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: 9

Answer

$\{(x,y)\in \mathbb{R}^{2}\ \ |\ \ $ $x^{2}-1\leq y\leq x^{2}+1$ $\}$
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Work Step by Step

The domain of f depends on the domain of $\cos^{-1}t$, which is the range of $\cos t,$ which is $[-1,1].$ So f will be defined only for those $(x,y)$ for which $1\leq y-x^{2}\leq 1,$ $x^{2}-1\leq y\leq x^{2}+1$ This is the region between the curves $\left\{\begin{array}{l} y=x^{2}-1\\ y=x^{2}+1 \end{array}\right..$ The curves themselves are included, so they are graphed with a solid line. Domain: $\{(x,y)\in \mathbb{R}^{2}\ \ |\ \ $ $x^{2}-1\leq y\leq x^{2}+1$ $\}$
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