## Thomas' Calculus 13th Edition

$a.\quad$ 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 $c=0,1,\sqrt{2},\sqrt{3}$....