Graph the level curves of the function $z=f(x,y)=e^{x^{2}+y^{2}}$ for $c=1, e, e^{4},$ and $e^{16}$.

Solution  Because $x^{2}+y^{2}\geq 0,$ it follows that $z\geq e^{0}=1$. The level curves satisfy the equation $e^{x^{2}+y^{2}}=c$ or $x^{2}+y^{2}=\ln c,$where $c\geq 1$. For $c=1,$ the level curve is the point $(0,0)$. If $c>1,$ the level curves are concentric circles. Figure 14 illustrates several level curves of $f$. A graph of the surface $z=e^{x^{2}+y^{2}}$ is given in Figure 15. Do you see how the graph evolved from the collection of its level curves?

Figure 14 Level curves of $z=e^{x^2+y^2}$.

Figure 15 The surface $z=e^{x^2+y^2}$.