Using Parametrizations to Calculate Line Integrals

Section 4.3 Using Parametrizations to Calculate Line Integrals

Motivating Questions

How can we use a parametrization of an oriented curve $C$ to calculate $\int_{C} F \cdot d r ?$
How does the parametrization chosen for an oriented curve $C$ alter the value of the line integral $\int_{C} F \cdot d r ?$
What can be said about the line integral of a vector field along two different oriented curves when the curves have the same starting point and same ending point?

We begin this section by taking a look at how we can calculate a line integral of a vector field along some line segments and use this calculation as inspiration to see how treating oriented curves as vector-valued functions will allow us to quickly turn a line integral of a vector field into a single variable integral.

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Preview Activity 4.3.1.

🔗

Let

F = ⟨ x y, y^{2} ⟩,

let

C_{1}

be the line segment from

(1, 1)

(4, 1),

let

C_{2}

be the line segment from

(4, 1)

(4, 3),

and let

C_{3}

be the line segment from

(1, 1)

(4, 3) .

Also let

C = C_{1} + C_{2} .

This vector field and the curves are shown in Figure 4.3.1.

described in detail following the image — Figure 4.3.1. A vector field $F$ and three oriented curves

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(a)

🔗

Every point along

C_{1}

has

y = 1 .

Therefore, along

C_{1},

the vector field

F

can be viewed purely as a function of

x .

In particular, along

C_{1},

we have

F (x, 1) = ⟨ x, 1 ⟩ .

Since every point along

C_{2}

has the same

x

-value, what (in terms of

y

only) is

F

along

C_{2} ?

Answer.

F = ⟨ 4 y, y^{2} ⟩

along

C_{2} .

Solution.

F = ⟨ 4 y, y^{2} ⟩

along

C_{2} .

🔗

(b)

🔗

Recall that

d r \approx Δ r,

and along

C_{1},

we have that

Δ r = Δ x i \approx d x i .

Thus,

d r = ⟨ d x, 0 ⟩ .

We know that along

C_{1},

F = ⟨ x, 1 ⟩ .

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(i)

🔗

Write

F \cdot d r

along

C_{1}

without using a dot product.

Answer.

x d x

Solution.

Along the path

C_{1},

F \cdot d r = ⟨ x, 1 ⟩ \cdot ⟨ d x, 0 ⟩ = x d x .

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(ii)

🔗

What interval of

x

-values describes

C_{1} ?

Answer.

1 \leq x \leq 4

Solution.

Since

C_{1}

is the line segment from

(1, 1)

(4, 1),

we have that

1 \leq x \leq 4 .

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(iii)

🔗

Write

\int_{C_{1}} F \cdot d r

as an integral of the form

\int_{a}^{b} f (x) d x

and evaluate the integral.

Answer.

\frac{15}{2}

Solution.

From the previous parts,

\int_{C_{1}} F \cdot d r = \int_{1}^{4} x d x = 8 - \frac{1}{2} = \frac{15}{2} .

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(c)

🔗

Use an analogous approach to write

\int_{C_{2}} F \cdot d r

as an integral of the form

\int_{c}^{d} g (y) d y

and evaluate the integral.

Answer.

\frac{26}{3}

Solution.

Along

C_{2},

d r = ⟨ 0, d y ⟩

and

1 \leq y \leq 3 .

Thus, we have

\int_{C_{2}} F \cdot d r = \int_{1}^{3} y^{2} d y = 9 - \frac{1}{3} = \frac{26}{3} .

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(d)

🔗

Use the previous parts and a property of line integrals to calculate

\int_{C} F \cdot d r

without having to evaluate any additional integrals.

Answer.

\frac{97}{6}

Solution.

\int_{C} F \cdot d r = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r = \frac{15}{2} + \frac{26}{3} = \frac{97}{6}

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(e)

🔗

Evaluating

\int_{C_{3}} F \cdot d r

takes more work at this stage, so let’s break the process into smaller pieces.

🔗

(i)

🔗

Find the slope-intercept (

y = m x + b

) form of the equation of the line containing the line segment

C_{3} .

Answer.

y = \frac{2}{3} x + \frac{1}{3}

Solution.

The slope is

(3 - 1) / (4 - 1) = 2 / 3 .

Using this with either point gives

b = 1 / 3,

so the slope-intercept form of the line’s equation is

y = \frac{2}{3} x + \frac{1}{3} .

