## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 16: Integrals and Vector Fields - Section 16.1 - Line Integrals - Exercises 16.1 - Page 943: 2

Graph(e)

#### Work Step by Step

Given $r(t)=i+j+t k$ Here, the $x$ and $y$ intercepts are both $1$ and $z$-varies from $-1$ to $+1$, and hence, it represents that any point on the graph is of the form of $(1,1,t)$ . Thus, the answer is Graph(e).

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