Q 58 - Calculus

Question

In Exercises $53-58$, determine the level surfaces $c=-3,-2,-1,0,1,2,3$ if they exist for the specified function.

$f\left(x,y,z\right)=\sqrt{1-\left(x^2+y^2+z^2\right)}$.

Step-by-Step Solution

Answer

The level surfaces are determined by the equation $x^2+y^2+z^2=1-c^2$.

For $c=-1,1$, $x^2+y^2+z^2=0$represents the origin of 3-D plane.

Also the level surface is sphere with equation :-

$x^2+y^2+z^2=1-c^2$, $c=0$.

For $c=-3,-2,2,3$ level surfaces are undefined. 

Step 1. Given Information

We have given the following function :-

$f\left(x,y,z\right)=\sqrt{1-\left(x^2+y^2+z^2\right)}$.

We have to determine level surfaces $c=-3,-2,-1,0,1,2,3$ for the given function.

Step 2. Determine level surfaces

The given function is :-

$f\left(x,y,z\right)=\sqrt{1-\left(x^2+y^2+z^2\right)}$.

We know that the level surfaces are determined as :-

$f\left(x,y,z\right)=c$.

That is we have :-

$$\begin{gathered} \sqrt{1-\left(x^2+y^2+z^2\right)}=c \\ \Rightarrow 1-\left(x^2+y^2+z^2\right)=c^2 \\ \Rightarrow x^2+y^2+z^2=1-c^2 \end{gathered}$$

This is the equation of sphere.

So the level surfaces are undefined for $c=-3,-2,2,3$ as radius of sphere cannot be negative.

Also for $c=-1,1$represents the origin of 3-D plane.

So that we have :-

The level surfaces are sphere with the equation $x^2+y^2+z^2=1-c^2$ for  $c=0$.