6.2 Line integrals (Page 7/20)

Calculus volume 3 Page 7 / 20

\int_{a_{1}}^{a_{n}} f (x) d x = \int_{a_{1}}^{a_{2}} f (x) d x + \int_{a_{1}}^{a_{3}} f (x) d x + \dots + \int_{a_{n - 1}}^{a_{n}} f (x) d x,

which is analogous to property iv.

Using properties to compute a vector line integral

Find the value of integral $\int_{C} F \cdot T d s,$ where C is the rectangle (oriented counterclockwise) in a plane with vertices $(0, 0), (2, 0), (2, 1), and (0, 1),$ and where $F = ⟨ x - 2 y, y - x ⟩$ ( [link] ).

A vector field in two dimensions. The arrows following roughly a 90-degree angle to the origin in quadrants 1 and 3 point to the origin. As the arrows deviate from this angle, they point away from the angle ad become smaller. Above, they point up and to the left, and below, they point down and to the right. A rectangle is drawn in quadrant 1 from 0 to 2 on the x axis and from 0 to 1 on the y axis. C_1 is the base, C_2 is the right leg, C_3 is the top, and C_4 is the left leg. — Rectangle and vector field for [link] .

Note that curve C is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let $C_{1}$ represent the side from $(0, 0)$ to $(2, 0),$ let $C_{2}$ represent the side from $(2, 0)$ to $(2, 1),$ let $C_{3}$ represent the side from $(2, 1)$ to $(0, 1),$ and let $C_{4}$ represent the side from $(0, 1)$ to $(0, 0)$ ( [link] ). Then,

\int_{C} F \cdot T d r = \int_{C_{1}} F \cdot T d r + \int_{C_{2}} F \cdot T d r + \int_{C_{3}} F \cdot T d r + \int_{C_{4}} F \cdot T d r .

We want to compute each of the four integrals on the right-hand side using [link] . Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse C counterclockwise):

\begin{matrix} C_{1} : ⟨ t, 0 ⟩, 0 \leq t \leq 2 \\ C_{2} : ⟨ 2, t ⟩, 0 \leq t \leq 1 \\ C_{3} : ⟨ 2 - t, 1 ⟩, 0 \leq t \leq 2 \\ C_{4} : ⟨ 0, 1 - t ⟩, 0 \leq t \leq 1. \end{matrix}

Therefore,

\begin{matrix} \int_{C_{1}} F \cdot T d r & = \int_{0}^{2} F (r (t)) \cdot r^{'} (t) d t \\ = \int_{0}^{2} ⟨ t - 2 (0), 0 - t ⟩ \cdot ⟨ 1, 0 ⟩ d t = \int_{0}^{1} t d t \\ = {[\frac{t^{2}}{2}]}_{0}^{2} = 2. \end{matrix}

Notice that the value of this integral is positive, which should not be surprising. As we move along curve C ₁ from left to right, our movement flows in the general direction of the vector field itself. At any point along C ₁ , the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along C ₁ , and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

\begin{matrix} \int_{C_{2}} F \cdot d r & = \int_{0}^{1} ⟨ 2 - 2 t, t - 2 ⟩ \cdot ⟨ 0, 1 ⟩ d t \\ = \int_{0}^{1} (t - 2) d t \\ = {[\frac{t^{2}}{2} - 2 t]}_{0}^{1} = - \frac{3}{2}, \end{matrix}

\begin{matrix} \int_{C_{3}} F \cdot T d s & = \int_{0}^{2} ⟨ (2 - t) - 2, 1 - (2 - t) ⟩ \cdot ⟨ −1, 0 ⟩ d t \\ = \int_{0}^{2} t d t = 2, \end{matrix}

and

\begin{matrix} \int_{C_{4}} F \cdot d r & = \int_{0}^{1} ⟨ −2 (1 - t), 1 - t ⟩ \cdot ⟨ 0, −1 ⟩ d t \\ = \int_{0}^{1} (t - 1) d t \\ = {[\frac{t^{2}}{2} - t]}_{0}^{1} = - \frac{1}{2} . \end{matrix}

Thus, we have $\int_{C} F \cdot d r = 2 .$

Got questions? Get instant answers now!

