Chapter 16.2, Problem 15E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Evaluate the line integral, where C is the given curve.15. ∫C z2 dx + x2 dy + y2 dz, C is the line segment from (1, 0, 0) to (4, 1, 2)

To determine

To Evaluate: The line integral C(z2)dx+(x2)dy+(y2)dz for the line segment from the point (1,0,0) to the point (4,1,2) .

Explanation

Given data:

The given curve C is a line segment from the point (1,0,0) to the point (4,1,2) .

Formula used:

Write the expression to find the parametric equations for a line through the point (x0,y0,z0) and parallel to the direction vector v=a,b,c .

x=x0+at,y=y0+bt,z=z0+ct (1)

Write the expression to find direction vector v=a,b,c for a line segment from the point (x0,y0,z0) to the point (x1,y1,z1) .

a,b,c=x1x0,y1y0,z1z0 (2)

Consider the points (x0,y0,z0) as (1,0,0) and (x1,y1,z1) as (4,1,2) .

Calculation of direction vector:

Substitute 4 for x1 , 1 for y1 , 2 for z1 , 1 for x0 , 0 for y0 , and 0 for z0 in equation (2),

a,b,c=41,10,20=3,1,2

Calculation of parametric equations of the curve:

Substitute 1 for x0 , 0 for y0 , 0 for z0 , 3 for a , 1 for b , 2 for c in equation (1),

x=1+(3)t,y=0+(1)t,z=0+(2)tx=1+3t,y=t,z=2t

Consider the limits of scalar parameter t are 0 to 1.

0t1

Find the expression (z2) as follows.

Substitute 2t for z in the expression (z2) ,

z2=(2t)2=4t2

Find dz as follows.

Differentiate on both sides of the expression z=2t .

ddt(z)=ddt(2t)dzdt=2dz=2dt

Find the expression (x2) as follows.

Substitute 1+3t for x in the expression (x2) ,

x2=(1+3t)2=1+9t2+6t=9t2+6t+1

Find dx as follows.

Differentiate on both sides of the expression x=1+3t .

ddt(x)=ddt(1+3t)dxdt=0+3dx=3dt

Find the expression (y2) as follows.

Substitute t for y in the expression (y2) .

y2=(t)2=t2

Find dy as follows.

Differentiate on both sides of the expression y=t

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