## Thomas' Calculus 13th Edition

$(x^2+y+z^2)=4$
Here, the level curve for $f(x,y,z)=\ln (x^2+y+z^2)$ has the form of $c=\ln (x^2+y+z^2)$ Also, $x=-1;y=2;z=1$ Then $c=\ln [(-1)^2+(2)+(1)^2]$ or,$c= \ln 4$ Now, $c=\ln (x^2+y+z^2) \implies \ln 4=\ln (x^2+y+z^2)$ or, $e^{(\ln 4)}=e^{[\ln (x^2+y+z^2)]}$ Hence, $(x^2+y+z^2)=4$