Answer
$\int_C (3y-x)dx+y^2 dy = \frac{916}{3}$
Work Step by Step
$x = 2t, y = 10t, dx = 2dt, dy = 10 dt$
$\therefore \int_C (3y-x)dx+y^2 dy\\
= \int_{t=0}^1 2(30t-2t)dt + 1000t^2dt\\
=[\frac{1000t^3}{3} - \frac{56t^2}{2}]_0^1 = \frac{916}{3}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 15 - Vector Analysis - 15.2 Exercises - Page 1063: 54
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
