#### Answer

See image.
.

#### Work Step by Step

$ a.\quad$
To sketch the surface $z=f(x,y)=\sqrt{x^{2}+y^{2}}$:
* note that z can not be negative, so the surface is on or above the xy plane
* in a plane $z=k$, the trace is a circle of radius $k$
* in the plane $x=0$, the trace is the graph of $z=|y|$
* in the plane $y=0$, the trace is the graph of $z=|x|$
This is the upper part of a (circular) cone.
$ b.\quad$
In the xy plane, we equate $f(x,y)$ with several values of c,
$z=\sqrt{x^{2}+y^{2}}=c\qquad $(so only nonnnegative c apply)
These are circles with radii $c.$
Take $c=0,1,2,3$....