Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.1 Functions of Several Variables - 14.1 Exercises - Page 940: 20

Answer

The domain is $$\mathcal{D}=\left\{(x,y)\left|-\frac{\pi}{2}\leq x+y\leq\frac{\pi}{2}\right.\right\}$$ and it is shown on the figure below
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Work Step by Step

The argument of the $\sin^{-1}$ function has to be from the segment $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ so we need that $x+y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, This can be rewritten as $$-\frac{\pi}{2}\leq x+y\leq\frac{\pi}{2}$$ and then as two inequalities: $$x+y\geq-\frac{\pi}{2}\Rightarrow y\geq-x-\frac{\pi}{2};$$ and $$x+y\leq\frac{\pi}{2}\Rightarrow y\leq-x+\frac{\pi}{2}.$$ Thus, the doman is the region bounded by two parallel lines (including them): $y=-x-\pi/2$ and $y=-x+\pi/2$. So we write for the domain $$\mathcal{D}=\left\{(x,y)\left|-\frac{\pi}{2}\leq x+y\leq\frac{\pi}{2}\right.\right\}$$ and it is presented in the figure below.
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