Scalar Line Integrals Along Parametrized Curves

Accumulating a Scalar Field Along a Curve

A scalar field assigns a number to each point in a region. Examples include temperature at each location, or the amount of material per unit length at each point of a wire. If we want to accumulate such values along a curve, simply integrating the field with respect to the parameter may not be enough: equal parameter intervals need not correspond to equal distances along the curve.

A scalar line integral accounts for both the field’s value and how quickly the parametrization travels along the curve. The field is evaluated at the current point, and its value is weighted by the curve’s speed.

Definition: Let $\mathbf r:[a,b]\to\mathbb R^n$ be a continuously differentiable parametrization, and let $f$ be a scalar-valued function continuous on the curve. The scalar line integral of $f$ along the parametrized curve is $$ \int_C f\,ds = \int_a^b f\bigl(\mathbf r(t)\bigr)\,\lVert\mathbf r'(t)\rVert\,dt. $$ Here $\lVert\mathbf r'(t)\rVert$ is the speed of the parametrization. The symbol $ds$ indicates accumulation with respect to distance along the curve.

The parameter interval $[a,b]$ is compact, and the stated continuity and differentiability assumptions ensure that $f(\mathbf r(t))$ and $\lVert\mathbf r'(t)\rVert$ are continuous. Thus the right-hand side is an ordinary definite integral. The curve need not be a graph, and the formula applies in two or more dimensions.

Why the Speed Appears

Imagine dividing the parameter interval into many short subintervals. On one such interval of width $\Delta t$, the curve travels a short distance that is approximately its speed at that point multiplied by $\Delta t$. If the field is approximately constant over that short piece, its contribution to the total is approximately the field value times that distance.

Consequently, a typical contribution has the form $\,f(\mathbf r(t))\lVert\mathbf r'(t)\rVert\Delta t$. Adding these contributions and taking the limit gives the definite integral in the definition. This explains the role of each factor: $f(\mathbf r(t))$ supplies the field value, while $\lVert\mathbf r'(t)\rVert$ converts parameter intervals into distance traveled. The detailed arc-length calculation is a separate step; here, the parametrized formula is the tool for evaluating the integral.

\text{field value at the point} \;\times\; \text{distance traveled per unit parameter} = f\bigl(\mathbf r(t)\bigr)\lVert\mathbf r'(t)\rVert

To use the formula, differentiate every component of $\mathbf r(t)$, find the magnitude of the resulting vector, substitute the parametrization into $f$, and integrate the product over the specified parameter interval.

Worked Examples

Worked Example: A Straight Parametrized Path

Evaluate $\int_C f\,ds$ for $f(x,y)=x+2y$, where $\mathbf r(t)=(t,2t)$ for $0\le t\le1$.

First evaluate the field along the curve. The coordinates are $x=t$ and $y=2t$, so

f\bigl(\mathbf r(t)\bigr)=t+2(2t)=5t.

Differentiate the parametrization and calculate its speed:

\mathbf r'(t)=(1,2), \qquad \lVert\mathbf r'(t)\rVert =\sqrt{1^2+2^2} =\sqrt{5}.

Substitute both quantities into the formula:

\int_C f\,ds = \int_0^1 (5t)\sqrt{5}\,dt = 5\sqrt{5}\left[\frac{t^2}{2}\right]_0^1 = \frac{5\sqrt{5}}{2}.

Check the integration: the derivative of $5\sqrt{5}\,t^2/2$ is $5\sqrt{5}\,t$, the integrand. The endpoint values are $5\sqrt{5}/2$ at $t=1$ and $0$ at $t=0$, confirming the result.

Worked Example: A Curved Path With a Convenient Integrand

Evaluate the integral for $f(x,y)=\sqrt{1+4y}$ along $\mathbf r(t)=(t,t^2)$, $0\le t\le1$. The field is defined here on a region containing the curve where $1+4y\ge0$.

Along the parametrization, $y=t^2$, so

f\bigl(\mathbf r(t)\bigr) = \sqrt{1+4t^2}.

The derivative and speed are

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

Multiplying the field value by the speed removes the square root:

\begin{aligned} \int_C f\,ds &=\int_0^1 \sqrt{1+4t^2}\sqrt{1+4t^2}\,dt\\ &=\int_0^1 (1+4t^2)\,dt\\ &=\left[t+\frac{4}{3}t^3\right]_0^1\\ &=1+\frac{4}{3} =\frac{7}{3}. \end{aligned}

For a check, the derivative of $t+\frac{4}{3}t^3$ is $1+4t^2$, the simplified integrand. Substitution of the endpoints gives $7/3-0=7/3$.

