Lesson 26: Line integrals of functions (17.2 - Part I)

Accessible transcription generated on 3/25/2026

Original Notes

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Lesson 26: Line integrals of functions (17.2 - Part I)

Review

1) What is a Vector Field?

A function that assigns a vector at each point.

eg: Wind, Acceleration due to gravity

\[ \vec{F} = \langle y, x \rangle, \quad \vec{F} = \langle z, x, -1 \rangle \]

2) What is a Gradient Field? Is every vector field a Gradient Field?

\[ \vec{F} = \vec{\nabla}\phi, \text{ for some } \phi, \qquad \text{NO} \]

Example of a non-gradient field: \[ \vec{F} = \langle y, -x \rangle \]

3) Is \( \vec{F} = \langle 2x, -1 \rangle \) a Gradient field?

\( \phi_x = 2x \leadsto \phi(x,y) = x^2 + K(y) \)
\( -1 = \phi_y = K'(y) \Rightarrow K(y) = -y + C \)
\( \phi(x,y) = x^2 - y + C \leadsto \vec{\nabla}\phi = \langle 2x, -1 \rangle = \vec{F} \)

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Recall: Definition of Definite Integral

A 2D Cartesian coordinate system showing a blue curve labeled y = f(x). The horizontal x-axis has two points marked, 'a' and 'b'. A shaded vertical rectangular strip with diagonal lines is shown under the curve at an arbitrary point between 'a' and 'b', representing a single rectangle in a Riemann sum. Dashed vertical lines extend from points 'a' and 'b' on the x-axis up to the curve to define the boundaries of the integration area.
Visual Description: A 2D Cartesian coordinate system showing a blue curve labeled y = f(x). The horizontal x-axis has two points marked, 'a' and 'b'. A shaded vertical rectangular strip with diagonal lines is shown under the curve at an arbitrary point between 'a' and 'b', representing a single rectangle in a Riemann sum. Dashed vertical lines extend from points 'a' and 'b' on the x-axis up to the curve to define the boundaries of the integration area.
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x \]

In this Riemann sum definition, \( f(x_i) \) is the height of the rectangular sub-intervals, and \( \Delta x \) is the width.

Generalize: Integrating \( f \) over a Line Segment on an Axis

Integrating a function can be generalized to three-dimensional space where the domain is a segment along one of the coordinate axes.

A 3D coordinate system with axes labeled x, y, and z. In the xz-plane, a blue curve is labeled z = f(x). A specific segment of the x-axis is highlighted in green. Above this green segment, a vertical rectangular strip, shaded purple with diagonal lines, extends upwards to meet the curve z = f(x). This illustrates the concept of integrating a function over a segment that lies specifically on the x-axis within a 3D context.
Visual Description: A 3D coordinate system with axes labeled x, y, and z. In the xz-plane, a blue curve is labeled z = f(x). A specific segment of the x-axis is highlighted in green. Above this green segment, a vertical rectangular strip, shaded purple with diagonal lines, extends upwards to meet the curve z = f(x). This illustrates the concept of integrating a function over a segment that lies specifically on the x-axis within a 3D context.
A 3D coordinate system with x, y, and z axes. In the yz-plane, a blue curve is labeled z = f(y). A segment of the y-axis is highlighted in green. A vertical rectangular strip, bounded by purple dashed lines and filled with purple diagonal shading, extends from the green segment on the y-axis up to the curve z = f(y). This represents the generalization of integration over a line segment located on the y-axis.
Visual Description: A 3D coordinate system with x, y, and z axes. In the yz-plane, a blue curve is labeled z = f(y). A segment of the y-axis is highlighted in green. A vertical rectangular strip, bounded by purple dashed lines and filled with purple diagonal shading, extends from the green segment on the y-axis up to the curve z = f(y). This represents the generalization of integration over a line segment located on the y-axis.

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Line integrals

Integrate \(f(x,y)\) over a curve \(C\) parametrized by \(\vec{r}(t) = \langle x(t), y(t) \rangle\), \(a \le t \le b\).

