Lecture 10: Path and Line Integrals

Path Integrals

Graph Description: A Cartesian coordinate system showing the x and y axes. A curve labeled \( y = f(x) \) is plotted above the x-axis. The area between the x-axis and the curve, bounded by vertical dashed lines at \( x = a \) and \( x = b \), is shaded to represent the area under the curve.

The area of the shaded region is given by the integral:

\[ \int_{a}^{b} f(x) dx \]

Question: Instead of walking on the x-axis, suppose we follow some other path \( c(t) \). How can we compute the area under a function above our path?

Definition: Paths in 3-D Space
A path in 3-D space is a function \( c: \mathbb{R} \rightarrow \mathbb{R}^3 \). We can think of the input (\( \mathbb{R} \)) as time and the output (\( \mathbb{R}^3 \)) as position.
Examples of Paths
  1. \( c(t) = (t, 2t^2, t+1) \) where \( 0 \le t \le 1 \).
  2. \( c(t) = (\sin t, \cos t, t) \) where \( 0 \le t \le 2\pi \). This path lies on the cylinder \( x^2 + y^2 = 1 \).
Important Properties

Given \( c(t) \), the tangent vector to the curve at any point is given by the derivative \( c'(t) \).

The unit tangent vector is the tangent vector normalized to a length of 1:

\[ \text{Unit tangent vector} = \frac{c'(t)}{||c'(t)||} \]

Reminder: Given a vector \( \vec{v} = (x, y, z) \), its length is \( ||\vec{v}|| = \sqrt{x^2 + y^2 + z^2} \).

Finding Unit Tangent Vectors

For the previously defined path examples:

Solution for Path 1:
Given \( c(t) = (t, 2t^2, t+1) \), the tangent vector is \( c'(t) = (1, 4t, 1) \).
The length is \( ||c'(t)|| = \sqrt{1^2 + (4t)^2 + 1^2} = \sqrt{2 + 16t^2} \).
The unit tangent vector is: \[ \frac{c'(t)}{||c'(t)||} = \left( \frac{1}{\sqrt{2 + 16t^2}}, \frac{4t}{\sqrt{2 + 16t^2}}, \frac{1}{\sqrt{2 + 16t^2}} \right) \]
Solution for Path 2:
Given \( c(t) = (\sin t, \cos t, t) \), the tangent vector is \( c'(t) = (\cos t, -\sin t, 1) \).
The length is \( ||c'(t)|| = \sqrt{\cos^2 t + (-\sin t)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \).
The unit tangent vector is: \[ \frac{c'(t)}{||c'(t)||} = \left( \frac{\cos t}{\sqrt{2}}, -\frac{\sin t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]

Computing Path Integrals

How do we perform the integral of a function \( f \) along a path \( c \)?

Along the x-axis, the integral is simply \( \int f(x) dx \). If we walk along \( c(t) \), the integral should be \( \int f(c(t)) d(c(t)) \). We know that the differential is \( d(c(t)) = ||c'(t)|| dt \).

Definition: Path Integral
The path integral of a function \( f \) along a path \( c(t) \) is given by: \[ \int_{c} f \, ds = \int f(c(t)) ||c'(t)|| dt \]
Example 1: Path Integral Computation

Given the function \( f(x,y,z) = x^2 + y^2 + z^2 \) and the path \( c(t) = (\sin t, \cos t, t) \) for \( 0 \le t \le 2\pi \), compute the path integral \( \int_c f \, ds \).

Solution:
First, evaluate the function along the path: \[ f(c(t)) = \sin^2 t + \cos^2 t + t^2 = 1 + t^2 \] Next, find the magnitude of the tangent vector: \[ ||c'(t)|| = ||(\cos t, -\sin t, 1)|| = \sqrt{2} \] Now, set up and compute the integral: \[ \int_{0}^{2\pi} (1 + t^2)\sqrt{2} \, dt = \left[ \sqrt{2}t + \frac{\sqrt{2}t^3}{3} \right]_{0}^{2\pi} = 2\sqrt{2}\pi + \frac{8\sqrt{2}\pi^3}{3} \]
Example 2: Path Integral Setup

Given \( f(x,y,z) = x^2 + y^2 + z^2 \) and \( c(t) = (t, 2t^2, t+1) \) for \( 0 \le t \le 1 \), set up the path integral.

