Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Section 14.1 - Functions of Several Variables - Exercises 14.1 - Page 787: 16

Answer

See image. .
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Work Step by Step

Equate $f(x,y)=c$ for each value of c $\sqrt{25-x^{2}-y^{2}}=c$ square both sides, $25-x^{2}-y^{2}=c^{2}$ $x^{2}+y^{2}=(25-c^{2})$ We have concentric circles about the origin with radii $\sqrt{25-c^{2}}$ $\sqrt{25-0^{2}}=5,$ $\sqrt{25-1^{2}}=\sqrt{24}=2\sqrt{6}$ $\sqrt{25-2^{2}}=\sqrt{21},$ $\sqrt{25-3^{2}}=4,$ $\sqrt{25-4^{2}}=3,$
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