A 3D coordinate system graph illustrating the geometric interpretation of a line integral. The diagram shows the x, y, and z axes. In the xy-plane, there is a green curve labeled C. Directly above it, a blue curve represents the function z = f(x,y). Vertical dashed lines connect the curve C in the xy-plane to the curve z = f(x,y), forming a vertical 'curtain' or 'fence' shape. A single thin vertical slice of this curtain is highlighted and shaded with diagonal purple lines to represent a differential element. This element's height is defined by the function value f(x,y) at a point along the curve, and its width is a small segment of the arc length of curve C. The visualization demonstrates that the line integral represents the area of this vertical surface above the curve.
Visual Description: A 3D coordinate system graph illustrating the geometric interpretation of a line integral. The diagram shows the x, y, and z axes. In the xy-plane, there is a green curve labeled C. Directly above it, a blue curve represents the function z = f(x,y). Vertical dashed lines connect the curve C in the xy-plane to the curve z = f(x,y), forming a vertical 'curtain' or 'fence' shape. A single thin vertical slice of this curtain is highlighted and shaded with diagonal purple lines to represent a differential element. This element's height is defined by the function value f(x,y) at a point along the curve, and its width is a small segment of the arc length of curve C. The visualization demonstrates that the line integral represents the area of this vertical surface above the curve.

\[ \sum_{i=1}^{n} \underbrace{f(x(t_i), y(t_i))}_{\text{height}} \underbrace{\Delta s}_{\substack{\text{width} \\ = \text{infinitesimal} \\ \text{arc length}}} \]

Notation:

\[ \int_C f \, ds = \lim_{n \to \infty} \sum_{i=1}^{n} f(x(t_i), y(t_i)) \Delta s \]

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Line integral of \(f\) over Curve \(C: \vec{r}(t) = \langle x(t), y(t) \rangle, a \le t \le b.\)

\[\int_C f \, ds = \lim_{n \to \infty} \sum_{i=1}^{n} f(x(t_i), y(t_i)) \Delta s\]

If we can write \(\Delta s\) as a function of \(t\), we can write the \(\Sigma\) as a definite integral.

Recall: Arc length function \[s(t) = \int_{a}^{t} |r'(u)| \, du\]

From the arc length function, we can derive the relationship between \(s\) and \(t\):

\[\frac{ds}{dt} = |r'(t)|\] \[ds = |r'(t)| \, dt\] \[\Delta s = |r'(t)| \Delta t\]

By substituting these values back into the limit definition, we obtain:

\[\int_C f \, ds = \lim_{n \to \infty} \sum_{i=1}^{n} f(x(t_i), y(t_i)) \Delta s = \lim_{n \to \infty} \sum_{i=1}^{n} f(x(t_i), y(t_i)) |r'(t_i)| \Delta t\]

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Derivation of the Line Integral Formula

\[ \int_{C} f \, ds = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x(t_i), y(t_i)) \Delta s = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \underbrace{(f(x(t_i), y(t_i)) |r'(t_i)|)}_{g(t_i)} \Delta t \] \[ = \int_{a}^{b} f(x(t), y(t)) |r'(t)| \, dt \]
Theorem: \[ \int_{C} f \, ds = \int_{a}^{b} f(r(t)) |r'(t)| \, dt \]

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eg â‘ : Evaluate \(\int_C x \, ds\), \(C\) is Parabola \(y = x^2\) on from \((0,0)\) to \((1,1)\)

A 2D Cartesian coordinate system graph showing a segment of the parabola y = x^2. The x and y axes intersect at the origin (0,0), marked with a green dot. A curve representing the parabola starts at this origin and extends upwards into the first quadrant, ending at the point (1,1), which is also marked with a green dot. This arc represents the path C used for the line integral.
Visual Description: A 2D Cartesian coordinate system graph showing a segment of the parabola y = x^2. The x and y axes intersect at the origin (0,0), marked with a green dot. A curve representing the parabola starts at this origin and extends upwards into the first quadrant, ending at the point (1,1), which is also marked with a green dot. This arc represents the path C used for the line integral.

parametrize: \(\vec{r}(t) = \langle x(t), y(t) \rangle\)

let \(x(t) = t\)
\(y(t) = (x(t))^2 = t^2\)
\(\vec{r}(t) = \langle t, t^2 \rangle, \ 0 \leq t \leq 1\)

\[\int_C x \, ds = \int_0^1 x(t) |\vec{r}'(t)| \, dt\]

\(x(t) = t\)
\(\vec{r}'(t) = \langle 1, 2t \rangle\)
\(|\vec{r}'(t)| = \sqrt{1 + 4t^2}\)

\[\int_C x \, ds = \int_0^1 t \cdot \sqrt{1 + 4t^2} \, dt\]

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\[ \int_C x \, ds = \int_0^1 t \cdot \sqrt{1 + 4t^2} \, dt \] Using u-substitution for the integral: \[ 1 + 4t^2 = u \] \[ 8t \, dt = du \] Updating the limits of integration: \[ t = 0 \implies u = 1 \] \[ t = 1 \implies u = 5 \] Substituting these values into the integral: \[ = \frac{1}{8} \int_{1}^{5} u^{1/2} \, du \] \[ = \frac{1}{8} \cdot \frac{2}{3} u^{3/2} \bigg|_1^5 \] \[ = \frac{1}{12} \left[ 5^{3/2} - 1 \right] \]

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Example 2

Evaluate \(\int_C y e^{x^2} ds\), where \(C\) is the line from \((0,0)\) to \((1,3)\).

