Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - Review - Exercises - Page 992: 2

Answer

${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ and $-1\leq x\leq 1$

Work Step by Step

Given: $f(x,y)=\sqrt {4-x^{2}-y^{2}}+\sqrt {1-x^{2}}$ Since we can only take the square root of non-negatives quantities, we want $4-x^{2}-y^{2}\geq 0$ This implies $x^{2}+y^{2}\leq 4$ and $1-x^{2}\geq 0$ This implies $x^{2}\leq 1$ or $-1\leq x\leq 1$ Therefore, the domain for given function is ${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ and $-1\leq x\leq 1$ Sketch the graph for the domain as shown below:
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