Solution:
The path integral is \( \int f(c(t)) ||c'(t)|| dt \). Substituting the components yields: \[ \int_{0}^{1} \left( t^2 + (2t^2)^2 + (t+1)^2 \right) \sqrt{2 + 16t^2} \, dt \]

Line Integrals

Motivation: Work
Recall that work is the product of force and distance: \( W = \vec{F} \cdot \vec{D} = ||\vec{F}|| ||\vec{D}|| \cos \theta \).
Diagram Description: Two vectors originating from the same point. Vector \( \vec{F} \) points horizontally to the right, and vector \( \vec{D} \) points diagonally upwards to the right. The angle between the two vectors is labeled \( \theta \).
Example: Constant Force Work Calculation

Given a force vector \( \vec{F} = (1, 1, 1) = \hat{i} + \hat{j} + \hat{k} \) and a displacement vector \( \vec{D} = (2, -3, 4) = 2\hat{i} - 3\hat{j} + 4\hat{k} \), calculate the work done.

Solution: \[ \text{Work} = \vec{F} \cdot \vec{D} = (1)(2) + (1)(-3) + (1)(4) = 2 - 3 + 4 = 3 \]

We can find the angle between \( \vec{F} \) and \( \vec{D} \) using the dot product formula:

\[ \cos \theta = \frac{\vec{F} \cdot \vec{D}}{||\vec{F}|| ||\vec{D}||} \]

What happens if \( \vec{F} \) and \( \vec{D} \) aren't constant, but depend upon some other variable \( t \)?

Diagram Description: A sketch illustrating the transition from a simple rectangular area (Area = length × height) to a line integral, showing a curved, irregular path \( c(t) \) and a varying function curve drawn above it to represent integrating a changing force along a non-linear path.
Definition: Line Integral of a Vector Field
Suppose we have a force applied to a particle moving along a path \( c(t) \). To compute the work, instead of the usual dot product \( \vec{F} \cdot c(t) \), we consider the integral of the dot product over the path: \[ \int_{c} \vec{F} \cdot dc = \int \vec{F}(c(t)) \cdot c'(t) dt \]

This expression is the line integral of the vector-valued function \( \vec{F} \) along the curve \( c(t) \).

Example 1: Line Integral Computation

Given a force field \( F(x,y,z) = x^2\hat{i} + yz\hat{j} + x\hat{k} = (x^2, yz, x) \) and a path \( c(t) = (t, 2t, 3t) \) for \( 0 \le t \le 1 \). Compute the line integral of \( F \) along \( c(t) \).

Solution:
First, find the derivative of the path: \[ c'(t) = (1, 2, 3) \] Next, evaluate the force field along the path: \[ F(c(t)) = t^2\hat{i} + (2t)(3t)\hat{j} + t\hat{k} = (t^2, 6t^2, t) \] Now, compute the dot product and integrate: \[ \int_{0}^{1} F(c(t)) \cdot c'(t) \, dt = \int_{0}^{1} \left( t^2(1) + 6t^2(2) + t(3) \right) dt \] \[ = \int_{0}^{1} (13t^2 + 3t) \, dt = \left[ \frac{13t^3}{3} + \frac{3t^2}{2} \right]_{0}^{1} = \frac{13}{3} + \frac{3}{2} = \frac{35}{6} \]

Alternative Notation for Line Integrals

When we have a path defined as \( c(t) = (x(t), y(t), z(t)) \), its derivative is \( c'(t) = (x'(t), y'(t), z'(t)) \).

For a vector field \( F = (F_1, F_2, F_3) \), the dot product is:

\[ F \cdot c'(t) = F_1 x'(t) + F_2 y'(t) + F_3 z'(t) \]

The line integral can thus be expanded as:

\[ \int_{c} F \cdot dc = \int F_1 x'(t) dt + \int F_2 y'(t) dt + \int F_3 z'(t) dt \]
Note on Differentials:
Since \( x'(t) dt = dx \), \( y'(t) dt = dy \), and \( z'(t) dt = dz \), we can reformulate the line integral as: \[ \int_{c} F \cdot dc = \int F_1 dx + \int F_2 dy + \int F_3 dz \]

Using this notation, a reformulation of the previous example's question is: Compute \( \int x^2 dx + \int yz dy + \int x dz \) along the curve \( c(t) = (t, 2t, 3t) \) for \( 0 \le t \le 1 \).

Try It Yourself

Compute the line integral:

\[ \int \cos z \, dx + \int e^x \, dy + \int e^y \, dz \]

along the curve \( (1, t, e^t) \) for \( 0 \le t \le 1 \).