Answer
See image.
.
Work Step by Step
$ a.\quad$
To sketch the surface $z=f(x,y)=\sqrt{x},$ note that:
* z is nonnegative (the surface is on or above the xy plane),
* x is nonnegative (the surface is above quadrants I and IV of the xy plane),
* y is ommited in the equation, so the trace in $y=0$ (the xz plane) will be translated along the y-axis (similar to unbounded cylinders).
The trace in the xz plane is $z=\sqrt{x}$, the upper half of the parabola $\ \ \ z^{2}=x$, which we translate along the y-axis into planes $y=k, k\in \mathbb{R}.$
$ b.\quad$
In the xy plane, we equate $f(x,y)$ with several values of c,
$z=\sqrt{x}=c\qquad $(so only nonnnegative c apply)
These are vertical lines $x=c^{2}.$
Take $c=0,1,\sqrt{2},\sqrt{3}$....
