial Space Curve

Evaluate \(\int_C f\,ds\) for \(\mathbf{r}(t)=(t,t^2,t)\), \(0\le t\le1\), with \(f(x,y,z)=x+z\).

Step 1: Differentiate the curve.

$$ \mathbf{r}'(t)=(1,2t,1). $$

Step 2: Find its speed.

$$ \lVert\mathbf{r}'(t)\rVert =\sqrt{1^2+(2t)^2+1^2} =\sqrt{2+4t^2}. $$

Step 3: Evaluate the field along the curve. Since \(x=t\) and \(z=t\),

$$ f(\mathbf{r}(t))=t+t=2t. $$

Step 4: Integrate.

$$ \begin{aligned} \int_C f\,ds &=\int_0^1 2t\sqrt{2+4t^2}\,dt. \end{aligned} $$

Use \(u=2+4t^2\), so \(du=8t\,dt\) and \(2t\,dt=\frac14\,du\). The bounds change from \(t=0\) to \(u=2\), and from \(t=1\) to \(u=6\). Therefore,

$$ \begin{aligned} \int_0^1 2t\sqrt{2+4t^2}\,dt &=\frac14\int_2^6 u^{1/2}\,du\\ &=\frac16\left[u^{3/2}\right]_2^6\\ &=\frac{6^{3/2}-2^{3/2}}{6}\\ &=\sqrt{6}-\frac{\sqrt{2}}{3}. \end{aligned} $$

Check: An antiderivative in the original parameter is \(\frac16(2+4t^2)^{3/2}\). Its derivative is

$$ \frac16\cdot\frac32(2+4t^2)^{1/2}\cdot 8t = 2t\sqrt{2+4t^2}, $$

which matches the integrand. Evaluating it at \(1\) and \(0\) gives the stated result.

When a Helix Uses a Nonuniform Parameter

A parametrization need not move along its curve at a constant speed. In particular, a helix can be described using a changing angle. The speed must then be computed from the actual coordinate functions in that parametrization, rather than borrowed from a different parametrization of the same geometric shape.

Worked Example: A Helix with Changing Angular Speed

Evaluate \(\int_C f\,ds\) for \(\mathbf{r}(t)=(2\cos(t^2),2\sin(t^2),t^2)\), \(1\le t\le2\), with \(f(x,y,z)=z\).

Step 1: Differentiate each coordinate.

$$ \mathbf{r}'(t)=(-4t\sin(t^2),\,4t\cos(t^2),\,2t). $$

Step 2: Calculate the speed.

$$ \begin{aligned} \lVert\mathbf{r}'(t)\rVert &=\sqrt{16t^2\sin^2(t^2)+16t^2\cos^2(t^2)+4t^2}\\ &=\sqrt{16t^2+4t^2}\\ &=\sqrt{20t^2}\\ &=2\sqrt{5}\,t. \end{aligned} $$

The final equality uses \(t\ge1\), so \(t\) is positive.

Step 3: Substitute into the field. The third coordinate is \(z=t^2\), hence \(f(\mathbf{r}(t))=t^2\).

Step 4: Integrate over the given parameter interval.

$$ \begin{aligned} \int_C f\,ds &=\int_1^2 t^2(2\sqrt{5}\,t)\,dt\\ &=2\sqrt{5}\left[\frac{t^4}{4}\right]_1^2\\ &=\frac{\sqrt{5}}{2}(16-1)\\ &=\frac{15\sqrt{5}}{2}. \end{aligned} $$

Check: Differentiating \(\frac{\sqrt{5}}{2}t^4\) gives \(2\sqrt{5}\,t^3\), exactly the integrand after substitution. The endpoint difference is \(\frac{\sqrt{5}}{2}(2^4-1^4)=\frac{15\sqrt{5}}{2}\).

A Common Speed Error

For the standard helix in the first example, a tempting but incorrect calculation is to use only the first two coordinate derivatives:

$$ \sqrt{(-2\sin t)^2+(2\cos t)^2}=2. $$

This is the speed of the projection onto the \(xy\)-plane, not the speed of the helix in three-dimensional space. The third coordinate has derivative \(z'(t)=1\), so the correct speed is

$$ \sqrt{(-2\sin t)^2+(2\cos t)^2+1^2}=\sqrt{5}. $$

If the incorrect speed \(2\) were used in the first example, the resulting integral would be \(2\left(4\pi+\frac{\pi^2}{2}\right)\), not \(\sqrt{5}\left(4\pi+\frac{\pi^2}{2}\right)\). The correction is to differentiate every coordinate and include every squared derivative under the square root.

A Reliable Calculation Process

Before evaluating the ordinary integral, check that the integrand contains both required factors: the field value along the curve and the speed. Keeping them separate at first makes substitutions easier to inspect.

1
Record the parameter interval.
Use the stated lower and upper parameter values for the curve segment being integrated.
2
Differentiate all coordinates.
For \(\mathbf{r}(t)=(x(t),y(t),z(t))\), calculate \(x'(t)\), \(y'(t)\), and \(z'(t)\).
3
Compute the speed.
Use \(\sqrt{(x')^2+(y')^2+(z')^2}\), simplifying only after all components have been included.
4
Substitute into the scalar field.
Replace each field coordinate by its corresponding coordinate function of \(t\).
5
Integrate and check.
Evaluate the resulting definite integral and verify the antiderivative or endpoint calculation.

The parameter alone does not determine how much distance is traveled. That is why a coordinate substitution without the speed factor is incomplete. For helices and other space curves alike, the integrand is the product \(f(\mathbf{r}(t))\lVert\mathbf{r}'(t)\rVert\).

Check Your Understanding

For each question, identify the field value along the curve and the speed before evaluating an integral.

  1. For \(\mathbf{r}(t)=(3\cos t,3\sin t,2t)\), what is \(\lVert\mathbf{r}'(t)\rVert\)?
  2. For the same curve, if \(f(x,y,z)=z\), what expression is \(f(\mathbf{r}(t))\)?
  3. Write the definite-integral formula for \(\int_C f\,ds\) when \(\mathbf{r}(t)=(t,t^2,2t)\), \(0\le t\le1\), and \(f(x,y,z)=x+y\).
  4. Why is the speed of a space curve not generally found by differentiating only its \(x\)- and \(y\)-coordinates?
  5. In the parametrization \(\mathbf{r}(t)=(2\cos(t^2),2\sin(t^2),t^2)\), why must the speed be calculated from these coordinate functions rather than assumed constant?

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