Worked Example: A Curve in Three Dimensions

Let $f(x,y,z)=\sqrt{5+16z}$, and let $\mathbf r(t)=(t,2t,4t^2)$ for $0\le t\le1$. Evaluate the scalar line integral along this curve.

The third coordinate on the curve is $z=4t^2$. Therefore,

f\bigl(\mathbf r(t)\bigr) = \sqrt{5+16(4t^2)} = \sqrt{5+64t^2}.

Differentiate each component and find the speed:

\mathbf r'(t)=(1,2,8t), \qquad \lVert\mathbf r'(t)\rVert = \sqrt{1^2+2^2+(8t)^2} = \sqrt{5+64t^2}.

The field value and speed are equal along this curve, so

\begin{aligned} \int_C f\,ds &=\int_0^1 \sqrt{5+64t^2}\sqrt{5+64t^2}\,dt\\ &=\int_0^1 (5+64t^2)\,dt\\ &=\left[5t+\frac{64}{3}t^3\right]_0^1\\ &=5+\frac{64}{3} =\frac{79}{3}. \end{aligned}

The derivative of $5t+\frac{64}{3}t^3$ is $5+64t^2$, as required. Evaluating at $t=1$ and $t=0$ gives $79/3-0=79/3$.

Signed Field Values and Cancellation

The scalar field need not be nonnegative. The integral records signed contributions: portions where $f$ is positive contribute positively, and portions where $f$ is negative contribute negatively. These contributions can cancel.

Worked Example: Positive and Negative Contributions

Let $f(x,y)=x$ and $\mathbf r(t)=(t,t^2)$ for $-1\le t\le1$. Along the curve, $f(\mathbf r(t))=t$. Its speed is

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

Thus the integral is

\int_{-1}^1 t\sqrt{1+4t^2}\,dt=0.

Indeed, $\sqrt{1+4t^2}$ has the same value at $t$ and $-t$, while the factor $t$ changes sign. The integrand is odd, so its integral over the symmetric interval is zero. Equivalently, the contributions from the two halves cancel. The curve’s speed is never zero, and the field is positive on the part with $t>0$ and negative on the part with $t<0$.

A Common Mistake: Omitting the Speed

A frequent error is to substitute the parametrization into $f$ and integrate only with respect to $t$. That calculation weights equal intervals of the parameter equally, even when the curve moves at different speeds. A scalar line integral instead weights each interval according to the distance traveled.

Pitfall: For the curve $\mathbf r(t)=(t,2t)$ and field $f(x,y)=x+2y$, omitting the speed gives $\int_0^1 5t\,dt=5/2$. But the speed is $\sqrt{5}$, so the correct integral is $\int_0^1 5t\sqrt{5}\,dt=5\sqrt{5}/2$. The omitted factor changes the result.

Before integrating, check that the integrand contains both required pieces: the field evaluated at the curve and the magnitude of the velocity vector. Also check that the norm is nonnegative. In particular, the speed is $\sqrt{(x'(t))^2+(y'(t))^2}$ for a plane curve and $\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}$ for a space curve.

Interpreting the Result

The units offer a useful check. If $f$ measures temperature, then $\int_C f\,ds$ has units of temperature times length. If $f$ is a density measured per unit length, the integral has units of total mass. Such interpretations depend on what the scalar field represents, but the same mathematical formula applies.

A scalar line integral is different from an ordinary integral over the parameter interval because the curve’s speed matters. It is also different from integrating a vector field by taking a dot product with the curve’s velocity: here, the integrand is a scalar field multiplied by speed. Keeping these integrands distinct prevents a common mix-up.

Check Your Understanding

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

For $f(x,y)=x+y$ and $\mathbf r(t)=(t,3t)$, $0\le t\le1$, what expression should be integrated with respect to $t$?
For $\mathbf r(t)=(\cos t,\sin t)$, what is $\lVert\mathbf r'(t)\rVert$?
Why is $\int_a^b f(\mathbf r(t))\,dt$ generally not the scalar line integral?
Can a scalar line integral be zero when the field is not zero everywhere on the curve? Explain how.
If the field has units of mass per unit length, what units does its scalar line integral have?