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(ii)

🔗

Just as we noticed that along

C_{1}

we always had

y = 1,

we now know how to express

y

in terms of

x

for all points along

C_{3} .

Use this to to express

F

as a vector purely in terms of

x

for the points along

C_{3} .

Answer.

F = ⟨ \frac{2}{3} x^{2} + \frac{1}{3} x, (\frac{2}{3} x + \frac{1}{3})^{2} ⟩

along

C_{3} .

Solution.

We know that along

C_{3},

y = \frac{2}{3} x + \frac{1}{3} .

Thus,

F (x, y) = F (x, \frac{2}{3} x + \frac{1}{3}) = ⟨ \frac{2}{3} x^{2} + \frac{1}{3} x, (\frac{2}{3} x + \frac{1}{3})^{2} ⟩

along

C_{3} .

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(iii)

🔗

We often think of the slope of a line as being

Δ y / Δ x .

Use this fact and the slope of the line containing

C_{3}

to express

Δ y

as a multiple of

Δ x .

Answer.

Δ y = \frac{2}{3} Δ x

Solution.

Since the line’s slope is

2 / 3,

we have

Δ y / Δ x = 2 / 3,

which implies that

Δ y = \frac{2}{3} Δ x

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(iv)

🔗

We may view

Δ r

⟨ Δ x, Δ y ⟩ .

Since

Δ x \approx d x

and

Δ y \approx d y,

write

d r

as a vector in terms of

d x .

You should be able to factor out

d x

from your result and write your answer in the form

⟨ a_{1}, a_{2} ⟩ d x .

Answer.

d r = ⟨ 1, \frac{2}{3} ⟩ d x

Solution.

d r = ⟨ d x, \frac{2}{3} d x ⟩ = ⟨ 1, \frac{2}{3} ⟩ d x

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(v)

🔗

Use the range of

x

-values covered by the line segment

C_{3}

to write

\int_{C_{3}} F \cdot d r

as a single-variable integral of the form

\int_{a}^{b} f (x) d x

and evaluate the integral.

Answer.

\frac{151}{6}

Solution.

From the coordinates of its endpoints, we know that

C_{3}

has

1 \leq x \leq 4 .

Combining the previous parts, we obtain

\int_{C_{3}} F \cdot d r = \int_{1}^{4} \frac{2}{3} x^{2} + \frac{1}{3} x + \frac{2}{3} (\frac{2}{3} x + \frac{1}{3})^{2} d x = \frac{151}{6} .

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(f)

🔗

Notice that

C

and

C_{3}

both start at

(1, 1)

and end at

(4, 3) .

How do the values of

\int_{C} F \cdot d r

and

\int_{C_{3}} F \cdot d r

compare?

Answer.

Different values.

Solution.

The results of the calculations for

\int_{C} F \cdot d r

and

\int_{C_{3}} F \cdot d r

are different.

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(g)

🔗

F

a gradient vector field? Why or why not?

Hint.

What would Clairaut’s Theorem tell you about a potential function

f

such that

F = \nabla f ?

Answer.

No.

Solution.

The vector field

F

is not a gradient vector field because there is no scalar-valued function

f

of two variables such that

\frac{\partial f}{\partial x} = x y

and

\frac{\partial f}{\partial y} = y^{2} .

According to Clairaut’s Theorem, if such a function

f

existed, then

\frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y}),

but

\frac{\partial}{\partial y} (x y) = y \neq \frac{\partial}{\partial x} (y^{2}) = 2 y .

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Subsection 4.3.1 Parametrizations in the Definition of $\int_{C} F \cdot d r$

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Preview Activity 4.3.1 has shown us that it is possible to evaluate line integrals without needing to resort to working with Riemann sums directly. However, the approaches taken there seem rather cumbersome to use for oriented curves that are not line segments. It was not critical that the paths in Preview Activity 4.3.1 were straight lines, but rather that the paths had a description where both

x

and

y

could be expressed in terms of one variable. Fortunately, parametrizing the oriented curve along which a line integral is calculated provides a powerful tool for evaluating line integrals.

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Suppose that

C

is an oriented curve traced out by the vector-valued function

r (t)

for

a \leq t \leq b,

and let

F

be a continuous vector field defined on a region containing

C .

We can divide the interval

[a, b]

into

n

sub-intervals, each of length

Δ t = (b - a) / n,

by letting

t_{i} = a + i Δ t

for

i = 0, 1, \dots, n .

