   Chapter 16.2, Problem 16E

Chapter
Section
Textbook Problem

Evaluate the line integral, where C is the given curve.16. ∫C (y + z) dx + (x + z) dy + (x + y) dz, C consists of line segments from (0, 0, 0) to (1, 0, 1) and from (1, 0, 1) to (0, 1, 2)

To determine

To Evaluate: The line integral C(y+z)dx+(x+z)dy+(x+y)dz for the line segment from the point (0,0,0) to (1,0,1) and from the point (1,0,1) to (0,1,2) .

Explanation

Given data:

The given curve C is a line segment from the point (0,0,0) to (1,0,1) and from the point (1,0,1) to (0,1,2) .

The evaluation of integral C(y+z)dx+(x+z)dy+(x+y)dz is the sum of evaluation of integral for the line segment from the point (0,0,0) to (1,0,1) and the evaluation of integral for the line segment from the point (1,0,1) to (0,1,2) .

Therefore, write the line integral C(y+z)dx+(x+z)dy+(x+y)dz as follows.

C(y+z)dx+(x+z)dy+(x+y)dz=[C1(y+z)dx+(x+z)dy+(x+y)dz+C2(y+z)dx+(x+z)dy+(x+y)dz] (1)

Here,

C1 is the line segment from the point (0,0,0) to (1,0,1) and

C2 is the line segment from the point (1,0,1) to (0,1,2) .

Parametric equations of C1 :

Consider the parametric equations such that the parameters must satisfy the line segment points (0,0,0) and (1,0,1) .

x=t,y=0,z=t,0t1

The parameters are satisfied for the points (0,0,0) and (1,0,1) , and for the limit of t .

Calculation of dx for the line segment C1 :

Differentiate on both sides of the expression x=t with respect to t as follows.

ddt(x)=ddt(t)dxdt=1dx=dt

Calculation of dy for the line segment C1 :

Differentiate on both sides of the expression y=0 with respect to t as follows.

ddt(y)=ddt(0)dydt=0dy=0

Calculation of dz for the line segment C1 :

Differentiate on both sides of the expression z=t with respect to t as follows.

ddt(z)=ddt(t)dzdt=1dz=dt

Evaluation of line integral C1(y+z)dx+(x+z)dy+(x+y)dz :

Substitute t for x , 0 for y , t for z , dt for dx , 0 for dy , dt for dz , 0 for lower limit, and 1 for upper limit in the line integral C1(y+z)dx+(x+z)dy+(x+y)dz as follows.

C1(y+z)dx+(x+z)dy+(x+y)dz=01(0+t)dt+(t+t)(0)+(t+0)dt=01(t)dt+0+(t)dt=01(2t)dt=[2(t22)]01

Simplify the expression as follows.

C1(y+z)dx+(x+z)dy+(x+y)dz=[t2]01==1

Parametric equations of C2 :

Consider the parametric equations such that the parameters must satisfy the line segment points (1,0,1) and (0,1,2)

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

In problems 1-16, solve each equation. 3. Solve

Mathematical Applications for the Management, Life, and Social Sciences

Given: m9=2x+17m11=5x94 Find: x

Elementary Geometry For College Students, 7e

True or False: The graph in question 3 has a vertical asymptote at x = 1.

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

In Exercises 15-22, use the laws of logarithms to solve the equation. log28=x

Finite Mathematics for the Managerial, Life, and Social Sciences 