bottom half of a sphere if a sphere is of the equation $r^2 = x^2 + y ^2 +z ^2$ and we want to find a hemisphere does  that just involve setting a limit to be half of the radius ?
For example if we know that the centre of the sphere sits at $(2,1,0)$ and the radius is $2$, then if we set the limit of z to be from 0 to 1 then that would be the top half of the sphere ?
$2^2 = (x-2)^2 + (y-1)^2 + (z)^2$ and the limit would be from $z = 0$  to  $z = 1$.
Is that right ?
 A: Due to the nature of a sphere, there is already only a limited range of values that can ever occur as a $z$ coordinate of one of its points.
To restrict the formula to only the upper or lower half of the sphere, you really only need to know the $z$ coordinate of the center of the sphere. The plane through that point cuts the sphere neatly in half.
So for a sphere with radius $2$ and center at $x = 2,$ $y = 1,$ and $z = 0,$ the lower half of the sphere consists of those points that simultaneously satisfy these two conditions:
$$2^2 = (x-2)^2 + (y-1)^2 + z^2,$$
$$z \le 0.$$
It is also true that all of these points satisfy $z \ge -2,$ but it is unnecessary to state that as a condition, because the equation $2^2 = (x-2)^2 + (y-1)^2 + z^2$
already implies that $z \ge -2.$
A: The center $(2, 1, 0)$ of the sphere sits in the $x$-$y$ plane, i.e., on the plane $z = 0$, with half the sphere above the plane, and half below. 
So for the bottom half of the sphere, you need $-2 \leq z \leq 0$, and for the top half, you need $0\leq z\leq 2$.