This subdivision of

[a, b]

then can be used to break

C

up into

n

pieces by letting

Δ r_{i} = r (t_{i + 1}) - r (t_{i})

for

i = 0, 1, \dots, n - 1

as we did in Figure 4.2.2. Now notice that

Δ r_{i} = r (t_{i + 1}) - r (t_{i}) = r (t_{i} + Δ t) - r (t_{i}) = (\frac{r (t_{i} + Δ t) - r (t_{i})}{Δ t}) Δ t .

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Looking at the definition of the line integral and substituting this new expression for

Δ r_{i}

gives

\int_{C} F \cdot d r = lim_{| Δ r_{i} | \to 0} \sum_{i = 0}^{n - 1} F (r_{i}) \cdot Δ r_{i} = lim_{| Δ r_{i} | \to 0} \sum_{i = 0}^{n - 1} F (r_{i}) \cdot (\frac{r (t_{i} + Δ t) - r (t_{i})}{Δ t}) Δ t .

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When evaluating the limit as

Δ t \to 0,

the expression in the parentheses will is

r^{'} (t_{i}) .

We then have a Riemann sum that reduces the evaluation of a line integral of a vector-valued function along an oriented curve to a definite integral of a function of one variable. Note that after evaluating the dot product,

F (r (t)) \cdot r^{'} (t)

is (scalar) function of

t .

We restate this result below for easy reference.

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Theorem 4.3.2.

🔗

Let

C

be a smooth, oriented curve traced out by the vector-valued function

r (t)

for

a \leq t \leq b .

F

is a continuous vector field defined near

C,

then

\int_{C} F \cdot d r = \int_{a}^{b} F (r (t)) \cdot r^{'} (t) d t .

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Example 4.3.3.

🔗

Let

F (x, y) = x i + y^{2} j

and let

C

be the quarter of the circle of radius

3

from

(0, 3)

(3, 0) .

This vector field and curve are shown in Figure 4.3.4. By properties of line integrals, we know that

\int_{C} F \cdot d r = - \int_{- C} F \cdot d r,

and we will use this property since

- C

is the usual clockwise orientation of a circle, meaning we can parametrize

- C

r (t) = ⟨ 3 \cos (t), 3 \sin (t) ⟩

for

0 \leq t \leq π / 2 .

🔗

To evaluate

\int_{- C} F \cdot d r

using this parametrization, we need to note that

F (r (t)) = ⟨ 3 \cos (t), 9 \sin^{2} (t) ⟩ and r^{'} (t) = ⟨ - 3 \sin (t), 3 \cos (t) ⟩ .

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Thus, we have

\begin{aligned} \int_{C} F \cdot d r & = - \int_{- C} F \cdot d r = - \int_{0}^{π / 2} ⟨ 3 \cos (t), 9 \sin^{2} (t) ⟩ \cdot ⟨ - 3 \sin (t), 3 \cos (t) ⟩ d t \\ = - \int_{0}^{π / 2} (- 9 \sin (t) \cos (t) + 27 \sin^{2} (t) \cos (t)) d t where u = \sin (t) \\ = - \int_{0}^{1} (- 9 u + 27 u^{2}) d u = - {[- \frac{9}{2} u^{2} + 9 u^{3}]}_{0}^{1} = - (- \frac{9}{2} + 9) = - \frac{9}{2} . \end{aligned}

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As the example above shows, Theorem 4.3.2 allows us to reduce the problem of calculating a line integral of a vector-valued function along an oriented curve to one of finding a suitable parametrization for the curve. Once we have such a parametrization, evaluating the line integral becomes evaluating a single-variable integral, which is something you have done many times before. The example also illustrates that using the properties of line integrals can allow us to use a more “natural” parametrization. You may find it interesting to use the parametrization

⟨ 3 \sin (t), 3 \cos (t) ⟩

for

0 \leq t \leq π / 2

to evaluate the line integral. Do you get the same result?

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Activity 4.3.2.

🔗

(a)

🔗

Find the work done by the vector field

F (x, y, z) = 6 x^{2} z i + 3 y^{2} j + x k

on a particle that moves from the point

(3, 0, 0)

to the point

(3, 0, 6 π)

along the helix given by

r (t) = ⟨ 3 \cos (t), 3 \sin (t), t ⟩ .

