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17.2 Line Integrals in a Vector Field

last time: line integral of scalar field \(\int_{C} f(x,y) ds\)

In a vector field, we evaluate / accumulate some component of the vector field along a path \(C\).

The diagram illustrates a path \(C\) (red curve) moving through a vector field \(\vec{F}\) (green arrows). At any point along the path, we consider the interaction between the field and the direction of travel.

A red path C in a 2D coordinate system surrounded by green vector field arrows F.

Along the path, we look at the dot product \(\vec{F} \cdot \vec{T}\), where \(\vec{T}\) is the unit tangent vector of the path.

A diagram showing a tangent vector T along a path C and its dot product with a vector field F.

\(\vec{F} \cdot \vec{T}\) is scalar, so the calculation of

\[ \int_{C} \vec{F} \cdot \vec{T} ds \]

is done just like with \(\int_{C} f(x,y) ds\).

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Look at \(\int_{C} \vec{F} \cdot \vec{T} ds\) more closely

let's parametrize \(C\) as \(\vec{r}(t)\), \(a \le t \le b\)

unit tangent \(\vec{T} = \frac{\vec{r}'(t)}{|\vec{r}'(t)|}\) and we know \(ds = |\vec{r}'(t)| dt\)

sub into \(\int_{C} \vec{F} \cdot \vec{T} ds\) we get

\[ \int_{C} \vec{F} \cdot \frac{\vec{r}'(t)}{|\vec{r}'(t)|} |\vec{r}'(t)| dt = \boxed{\int_{C} \vec{F} \cdot \vec{r}' dt} \]

another equivalent expression: \(\vec{r}' dt = \frac{d\vec{r}}{dt} dt = d\vec{r}\)

so,

\[ \int_{C} \vec{F} \cdot d\vec{r} \]

Common application: work in moving along \(C\) in a force vector field \(\vec{F}\).

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Example: Line Integral of a Vector Field

Given the vector field:

\[ \vec{F} = \langle xy, y-x \rangle \]

Let \( C \) be the line segment from \( (0, 1) \) to \( (2, 4) \).

Calculate the line integral:

\[ \int_C \vec{F} \cdot \vec{T} ds \]

First Step: Parametrize \( C \)

Here, it's a line:

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

Where:

  • \( \vec{v} = \langle 2, 3 \rangle \)
  • \( \vec{r}_0 = \langle 0, 1 \rangle \)
\[ \vec{r}(t) = \langle 0, 1 \rangle + t \langle 2, 3 \rangle = \langle 2t, 1+3t \rangle \quad 0 \le t \le 1 \]

Remember, the parametrization of \( C \) is NOT unique.

So, here:

\[ \vec{r}(t) = \langle \underbrace{2t}_{x}, \underbrace{1+3t}_{y} \rangle \quad 0 \le t \le 1 \]
\[ \int_C \vec{F} \cdot \vec{T} ds = \underbrace{\int_C \vec{F} \cdot \vec{r}' dt}_{\text{seems easiest to use}} = \int_C \vec{F} \cdot d\vec{r} \]

Calculating the components:

\[ \vec{r}' = \langle 2, 3 \rangle, \quad \vec{F} = \langle xy, y-x \rangle = \langle (2t)(1+3t), 1+3t-2t \rangle \]
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Continuing the calculation:

\[ \int_C \vec{F} \cdot \vec{r}' dt = \int_0^1 \langle 6t^2 + 2t, t + 1 \rangle \cdot \langle 2, 3 \rangle dt \]
\[ = \int_0^1 (12t^2 + 4t + 3t + 3) dt \]
\[ = \dots = \boxed{\frac{25}{2}} \]

One Possible Interpretation:

Work done in moving from \( (0, 1) \) to \( (2, 4) \) along a line in a force vector field:

\[ \vec{F} = \langle xy, y-x \rangle \]
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If \( C \) is a closed loop (start and end are the same), then

\[ \int_{C} \vec{F} \cdot \vec{T} ds = \int_{C} \vec{F} \cdot \vec{r}' dt = \int_{C} \vec{F} \cdot d\vec{r} \]

is also called the circulation of \( \vec{F} \) on \( C \).

Example

\( \vec{F} = \langle x, y \rangle \)

\( C \): \( (0,0) \) to \( (1,0) \) along a line segment

then \( (1,0) \) to \( (0,1) \) along another line segment

then \( (0,1) \) to \( (0,0) \) along another line segment

Calculate \( \int_{C} \vec{F} \cdot \vec{T} ds \)

A coordinate graph showing a closed triangular path C with segments C1, C2, and C3 and vector field F arrows.

