## Calculus: Early Transcendentals 8th Edition

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