Answer.

324 π

Solution.

r (t) = ⟨ 3 \cos (t), 3 \sin (t), t ⟩

for

0 \leq t \leq 6 π,

r^{'} (t) = ⟨ - 3 \sin (t), 3 \cos (t), 1 ⟩,

F (t) = ⟨ 54 t (\cos (t))^{2}, 27 (s \sin (t))^{2}, 3 \cos (t) ⟩

and

\int_{C} F \cdot d r = \int_{0}^{6 π} - 162 t (\cos (t))^{2} \sin (t) + 81 (\sin (t))^{2} \cos (t) + 3 \cos (t) d t = 324 π

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(b)

🔗

Let

F (x, y) = ⟨ 0, x ⟩ .

Let

C

be the closed curve consisting of the top half of the circle of radius

2

centered at the origin and the portion of the

x

-axis from

(2, 0)

(- 2, 0),

oriented clockwise. Find the circulation of

F

around

C .

Answer.

- 2 π

Solution.

Let

C_{1}

be given by

r (t) = ⟨ 2 \cos (- t), 2 \sin (- t) ⟩

where

- π \leq t \leq 0 .

This means that

F (t) = ⟨ 0, 2 \cos (- t) ⟩

and

\int_{C_{1}} F \cdot d r = - 4 \int_{- π}^{0} (\cos (t))^{2} d t = - 2 π .

Let

C_{2}

be given by

r (t) = ⟨ t, 0 ⟩

where

- 2 \leq t \leq 2 .

This means that

F (t) = ⟨ 0, t ⟩

and

\int_{C_{2}} F \cdot d r = \int_{- 2}^{2} 0 d t = 0 .

Thus the line integral over

C_{1}

then

C_{2}

will give a result of

- 2 π .

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The next activity has you investigate the line integrals of a fixed vector field along different curves. This will be an important, recurring theme as we study a variety of different integrals and vector fields in this chapter.

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Activity 4.3.3.

🔗

Let

F (x, y) = ⟨ y^{2}, 2 x y + 3 ⟩ .

🔗

(a)

🔗

Let

C_{1}

be the portion of the graph of

y = 2 x^{3} + 3 x^{2} - 12 x - 15

from

(- 2, 5)

(3, 30) .

Calculate

\int_{C_{1}} F \cdot d r .

Answer.

2825

Solution.

F = ⟨ y^{2}, 2 x y + 3 ⟩

and the curve

C_{1}

is traced out by

r (t) = ⟨ t, 2 t^{3} + 3 t^{2} - 12 t - 15 ⟩

for

- 2 \leq t \leq 3,

then

\int_{C_{1}} F \cdot d r = \int_{- 2}^{3} 28 t^{6} + 72 t^{5} - 195 t^{4} - 528 t^{3} + 180 t^{2} + 738 t + 189 d t = 2825

🔗

(b)

🔗

Let

C_{2}

be the line segment from

(- 2, 5)

(3, 30) .

Calculate

\int_{C_{2}} F \cdot d r .

Answer.

2825

Solution.

In this part, the curve

C_{2}

is traced out by

r (t) = ⟨ - 2 + 5 t, 5 + 25 t ⟩

for

0 \leq t \leq 1 .

\int_{C_{2}} F \cdot d r = \int_{0}^{1} ⟨ (5 + 25 t)^{2}, 2 (- 2 + 5 t) (5 + 25 t) + 3 ⟩ \cdot ⟨ 5, 25 ⟩ d t = 2825

🔗

(c)

🔗

Let

C_{3}

be the circle of radius

3

centered at the origin, oriented counterclockwise. Calculate

\oint_{C_{3}} F \cdot d r .

Answer.

0

Solution.

In this part, the curve

C_{3}

is traced out by

r (t) = ⟨ 3 \cos (t), 3 \sin (t) ⟩

0 \leq t \leq 2 π .

\int_{C_{3}} F \cdot d r = \int_{0}^{2 π} ⟨ 9 (\sin (t))^{2}, 2 \cos (t) \sin (t) + 3 ⟩ \cdot ⟨ - 3 \sin (t), 3 \cos (t) ⟩ d t = 0

🔗

(d)

🔗

To connect the previous parts of this activity, use a graphing utility to plot the curves

C_{1}

and

C_{2}

on the same axes.

🔗

(i)

🔗

What type of curve is

C_{1} - C_{2} ?

