Integrals of a Vector Field

Section 7.4 Integrals of a Vector Field

We now study integrals along curves of vector valued functions. If

f

is a function defined on

D \subset R^{N}

R^{N}

we can “attach” the image

f (x)

of every point

x \in D

x .

As a result we get what we call a vector field. There are many physical interpretations of vector fields. For instance

f (x)

is the velocity of a fluid at the point

x,

or the force acting on a particle. In the latter case we talk about a `force field.’ Examples of force fields are the electric, magnetic or gravitational fields. As an example of a vector field consider

f (x, y) := (x y, x^{2} - y^{2})

shown in Figure 7.13.

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Figure 7.13. The vector field

f (x, y) := (x y, x^{2} - y^{2})

R^{2} .

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To motivate the definition of line integrals we consider

f

to be a force field. We want to estimate the work done moving along the curve from one end to the other. Quite obviously it matters in which direction we move. Hence we look at a positively oriented curve as defined at the end of Section 7.1, and assume that a particle travels in the positive direction from one end of the curve to the other. Recall that in physics work is defined by the force times the distance travelled. Only the component in the direction travelled counts. To estimate the work done moving along

C

we partition

C

n

points

P_{0}

P_{n} .

Denote the distance along

C

from

P_{i - 1}

P_{i}

Δ s_{i} .

f

is a continuous vector field, then

f (P)

is almost constant between

P_{i - 1}

and

P_{i}

provided

Δ s_{i}

is sufficiently small. Moreover, the segment of the curve between

P_{i - 1}

and

P_{i}

is almost on the tangent to

C

P_{i - 1} .

The situation is depicted in Figure 7.14.

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Figure 7.14. Work done moving along a small segment of the curve $C .$
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The component of

f (P_{i - 1})

into the direction of the tangent is given by

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f (P_{i - 1}) \cdot τ (P_{i - 1}),

where

τ (P_{i - 1})

denotes the unit tangent vector at

P_{i - 1}

C

as shown in Figure 7.14 we can rewrite this component as a scalar product

f (P_{i - 1}) \cdot τ (P_{i - 1})

(see Remark 1.9). Hence the work done travelling from

P_{i - 1}

P_{i}

is approximately

f (P_{i - 1}) \cdot τ (P_{i - 1}) Δ s_{i} .

The total work moving along the curve is therefore close to

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\sum_{i = 1}^{n} f (P_{i - 1}) \cdot τ (P_{i - 1}) Δ s_{i} .

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The above sum looks like a Riemann sum for the scalar valued function

f \cdot τ

along the curve

C .

Using the theory from the previous section the limit of the sum as

max Δ s_{i}

tends to zero is

\int_{C} f \cdot τ d s .

γ (t),

t \in [a, b],

is a regular parametrisation consistent with the orientation of

C

then the positive unit tangent vector

τ (t)

γ (t)

is given by (7.1). Hence applying Definition 7.10 to

f \cdot τ

we get

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\begin{aligned} \int_{C} f \cdot τ d s & = \int_{a}^{b} f (γ (t)) \cdot \frac{γ^{'} (t)}{‖ γ^{'} (t) ‖} ‖ γ^{'} (t) ‖ d t \\ = \int_{a}^{b} f (γ (t)) \cdot γ^{'} (t) d t . \end{aligned}

Setting

x_{i} = γ_{i} (t)

we can formally write

d x_{i} = γ_{i}^{'} (t) d t .

Hence, if

x = (x_{1}, \dots, x_{N})

and

γ = (γ_{1}, \dots, γ_{N})

as usual, we see that formally

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\begin{aligned} \int_{C} f \cdot τ d s & = \int_{a}^{b} \sum_{i = 1}^{N} f_{i} (γ (t)) γ_{i}^{'} (t) d t \\ = \int_{C} \sum_{i = 1}^{N} f_{i} (x) d x_{i} \\ = \int_{C} f (x) \cdot d x . \end{aligned}

The last expression is obtained by formally taking the scalar product of

f (x)

with

d x := (d x_{1}, \dots, d x_{N}) .

The above motivates the following definition.

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Definition 7.15. Line integral of vector fields.

Let

C

be a smooth positively oriented curve, and

γ (t),

t \in [a, b],

a regular parametrisation of

C

consistent with the orientation of

C .

