Explanation Given: The line integral is ∫ C x y z d s . The C is the line segment from ( 0 , 0 , 0 ) to ( 1 , 2 , 3 ) . Result used: (1)“Suppose f be continuous function on a region containing a smooth curve C : r ( t ) = 〈 x ( t ) , y ( t ) , z ( t ) 〉 in the interval 0 ≤ t ≤ 2 π . Then ∫ C f d s = ∫ a b f ( x ( t ) , y ( t ) ) | r ′ ( t ) | d t ”. (2)“The parameterized form of the line segment from ( x 0 , y 0 , z 0 ) to ( x 1 , y 1 , z 1 ) is r ( t ) = ( 1 − t ) ( x o , y 0 , z 0 ) + t ( x 1 , y 1 , z 1 ) for 0 ≤ t ≤ 1 . Therefore r ( t ) = 〈 ( 1 − t ) x 0 + t x 1 , ( 1 − t ) y 0 + t y 1 , ( 1 − t ) z 0 + t z 1 〉 ”. Calculation: The line integral is ∫ C x y z d s . Given that the line segment from ( 0 , 0 , 0 ) to ( 1 , 2 , 3 ) where ( x 0 , y 0 , z 0 ) = ( 0 , 0 , 0 ) and ( x 1 , y 1 , z 1 ) = ( 1 , 2 , 3 ) . So by above result, obtain the value of r ( t ) . r ( t ) = ( 1 − t ) ( 0 , 0 , 0 ) + t ( 1 , 2 , 3 ) = 〈 [ ( 1 − t ) ( 0 ) + t ( 1 ) ] , [ ( 1 − t ) ( 0 ) + t ( 2 ) ] , [ ( 1 − t ) ( 0 ) + t ( 3 ) ] 〉 = 〈 t , 2 t , 3 t 〉 For 0 ≤ t ≤ 1 , x ( t ) = t , y ( t ) = 2 t and z ( t ) = 3 t Therefore, the parametric equation of the line segment is r ( t ) = 〈 t , 2 t , 3 t 〉 for 0 ≤ t ≤ 1

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 14.2, Problem 27E

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To convert: The line integral ∫Cxyz ds into an ordinary integral and evaluate the value.

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Chapter 14.2, Problem 26E

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Chapter 14 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The line integral ∫ C x y z d s into an ordinary integral and evaluate the value.

Solution Summary: The author evaluates the value of the line integral displaystyleundersetCintxyzds.

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Chapter 14 Solutions