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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
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=[1202]=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)

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Chapter 16 Solutions

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Sect-16.1 P-11ESect-16.1 P-12ESect-16.1 P-13ESect-16.1 P-14ESect-16.1 P-15ESect-16.1 P-16ESect-16.1 P-17ESect-16.1 P-18ESect-16.1 P-21ESect-16.1 P-22ESect-16.1 P-23ESect-16.1 P-24ESect-16.1 P-25ESect-16.1 P-26ESect-16.1 P-29ESect-16.1 P-30ESect-16.1 P-31ESect-16.1 P-32ESect-16.1 P-33ESect-16.1 P-34ESect-16.1 P-35ESect-16.1 P-36ESect-16.2 P-1ESect-16.2 P-2ESect-16.2 P-3ESect-16.2 P-4ESect-16.2 P-5ESect-16.2 P-6ESect-16.2 P-7ESect-16.2 P-8ESect-16.2 P-9ESect-16.2 P-10ESect-16.2 P-11ESect-16.2 P-12ESect-16.2 P-13ESect-16.2 P-14ESect-16.2 P-15ESect-16.2 P-16ESect-16.2 P-17ESect-16.2 P-18ESect-16.2 P-19ESect-16.2 P-20ESect-16.2 P-21ESect-16.2 P-22ESect-16.2 P-23ESect-16.2 P-24ESect-16.2 P-25ESect-16.2 P-26ESect-16.2 P-31ESect-16.2 P-32ESect-16.2 P-33ESect-16.2 P-34ESect-16.2 P-35ESect-16.2 P-36ESect-16.2 P-37ESect-16.2 P-38ESect-16.2 P-39ESect-16.2 P-40ESect-16.2 P-41ESect-16.2 P-42ESect-16.2 P-43ESect-16.2 P-44ESect-16.2 P-45ESect-16.2 P-46ESect-16.2 P-47ESect-16.2 P-48ESect-16.2 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