Calculate line integral $\int_{C} F \cdot d r,$ where F is vector field $⟨ y^{2}, 2 x y + 1 ⟩$ and C is a triangle with vertices $(0, 0),$ $(4, 0),$ and $(0, 5),$ oriented counterclockwise.

Got questions? Get instant answers now!

Applications of line integrals

Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.

Suppose that a piece of wire is modeled by curve C in space. The mass per unit length (the linear density) of the wire is a continuous function $ρ (x, y, z) .$ We can calculate the total mass of the wire using the scalar line integral $\int_{C} ρ (x, y, z) d s .$ The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by $ρ (x *, y *, z *) Δ s$ for some point $(x *, y *, z *)$ in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral $\int_{C} ρ (x, y, z) d s .$

Calculating the mass of a wire

Calculate the mass of a spring in the shape of a curve parameterized by $⟨ t, 2 cos t, 2 sin t ⟩,$ $0 \leq t \leq \frac{π}{2},$ with a density function given by $ρ (x, y, z) = e^{x} + y z$ kg/m ( [link] ).

A three dimensional diagram. An increasing, then slightly decreasing concave down curve is drawn from (0,2,0) to (pi/2, 0, 2). The arrow on the curve is pointing to the latter endpoint. — The wire from [link] .

To calculate the mass of the spring, we must find the value of the scalar line integral $\int_{C} (e^{x} + y z) d s,$ where C is the given helix. To calculate this integral, we write it in terms of t using [link] :

\begin{matrix} \int_{C} e^{x} + y z d s & = \int_{0}^{π / 2} ((e^{t} + 4 cos t sin t) \sqrt{1 + {(−2 cos t)}^{2} + {(2 sin t)}^{2}}) d t \\ = \int_{0}^{π / 2} ((e^{t} + 4 cos t sin t) \sqrt{5}) d t \\ = \sqrt{5} {[e^{t} + 2 {sin}^{2} t]}_{t = 0}^{t = π / 2} \\ = \sqrt{5 (e^{π / 2} + 1)} . \end{matrix}

Therefore, the mass is $\sqrt{5} (e^{π / 2} + 1)$ kg.

Got questions? Get instant answers now!

Questions & Answers

I'm interested in biological psychology and cognitive psychology

Tanya Reply

what does preconceived mean

sammie Reply

physiological Psychology

Nwosu Reply

How can I develope my cognitive domain

Amanyire Reply

why is communication effective

Dakolo Reply

Communication is effective because it allows individuals to share ideas, thoughts, and information with others.

effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.

it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.

miss

Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment

Wekolamo Reply

please i need answer

Wekolamo

because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).

Manix Reply

Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!

ARC

A child is a member of community not society elucidate ?

JESSY Reply

Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?

Simon Reply

compare and contrast skinner's perspective on personality development on freud

namakula Reply

Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning

war

explain how nature and nurture affect the development and later the productivity of an individual.

Amesalu Reply

nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.

Samuel

I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills

Zyryn Reply

good👍

Jonathan

and having a good philosophy of the world is like a sandwich and a peanut butter 👍

Jonathan

generally amnesi how long yrs memory loss

Kelu Reply

interpersonal relationships

Abdulfatai Reply

What would be the best educational aid(s) for gifted kids/savants?

Heidi Reply

treat them normal, if they want help then give them. that will make everyone happy

Saurabh

Got questions? Join the online conversation and get instant answers!

Jobilize.com Reply

<< Chapter < Page Page > Chapter >>

Practice Key Terms 8

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask

	1 AP 01 Human Body Anatomy Physiology MCQ By OpenStax Start Quiz
©flickr: Eduardo	Nursing Education NU589 Program Outcomes By Karen Gowdey Start Quiz
	NCE Ch 11 Counseling Families, Diagnosis... By Anh Dao Start Quiz
	1 AP Key Terms 01 Human Body Anatomy Physiology By OpenStax Start Key Terms
	Interventionism Mixed Economy By Robert Murphy Start Test
	14 Dr Landholt Large Animal Medicine-GI quiz By Brooke Delaney Start Exam
	6 Physiotherapy Flashcards Set 6 By Rhodes Start Flashcards
	Anthropology Life Cycle By Richley Crapo Start Assignment
	2 Week 2 Social Psych By Yacoub Jayoghli Start Quiz
	9 Arts Society: Theater 9 Exam By Jonathan Long Start Exam