A Cartesian coordinate plane graph showing the path of integration C. A solid line segment starts at the origin, labeled (0,0), and ends at a point in the first quadrant labeled (1,3). Both endpoints are marked with solid dots. The x and y axes are clearly drawn as intersecting perpendicular lines.
Visual Description: A Cartesian coordinate plane graph showing the path of integration C. A solid line segment starts at the origin, labeled (0,0), and ends at a point in the first quadrant labeled (1,3). Both endpoints are marked with solid dots. The x and y axes are clearly drawn as intersecting perpendicular lines.

Slope: 3
Equation: \(y = 3x\)

Parametrize: \(\vec{r}(t) = \langle t, 3t \rangle\), \(0 \le t \le 1\)

OR

Vector form of Line:

\[ \vec{r}(t) = \vec{r}_0 + t \vec{v} \]

Where \(\vec{r}_0 = \langle 0,0 \rangle\) is a point on the line and \(\vec{v} = \langle 1,3 \rangle\) is the direction vector.

\[ \vec{r}(t) = \langle 0,0 \rangle + t \langle 1,3 \rangle = t \langle 1,3 \rangle \] \[ \vec{r}'(t) = \langle 1, 3 \rangle \implies |\vec{r}'(t)| = \sqrt{10} \]

The line integral is evaluated as follows:

\[ \int_C y e^{x^2} ds = \int_0^1 y(t) e^{(x(t))^2} |\vec{r}'(t)| dt = \int_0^1 3t e^{t^2} \cdot \sqrt{10} dt \]

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Line Integral Calculation

\[ \int_C y e^{x^2} \, ds = \int_0^1 y(t) e^{(x(t))^2} |r'(t)| \, dt = \int_0^1 3t \cdot e^{t^2} \cdot \sqrt{10} \, dt \]

To solve the integral, we perform a substitution:

Substitution:
\( u = t^2 \)
\( du = 2t \, dt \)
When \( t = 0 \rightsquigarrow u = 0 \)
When \( t = 1 \rightsquigarrow u = 1 \)

Applying the substitution: \[ = \frac{1}{2} \int_0^1 3 \cdot \sqrt{10} \cdot e^u \, du \] \[ = \frac{3\sqrt{10}}{2} \left[ e^1 - e^0 \right] \]

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Line Integrals in 3D

Integral of \(f(x, y, z)\), over curve \(C\) parameterized by \(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\), \(a \le t \le b\)

\[ \int_C f \, ds = \int_a^b f(x(t), y(t), z(t)) |\vec{r}'(t)| \, dt = \int_a^b f(\vec{r}(t)) |\vec{r}'(t)| \, dt \]

Physical Meaning:

\(f = 1 \implies \int_C 1 \, ds = \int_a^b |\vec{r}'(t)| \, dt = \text{Arc Length}\)

Suppose \(f = \rho\) represents density \(\implies \int_C \rho \, ds = \text{Mass of the curve}\)

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Example: Find the mass of a helix

Find the mass of the helix defined by \(\vec{r}(t) = \langle 5 \cos t, 5 \sin t, t \rangle\), for \(0 \le t \le 3\pi\), with a density function \(\rho(x, y, z) = x + z\).

The mass is calculated using the line integral of the density over the curve \(C\):

\[ \text{Mass} = \int_{C} \rho \, ds = \int_{0}^{3\pi} \rho(\vec{r}(t)) |\vec{r}'(t)| \, dt \]

First, we determine the derivative of the position vector \(\vec{r}(t)\):

\[ \vec{r}'(t) = \langle -5 \sin t, 5 \cos t, 1 \rangle \]

Next, we calculate the magnitude of the velocity vector (the speed):

\[ |\vec{r}'(t)| = \sqrt{25 \sin^2 t + 25 \cos^2 t + 1} = \sqrt{26} \]

Now, we substitute the components of \(\vec{r}(t)\) into the density function \(\rho(x, y, z) = x + z\), where \(x = 5 \cos t\) and \(z = t\), and set up the integral:

\[ \text{Mass} = \int_{0}^{3\pi} (5 \cos t + t) \sqrt{26} \, dt \]

Evaluating the integral gives:

\[ = \left( 5 \sin t + \frac{t^2}{2} \right) \sqrt{26} \bigg|_{0}^{3\pi} \] \[ = \frac{9\pi^2}{2} \sqrt{26} \]