Properties of vector line integrals (continued)

 In the last post, we showed the properties of vector line integrals. Let's use them to compute a vector line integral. 

Solution

Let's draw the vector field and represent the domain curve

As seen in the figure, the domain curve is the sum of the four sides of a rectangle: C₁ from (0,0) to (2,0), C₂ from (2,0) to (2,1), C₃ from (2,1) to (0,1), C₄ from (1,0) to (0,0).

The pattern

1. Start at the first vertex of the side, in the counterclockwise order (0,0) → (2,0) → (2,1) → (0,1) → (0,0).
2. Keep one coordinate fixed, since the side is horizontal or vertical.
3. Let the other coordinate change at speed 1, increasing or decreasing depending on the direction of travel.
4. Let t run from 0 to the length of the side, so that t measures distance traveled.

The four sides

Side	From → to	Parametrization	Range of t	r′(t)
C₁ (bottom)	(0,0) → (2,0)	⟨t, 0⟩	0 ≤ t ≤ 2	⟨1, 0⟩
C₂ (right)	(2,0) → (2,1)	⟨2, t⟩	0 ≤ t ≤ 1	⟨0, 1⟩
C₃ (top)	(2,1) → (0,1)	⟨2 − t, 1⟩	0 ≤ t ≤ 2	⟨−1, 0⟩
C₄ (left)	(0,1) → (0,0)	⟨0, 1 − t⟩	0 ≤ t ≤ 1	⟨0, −1⟩

Interpretation

In the table above we find the parametrization for each side r(t), the range of the parameter and r'(t)

• C₁: y stays at 0 and x grows from 0 to 2. At t = 0 you are at (0,0), and at t = 2 you are at (2,0).
• C₂: x stays at 2 and y grows from 0 to 1. The side has length 1, so t stops at 1.
• C₃: y stays at 1, but x must go from 2 down to 0. Writing 2 − t does this, since at t = 0 it gives 2 and at t = 2 it gives 0. Using ⟨t, 1⟩ would trace the same segment in the wrong direction.
• C₄: x stays at 0 and y goes from 1 down to 0, so 1 − t with t from 0 to 1.

According to the 4th property of vector line integral, we have:

Let's calculate each integral separately:

Let's calculate F(r(t)) and substitute r'(t):

 

Let's do the same for the other integrals:

Adding all the 4 integrals, we finally have: 

Practice

Properties of vector line integrals

 Reversing orientation 

We continue with the same example as previously where we consider the work done by the wind in helping a bird fly along a certain path in a definite direction. Let's say that the bird flies in the opposite direction. What happens to the work done by the wind when the bird flies in the opposite direction. Let's compare the 2 works by comparing two vector line integrals.

In the previous example the value of the integral ∫꜀ F·dr where C is a semicircle parameterized by r(t) = <cost, sint> where 0 ≤ t  ≤ ℼ and F = <-y, x> is ℼ. In this example the direction of travel and the wind is illustrated by the following figure:

The direction of travel is the same as the direction of the field.

Let's consider a second example ∫꜀ F·dr where C is a semicircle parameterized by r(t) = <cos (t + ℼ), sint> where 0 ≤ t  ≤ ℼ and F = <-y, x> is ℼ.

The direction of travel and the wind are illustrated in the following figure:

The direction of travel is the opposite of the direction of the field

Let's evaluate this integral:

This is the same integral as previously except the direction of travel reverses. The parameterization starts r(0) = <-1,0> and ends at r(ℼ) = <1,0>. Let's calculate the integral:

The sign of the integral ∫꜀ F·dr changes when the orientation is reversed. Let C be an oriented curve. Let -C be the same curve with the orientation reversed. Based on the previous 2 examples, then:

The reverse orientation of the curve changes the sign of a line integral. This is one of the properties of of the vector line integrals. Other properties of the vector line integrals are included in the following:

 

What is a vector line integral and how to evaluate it?

