Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1022: 2


${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ 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:
Small 1524685784
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.