   Chapter 16.2, Problem 11E

Chapter
Section
Textbook Problem

Evaluate the line integral, where C is the given curve.11. ∫C xeyz ds, C is the line segment from (0, 0, 0) to (1, 2, 3)

To determine

To Evaluate: The line integral C(xeyz)ds for the line segment from the point (0,0,0) to the point (1,2,3) .

Explanation

Given data:

The given curve C is a line segment from the point (0,0,0) to the point (1,2,3) .

Formula used:

Write the expression to evaluate the line integral for a function f(x,y,z) along the curve C .

Cf(x,y,z)ds=abf(x(t),y(t),z(t))(dxdt)2+(dydt)2+(dzdt)2dt (1)

Here,

a is the lower limit of the curve C and

b is the upper limit of the curve C .

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 (2)

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 (3)

Write the required differential formulae to evaluate the given integral.

ddttn=ntn1

Consider the points (x0,y0,z0) as (0,0,0) and (x1,y1,z1) as (1,2,3) .

Calculation of direction vector:

Substitute 1 for x1 , 2 for y1 , 3 for z1 , 0 for x0 , 0 for y0 , and 0 for z0 in equation (3),

a,b,c=10,20,30=1,2,3

Calculation of parametric equations of the curve:

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

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

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

0t1

Find the expression (xeyz) as follows

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