## University Calculus: Early Transcendentals (3rd Edition)

$4=x^2+y^2+z^2$
Remember that the level curve for $f(x,y,z)=\sqrt {x^2+y^2+z^2}$ has the form of $c=\sqrt {x^2+y^2+z^2}$ ...(1) Here, we have $x=1,y=-1,z=\sqrt 2$ Then $c=\sqrt {(1)2+(-1)^2+(\sqrt 2)^2} \implies c=2$ Thus, equation (1), becomes: $2=\sqrt {x^2+y^2+z^2}$ Hence, $4=x^2+y^2+z^2$