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


$\bf {Graph(e)}$

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 $\bf {Graph(e)}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.