Q 50. - Calculus

Question

Determine the domains of the functions in Exercises 47–56, and find where the functions are continuous. 

$f(x,y,z)=\ln (x^2+y^2+z^2)$

Step-by-Step Solution

Answer

The function $f(x,y,z)=\ln (x^2+y^2+z^2)$ is continuous on the set $\left\{\left(x,y,z\right)\in f^3:x^2+y^2+z^2>0\right\}$.

Step 1. Given information.

We have given expression: $f(x,y,z)=\ln (x^2+y^2+z^2)$

Step 2: Determine the domains of the functions.

Consider the function: $g:f^3\to f$.Then the domain of the function $g$ is defined as

$Domain\left(g\right)=\left\{\left(x,y,z\right)\in f^{3 }:g(x,y,z)\right\}$

Since the domain of the natural logarithm $\ln \left(x\right)$ is the set $\left\{x\in f:x>0\right\}$.

$ln(x^2+y^2+z^2)$ is defined on the set$\left\{\left(x,y,z\right)\in f^3:x^2+y^2+z^2>0\right\}$.

Here from the above statement, the domain of the function is   $f(x,y,z)=\ln (x^2+y^2+z^2)$ is

$Domain \left(f\right)=\left\{\left(x,y,z\right)\in f^3:x^2+y^2+z^2>0\right\}$ 

Step 3. To find continuous of the function.

Since the natural logarithm is continuous on the  entire domain, therefore the function$f(x,y,z)=\ln (x^2+y^2+z^2)$ is continuous on the set $\left\{\left(x,y,z\right)\in f^3:x^2+y^2+z^2>0\right\}$