Then for every continuous vector field

f

C

we define

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\begin{aligned} \int_{C} f \cdot d x & := \int_{C} f \cdot τ d s \\ = \int_{a}^{b} f (γ (t)) \cdot γ^{'} (t) d t \end{aligned}

The integral over a piecewise smooth curve is defined to be the sum over the integrals over the smooth parts of the curve.

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As we integrate the tangential component along the curve we also talk about the circulation of

f

along the curve

C

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Insight 7.16. Physical interpretation.

f

is a force field, then, as shown before,

\int_{C} f \cdot d x

is the work done by a particle moving along

C .

f

describes the motion of a medium (fluid or gas) then the same arguments as above show that

\int_{C} f \cdot d x

is the net flow along the curve

C .

C

is a closed curve we talk about the circulation of the field along

C .

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In the special case of a function of three variables,

x, y, z

and a vector field

f = (P, Q, R)

the above reads

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\begin{aligned} \int_{C} P d x & + Q d y + R d z \\ = \int_{a}^{b} P x^{'} (t) + Q y^{'} (t) + R z^{'} (t) d t \end{aligned}

(x (t), y (t), z (t)),

t \in [a, b],

is a parametrisation of

C

consistent with the orientation of

C .

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Remark 7.17.

We saw that it matters how the curve is oriented. As the opposite orientation of a curve

C

is given by reversing the sign of the unit tangent vectors we have that

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\int_{- C} f \cdot d x = - \int_{C} f \cdot d x .

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Remark 7.18.

C

is a closed curve then we often write

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\oint_{C} f \cdot d x := \int_{C} f \cdot d x .

The circle is really redundant but often useful to remind us that

C

is a closed curve.

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Example 7.19.

Consider the curve

C

shown in Figure 7.20. The arc is a half circle. Compute the line integral

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\int_{C} 2 x y d x + x^{2} d y .

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Figure 7.20. A piecewise smooth positively oriented closed curve.

Answer.

The integral is

0 .

Solution.

We compute the integral over the smooth parts,

C_{1}

and

C_{2},

C

separately. We first look at

C_{1} .

The function

γ (t) := (t, t),

t \in [- 1, 1]

is a parametrisation consistent with the orientation of

C_{1} .

Hence

γ^{'} (t) = (1, 1)

and therefore

\begin{aligned} \int_{C_{1}} 2 x y d x + x^{2} d y & = \int_{- 1}^{1} 2 γ_{1} (t) γ_{2} (t) γ_{1}^{'} (t) + (γ_{1} (t))^{2} γ_{2}^{'} (t) d t \\ = \int_{- 1}^{1} 2 t^{2} + t^{2} d t \\ = \int_{- 1}^{1} 3 t^{2} d t = [t^{3}]_{- 1}^{1} = 1 - (- 1) = 2 . \end{aligned}

Next we compute the integral over

C_{2} .

A regular parametrisation, consistent with the orientation, is given by

γ (t) := \sqrt{2} (\cos t, \sin t),

t \in [π / 4, 5 π / 4] .

We have

γ^{'} (t) = \sqrt{2} (- \sin t, \cos t),

and so

\begin{aligned} \int_{C_{2}} 2 x y d x + x^{2} d y & = \int_{π / 4}^{5 π / 4} 2 γ_{1} (t) γ_{2} (t) γ_{1}^{'} (t) + (γ_{1} (t))^{2} γ_{2}^{'} (t) d t \\ = \int_{π / 4}^{5 π / 4} 2 (\sqrt{2} \cos t) (\sqrt{2} \sin t) (- \sqrt{2} \sin t) + 2 \cos^{2} t \sqrt{2} \cos t d t \\ = \int_{π / 4}^{5 π / 4} - 4 \sqrt{2} \sin^{2} t \cos t + 2 \sqrt{2} \cos t (1 - \sin^{2} t) d t \\ = \int_{π / 4}^{5 π / 4} - 6 \sqrt{2} \sin^{2} t \cos t + 2 \sqrt{2} \cos t d t \\ = [- 2 \sqrt{2} \sin^{3} t + 2 \sqrt{2} \sin t]_{π / 4}^{5 π / 4} = - 2 . \end{aligned}

Hence we have

\begin{aligned} \int_{C} 2 x y d x + x^{2} d y & = \int_{C_{1}} 2 x y d x + x^{2} d y + \int_{C_{2}} 2 x y d x + x^{2} d y \\ = 2 - 2 = 0 . \end{aligned}

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