Answer.

Closed

Solution.

Since the curves have the same endpoints but

- C_{2}

is oriented oppositely,

C_{1} - C_{2}

forms a closed curve.

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(ii)

🔗

What is the value of

\oint_{C_{1} - C_{2}} F \cdot d r ?

Answer.

0

Solution.

Since

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r,

we have that

\oint_{C_{1} - C_{2}} F \cdot d r = \int_{C_{1}} F \cdot d r - \int_{C_{2}} F \cdot d r = 0 .

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(iii)

🔗

What does your answer to part c allow you to say about the value of the line integral of

F

along the top half of

C_{3}

compared to the line integral of

F

from

(3, 0)

(- 3, 0)

along the bottom half of the circle of radius

3

centered at the origin?

Answer.

They are equal.

Solution.

If we reverse the orientation of the bottom half, these curves combine to form

C_{3} .

Thus, the sum of the two line integrals is

0,

and so when changing the orientation of the bottom portion of the circle, we get that the line integrals are equal.

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Subsection 4.3.2 Alternative Notation for Line Integrals

🔗

In contexts where the fact that the quantity we are measuring via a line integral is best measured via a dot product (such as calculating work), the notation we have used thus far for line integrals is fairly common. However, sometimes the vector field is such that the units on

x,

y,

and

z

are not distances. In this case, a dot product may not have a physical meaning, and an alternative notation using differentials can be common. Specifically, if

F (x, y, z) = F_{1} (x, y, z) i + F_{2} (x, y, z) j + F_{3} (x, y, z) k,

then

\int_{C} F \cdot d r = \int_{C} ⟨ F_{1}, F_{2}, F_{3} ⟩ \cdot ⟨ d x, d y, d z ⟩ = \int_{C} F_{1} d x + F_{2} d y + F_{3} d z .

🔗

A line integral in the form of

\int_{C} F_{1} d x + F_{2} d y + F_{3} d z

is called the differential form of a line integral.

🔗

(If

F

is a vector field in

R^{2},

the

F_{3} d z

term is omitted.) As a concrete example, if

F (x, y, z) = ⟨ x^{2} y, z^{3}, x \cos (z) ⟩

and

C

is some oriented curve in

R^{3},

then

\int_{C} F \cdot d r = \int_{C} x^{2} y d x + z^{3} d y + x \cos (z) d z .

🔗

It is important to recognize that the integral on the right-hand side is still a line integral and must be evaluated using techniques for evaluating line integrals. We cannot simply try to treat it as if it were a definite integral of a function of one variable. Because the notation

\int_{C} F \cdot d r

provides a reminder that this is a line integral and not a definite integral of the types calculated earlier in your study of calculus, we will only use the vector notation for line integrals in the body of the text. However, some exercises may require use of the differential form.

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Subsection 4.3.3 Independence of Parametrization for a Fixed Curve

🔗

Up to this point, we have chosen whatever parametrization of an oriented curve

C

came to mind, and our argument for how we can use parametrizations to calculate line integrals did not depend on the specific choice of parametrization. However, it is not immediately obvious that different parametrizations don’t result in different values of the line integral. Our next example explores this question.

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Example 4.3.5.

🔗

Let

F = x i .

We consider two different oriented curves from

(0, 1)

(3, 3) .

The first oriented curve

C

travels horiznotally to

(3, 1)

and then proceeds vertically to

(3, 3) .

The second oriented curve

C_{3}

is the line segment from

(0, 1)

(3, 3) .

Notice that, as depicted in Figure 4.3.6, we can break

C

up into two oriented curves

C_{1}

(the horiznotal portion) and

C_{2}

(the vertical portion) so that

C = C_{1} + C_{2} .

🔗

We first note that since

F

is orthogonal to

C_{2},

\int_{C_{2}} F \cdot d r = 0;

therefore

\int_{C} F \cdot d r = \int_{C_{1}} F \cdot d r .

We can parametrize

C_{1}

t i + j

for

0 \leq t \leq 3

(

t

is treated like the coordinate

x

), which leads to

\int_{C_{1}} F \cdot d r = \int_{0}^{3} ⟨ t, 0 ⟩ \cdot ⟨ 1, 0 ⟩ d t = \int_{0}^{3} t d t = \frac{9}{2} .

🔗

Thus,

\int_{C} F \cdot d r = 9 / 2 .

