Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.1 - Functions of Several Variables - 14.1 Exercise: 11

Answer

a: 3 b: $x, y, z \geq 0$ and $ x^2, y^2, z^2 \lt 4$

Work Step by Step

a: Evaluate for $ f(1,1,1)$: $f(x, y, z) = \sqrt x + \sqrt y + \sqrt z + ln(4- x^2 - y^2 -z^2)$ $f(1, 1, 1) = \sqrt 1 + \sqrt 1 + \sqrt 1 + ln(4- 1^2 - 1^2 -1^2) = 3 + ln(1) = 3$ b: Because a negative square root like $\sqrt -a $ isn't possible, $ x, y$ and $z$ have to be equal to or larger than $0$, so: $x, y, z \geq 0$ Because negative natural logarithms like $ln(-a)$ and the natural logarithm of 0 ($ln(0)$) aren't possible, the ln term must be larger than $0$ so: $4- x^2 - y^2 -z^2 \gt 0$ $-x^2 - y^2 -z^2 \gt -4$ $x^2 + y^2 +z^2 \lt 4$ Now we recognize the formula for a sphere, so the radius is $\sqrt 4 = 2$ As we learned earlier on, x, y and z have to also be larger or equal to 0 so geometrically this gives us the piece of a sphere with radius 2 in the first quadrant.
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