\( y = 1 - x \)
\( x = 1 - y \)

We need to parametrize the segments separately:

  • \( C_1: \vec{r}(t) = \langle t, 0 \rangle \quad 0 \le t \le 1 \)
  • \( C_2: \vec{r}(t) = \langle 1-t, t \rangle \quad 0 \le t \le 1 \)
  • \( C_3: \vec{r}(t) = \langle 0, 1-t \rangle \quad 0 \le t \le 1 \)
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\( \int_{C} \vec{F} \cdot \vec{r}' dt \) on \( C_1, C_2, C_3 \) then add up

\( C_1 \):

\( \vec{F} = \langle x, y \rangle \)

\( \vec{r}(t) = \langle t, 0 \rangle \quad 0 \le t \le 1 \)

\( \vec{F} = \langle t, 0 \rangle \)

\( \vec{r}' = \langle 1, 0 \rangle \)

\[ \int_{C_1} \vec{F} \cdot \vec{r}' dt = \int_{0}^{1} \langle t, 0 \rangle \cdot \langle 1, 0 \rangle dt = \int_{0}^{1} t dt = \frac{1}{2} \]

\( C_2 \):

\( \vec{F} = \langle x, y \rangle \)

\( \vec{r}(t) = \langle 1-t, t \rangle \quad 0 \le t \le 1 \)

\( \vec{F} = \langle 1-t, t \rangle \)

\( \vec{r}' = \langle -1, 1 \rangle \)

\[ \int_{C_2} \vec{F} \cdot \vec{r}' dt = \int_{0}^{1} \langle 1-t, t \rangle \cdot \langle -1, 1 \rangle dt = \int_{0}^{1} (2t - 1) dt = 0 \]

\( C_3 \):

\( \vec{F} = \langle x, y \rangle \)

\( \vec{r}(t) = \langle 0, 1-t \rangle \quad 0 \le t \le 1 \)

\( \vec{F} = \langle 0, 1-t \rangle \)

\( \vec{r}' = \langle 0, -1 \rangle \)

\[ \int_{C_3} \vec{F} \cdot \vec{r}' dt = \int_{0}^{1} \langle 0, 1-t \rangle \cdot \langle 0, -1 \rangle dt = \int_{0}^{1} (t - 1) dt = -\frac{1}{2} \]
\[ \int_{C} \vec{F} \cdot \vec{r}' dt = \int_{C_1} \vec{F} \cdot \vec{r}' dt + \int_{C_2} \vec{F} \cdot \vec{r}' dt + \int_{C_3} \vec{F} \cdot \vec{r}' dt = 0 \]
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Line Integrals: Differential Form

Another equivalent form of

\[ \int_C \vec{F} \cdot \vec{T} \, ds = \int_C \vec{F} \cdot \vec{r}' \, dt = \int_C \vec{F} \cdot d\vec{r} \]

If \( C : \vec{r}(t) = \langle x(t), y(t) \rangle \) for \( a \le t \le b \), then:

\[ \frac{d\vec{r}}{dt} = \vec{r}' = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle \] \[ \vec{r}' \, dt = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle dt = \langle dx, dy \rangle \]

Let \( \vec{F} = \langle f, g \rangle \), then:

\[ \vec{F} \cdot \vec{r}' \, dt = \langle f, g \rangle \cdot \langle dx, dy \rangle = f \, dx + g \, dy \]

So,

\[ \int_C \vec{F} \cdot \vec{T} \, ds = \int_C f \, dx + g \, dy \]

where \( \vec{F} = \langle f, g \rangle \) and \( \vec{r} = \langle x, y \rangle \)

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Flux Integrals

If we replace \( \vec{T} \) in \( \int_C \vec{F} \cdot \vec{T} \, ds \) with the component of \( \vec{F} \) that is perpendicular to the path instead, \( \vec{F} \cdot \vec{N} \).

\( \vec{N} \) is the unit vector normal to \( C \).

Vector diagram showing tangent vector T and normal vector N perpendicular to a path C.

\( \int_C \vec{F} \cdot \vec{T} \, ds \) becomes \( \int_C \vec{F} \cdot \vec{N} \, ds \) → "flux integral"

\( \vec{F} \cdot \vec{N} \) is "flowing through" \( C \). (This integral accumulates all the things flowing through the shape \( C \)).

Graph showing a curve C with multiple normal vectors F dot N pointing outward across the boundary.

There are two possible \( \vec{N} \): right of \( \vec{T} \) or left of \( \vec{T} \).

Diagram showing a closed circular path C with normal vector N pointing outward.
Diagram showing tangent vector T and two possible normal vectors N to the left and right.

Convention:

  • If \( C \) is closed, choose \( \vec{N} \) such that it points out.
  • If \( C \) is not closed, choose \( \vec{N} \) to the right of \( \vec{T} \).