🔗

Now we look at

\int_{C_{3}} F \cdot d r,

but we parametrize

C_{3}

in a nonstandard way by letting

r (t) = ⟨ 3 \sin (t), 1 + 2 \sin (t) ⟩

for

0 \leq t \leq \frac{π}{2} .

(You should use a graphing utility to plot this parametrization to help convince yourself that it really does give

C_{3} .

) This gives

r^{'} (t) = ⟨ 3 \cos (t), 2 \cos (t) ⟩,

and

\int_{C_{3}} F \cdot d r = \int_{0}^{π / 2} ⟨ 3 \sin (t), 0 ⟩ \cdot ⟨ 3 \cos (t), 2 \cos (t) ⟩ d t = \int_{0}^{π / 2} 9 \sin (t) \cos (t) d t = \frac{9}{2}

🔗

In the next activity, you are asked to consider the more typical parametrization of

C_{3}

and verify that using it gives the same value for the line integral.

🔗

It’s also worth observing here that

\int_{C} F \cdot d r = \int_{C_{3}} F \cdot d r,

so at least two (very different) paths from

(0, 1)

(3, 3)

give the same value of the line integral here. The next section will further investigate when line integrals over different paths (with the same initial point and final point) will evaluate to the same value.

🔗

As promised, the final activity of this section asks you to look at another parametrization of the curve

C_{3}

from the previous example. It also asks you to look at two different oriented curves between a pair of points, similarly to what you did in Activity 4.3.3.

🔗

Activity 4.3.4.

🔗

(a)

🔗

The typical parametrization of the line segment from

(0, 1)

(3, 3)

(the oriented curve

C_{3}

in Example 4.3.5) is

r (t) = ⟨ 3 t, 1 + 2 t ⟩

where

0 \leq t \leq 1 .

Use this parametrization to calculate

\int_{C_{3}} F \cdot d r

for the vector field

F = x i

and compare your answer to the result of Example 4.3.5.

Answer.

\frac{9}{2}

Solution.

Using the parametrization given the line integral becomes

\int_{0}^{1} ⟨ 3 t, 0 ⟩ \cdot ⟨ 3, 2 ⟩ d t = \int_{0}^{1} 9 t d t = \frac{9}{2}

This is the same result as in Example 4.3.5.

🔗

(b)

🔗

Calculate

\int_{C} ⟨ (3 x y + e^{z}), x^{2}, (4 z + x e^{z}) ⟩ \cdot d r

where

C

is the oriented curve consisting of the line segment from the origin to

(1, 1, 1)

followed by the line segment from

(1, 1, 1)

(0, 0, 2) .

Answer.

2 e^{2} - 2 e + 8

Solution.

Let

r_{1} (t) = ⟨ t, t, t ⟩

where

0 \leq t \leq 1

and

r_{2} (t) = ⟨ 1 - t, 1 - t, 1 + t ⟩

where

0 \leq t \leq 1 .

Our line integral can be computed with

\int_{C_{1}} F \cdot d r = \int_{0}^{1} ⟨ 3 t^{2} + e^{t}, t^{2}, 4 t + t e^{t} ⟩ \cdot ⟨ 1, 1, 1 ⟩ d t = e + \frac{10}{3}

\int_{C_{2}} F \cdot d r = \int_{0}^{1} ⟨ 3 (1 - t)^{2} + e^{1 - t}, (1 - t)^{2}, 4 (1 + t) + (1 - t) e^{1 + t} ⟩ \cdot ⟨ - 1, - 1, 1 ⟩ d t = 2 e^{2} - 3 e + \frac{14}{3}

Thus the line integral over

C_{1}

then

C_{2}

2 e^{2} - 2 e + 8

🔗

(c)

🔗

Calculate

\int_{C^{'}} ⟨ 3 x y + e^{z}, x^{2}, 4 z + x e^{z} ⟩ \cdot d r

where

C^{'}

is the line segment from

(0, 0, 0)

(0, 0, 2) .

Answer.

16

Solution.

The line segment from

(0, 0, 0)

(0, 0, 2)

can be given by

r (t) = ⟨ 0, 0, t ⟩

with

0 \leq t \leq 2 .

Note that

d r (t) = ⟨ 0, 0, 1 ⟩ d t,

\int_{C^{'}} F \cdot d r = \int_{0}^{2} 4 t d t = 16

🔗

(d)

🔗

Is the vector field you considered in the previous two parts a gradient vector field? Why or why not? How does this compare to the vector field

F

of Activity 4.3.3?

