Answer
False.
Work Step by Step
False, if $\sqrt{x}=f(x)$ determines the radius of the area $A(x),$ and then $A(x)=\pi(\sqrt{x})^{2}=\pi x$ and hence the area isn't a quadratic function.
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 363: 23
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.
