Answer
16
Work Step by Step
A(x)=$\frac{(diagonal)^{2}}{2}$
=$\frac{\sqrt x - (-\sqrt x))^{2}}{2}$
= 2x;
a=0,b=4
v=$\int^b_a A(x)$
=$\int^4_a 2xdx$
=$[x^2]^4_0$
=16
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 321: 1
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