Answer.

Solution.

No. One way to observe this would be to check the partial derivative with respect to

y

of the

i

component and compare it to the partial derivative with respect to

x

of the

j

component. Since they are not equal, Clairaut’s Theorem tells us that this vector field is not a gradient vector field.

🔗

Although we have not given a proof or even an intuitive argument, the phenomenon you observed in part a of Activity 4.3.4 is not particular to this curve or this vector field. The value of

\int_{C} F \cdot d r

does not depend on the parametrization of

C

used to calculate the line integral when using Theorem 4.3.2. We will take this for granted in the remainder of the chapter. The next section will engage in a serious exploration of when a vector field has the property that the line integral from a point

P

to a point

Q

does not depend not he choice of oriented path between them.

🔗

Subsection 4.3.4 Summary

🔗

Line integrals of vector fields along oriented curves can be evaluated by parametrizing the curve in terms of $t$ and then calculating the integral of $F (r (t)) \cdot r^{'} (t)$ on the interval $[a, b] .$
The parametrization chosen for an oriented curve $C$ when calculating the line integral $\int_{C} F \cdot d r$ using the formula $\int_{a}^{b} F (r (t)) \cdot r^{'} (t) d t$ does not impact the value of the line integral.
If $C_{1}$ and $C_{2}$ are different paths from $P$ to $Q,$ it is possible for $\int_{C_{1}} F \cdot d r$ to have a different value to $\int_{C_{2}} F \cdot d r .$

🔗

Compute

\int_{C_{1}} F \cdot d r

when

F = ⟨ x^{2}, x y ⟩

and

C_{1}

is the line segment from

(0, 0)

(2, 2) .

🔗

(b)

🔗

Compute

\int_{C_{2}} F \cdot d r

when

F = ⟨ x^{2}, x y ⟩

and

C_{2}

is the line segment from

(2, 2)

(0, 0) .

🔗

2.

🔗

If the wind in a region of space is given by

F = ⟨ y + z, z - x, - z ⟩

and a helicopter flies along the path given by

r (t) = ⟨ 10 \sin (t), 10 \cos (t), (10 - t)^{2} ⟩

0 \leq t \leq 4 π .

Calculate the work done by the wind on the helicopter.

Hint.

Set up your integral carefully and then use either integration by parts or an algebraic solver to compute the definite integral.

🔗

3.

🔗

Let

C_{3}

be the circle of radius

7

centered at the origin traveled counterclockwise. Compute

\int_{C_{3}} ⟨ M, N ⟩ \cdot d r

when:

🔗

(a)

🔗

⟨ M, N ⟩ = ⟨ x, y ⟩

🔗

(b)

🔗

⟨ M, N ⟩ = ⟨ - y, x ⟩

🔗

(c)

🔗

⟨ M, N ⟩ = ⟨ 3, x ⟩

🔗

4.

🔗

Let

C_{4}

be the curve given by traveling along the path given by

y = x^{3} - x

on the surface given by

z = x y

x

goes from

- 1

2 .

What is the work done by

⟨ x, z, x + y ⟩ ?

Hint.

Parametrize y in terms of x first, then use that relationship to give z as a function x.

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Section 4.3 Using Parametrizations to Calculate Line Integrals

Motivating Questions

Preview Activity 4.3.1.

(a)

(b)

(i)

(ii)

(iii)

(c)

(d)

(e)

(i)

(ii)

(iii)

(iv)

(v)

(f)

(g)

Subsection 4.3.1 Parametrizations in the Definition of ∫CF⋅dr

Theorem 4.3.2.

Example 4.3.3.

Activity 4.3.2.

(a)

(b)

Activity 4.3.3.

(a)

(b)

(c)

(d)

(i)

(ii)

(iii)

Subsection 4.3.2 Alternative Notation for Line Integrals

Subsection 4.3.3 Independence of Parametrization for a Fixed Curve

Example 4.3.5.

Activity 4.3.4.

(a)

(b)

(c)

(d)

Subsection 4.3.4 Summary

Exercises 4.3.5 Exercises

1.

(a)

(b)

2.

3.

(a)

(b)

(c)

4.

Subsection 4.3.1 Parametrizations in the Definition of $\int_{C} F \cdot d r$