Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Practice Exercises - Page 36: 2

Answer

$ A=\sqrt[3] {36\pi V^2}$

Work Step by Step

Let $ r $ be the radius of a sphere; the surface area is $ A=4\pi r^2$ and $ r=\sqrt {\frac{A}{4\pi}}$ The volume of the sphere is given by $ V=\frac{4}{3}\pi r^3$ and $ r=\sqrt[3] {\frac{3V}{4\pi}}$ We have $\sqrt {\frac{A}{4\pi}}=\sqrt[3] {\frac{3V}{4\pi}}$ and $ A=\sqrt[3] {36\pi V^2}$
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