University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 15 - Section 15.1 - Line Integrals - Exercises - Page 826: 2

Answer

$\bf {Graph(e)}$

Work Step by Step

Since, we have $r(t)=i+j+t k$ This shows us that any point on the graph has the form of $(1,1,t)$ . Thus, the x and y intercepts are both $1$ and z-varies from $-1$ to $+1$ Hence, the required answer is $\bf {Graph(e)}$
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