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: 11

Answer

(a) $f(1,1,1)=3$. (b) The domain is $$\mathcal{D}=\{(x,y,z)|x,y,z\geq0, x^2+y^2+z^2<2^2\}.$$ Geometrically, it is the interior of the slice of the ball (one eight of the ball situated in the first octant) with the center at the origin and the radius of $2$.

Work Step by Step

(a) By direct substitution $$f(1,1,1)=\sqrt{1}+\sqrt{1}+\sqrt{1}+\ln(4-1^2-1^2-1^2)=3+\ln1=3+0=3.$$ (b) For the domain we need that the arguments of square roots to be non negative which gives $x,y,z\geq0.$ Also, the argument of the $\ln$ has to be positive so we have $4-x^2-y^2-z^2=4-(x^2+y^2+z^2)>0$, which can be rewritten as $$x^2+y^2+z^2<2^2.$$ This inequality describes the interior of the ball with the center at the origin and with the radius of $2$. With the addition that $x,y,z$ have to be nonnegative we get that the domain is $$\mathcal{D}=\{(x,y,z)|x,y,z\geq0, x^2+y^2+z^2<2^2\}.$$ Geometrically, it is the interior of the slice of the ball (one eight of the ball situated in the first octant) with the center at the origin and the radius of $2$.
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