 In a recent post, we defined a line integral and showed how to evaluate it. We considered a function over its curve domain. We divided the domain curve into small pieces, took a point P inside each piece, evaluated it by the scalar function and multiply f(P) by the length of a piece Δs. We added all the products f(P)Δs, let Δs shrink to zero and took the limit of the sum. The result is the scalar line integral. Now the function considered is a vector field, can we use the same method used for the scalar line integral? Let's make the following considerations. A scalar line integral add up quantities like density or temperature along a curve. A vector field evaluates how much a field pushes in the direction of travel along a curve. We will make this consideration in evaluating the vector Mathematics and Physics entangle. Let's consider a physical problem to calculate the vector line integral. We consider the travel of a bird along a definite path and we want to evaluate how much the wind helps it move along that path.

Let 

be the vector field that represents the force of the wind that moves the bird along the chosen path. This path is represented by a parameterized curve r(t) such that a ≤ t ≤ b. The wind is moving in a positive direction which is the direction given by the curve. The opposite direction would be negative.

 Let P₀, P₁, P₂, ......... Pₙ divide the curve into n pieces. The distance from Pi-1 to Pi is represented by Δsᵢ. Let P*ᵢ be a point  between those the correspond force is F(P*ᵢ). If the distance Δsᵢ is small, when the bird is moving from Pi-1 to Pi it's moving in the direction of the tangent T(P*i). The work done by the force of the wind is given by: F(P*i).(Δsᵢ.T(P*i)). The total work done along C is given by:

Let the arc length of C be arbitrarily small as n→∞, the work done by the wind in moving the bird along its path is represented by:

 

Definition

The vector line integral of a vector field F along a smooth curve is given by:

The unit tangent vector is given by T = r'(t)/𐽕r'(t)𐽕 and ds = 𐽕r'(t)𐽕.dt. Let's calculate F.Tds by substituting T and ds:

Then the vector line integral becomes:

Example

Solution

Let's use the formula: 

Let's calculate F(r(t)) and r'(t):

Let's substitute F(r(t)) and r'(t) in the formula:

Practice

 Find the value of the integral ∫_C F · dr, where C is the semicircle of radius 2 parameterized by r(t) = ⟨2 cos t, 2 sin t⟩, 0 ≤ t ≤ π, and F = ⟨−y, x⟩. 

Scalar line integral: Independence of parameterization

 Objective: Show that the value of a line integral doesn't change if the parameterization of the curve changes. 

In the previous post we calculated the value of a line integral. What if the parameterization of the curve changes? Let's use the same example and change the parameterization of the curve?

Example

  Solution  

Let's apply the formula to evaluate a line integral:

Let's calculate f(r(t)) by substituting x, y and z in the function:

Let's calculate the radical expression:

Let's substitute f(x,y.z), f(r(t)) and the radical expression in the formula:

We obtain the same result as in the previous example. Scalar line integrals are independent of parameterization. 

Practice

What is a scalar line integral and how to evaluate it?

 Objective: Evaluate a scalar line integral

Let's consider a simple integral of the form: 

The domain of integration [a b] is a line segment that can be considered as a curve in the x-axis even though it doesn't have a curvature. Let's say that we want to integrate a function over a domain of integration that is not a straight line. The resulting integral is a line integral. A line integral is a type of integral with a domain of integration that can be a line or a curve. The line integral can be the integral of a multivariable function or the the integral of a vector field over a domain of integration that is a curve. In the case of a multivariable function the line integral is called a scalar line integral. If the function is a vector field the line integral is called a vector line integral.

How to evaluate a scalar line integral?   

We divide the curve C into small pieces of curve C₁, C₂, C₃,......Cₙ by the points P₀, P₁, P₂....Pₙ. Each piece has a length Δs₁, Δs₂, Δs₃.....Δsₙ.  Let choose a point P*ᵢ inside the piece Δsᵢ and evaluate f at P we have f(P). Multiplying f(P*i) by Δsᵢ we have the product f(P*i)Δs. Add all the products and let the arc length of the pieces shrink to zero by taking a limit we have:

Calculating the integral using this formula would be cumbersome. The following theorems provide better ways to evaluate a scalar line integral.

Theorem I

Theorem II

Let f be a continuous function with a domain that includes a smooth curve C with parameterization:

Let's substitute: 

in the previous formula. Then we have a definite expression to calculate the line integral:

Example

Solution

Let's use the formula in theorem II : 

Let's calculate f(r,t) and the radical expression:

Let's substitute the given scalar line integral and these 2 expressions in the formula:

Evaluating the integral on the right we have:

Finally the value of the given integral is:

Practice