Explicit function definition with the normal vector included

The formula

$$ mathbf{n} = left( -frac{partial f}{partial x}, -frac{partial f}{partial y}, 1 right)$$

gives a valid normal vector $mathbf{n}$ to the surface $z = f(x,y)$ only where the function $f$ is differentiable.

In the hemisphere example, for the points where $x^2+y^2=1$ and $z=0$, $f$ won't be differentiable, since the tangent plane to the surface is vertical.

We can still use the formula for points where $x^2+y^2<1$ and $z>0$. Since $f(x,y) = sqrt{1-x^2-y^2}$:

$$ mathbf{n} = left( frac{x}{sqrt{1-x^2-y^2}}, frac{y}{sqrt{1-x^2-y^2}}, 1 right) $$

The limit of this formula as $x^2+y^2 to 1^{-}$ isn't helpful: it doesn't exist, but informally we can say it approaches $(pm infty, pm infty, 1)$.

But a normal vector isn't unique, since multiplying it by a scalar gives another normal vector. So let's try finding a unit normal vector:

$$ begin{align*} | mathbf{n} |^2 &= frac{x^2}{1-x^2-y^2} + frac{y^2}{1-x^2-y^2} + 1^2 = frac{1}{1-x^2-y^2} \ | mathbf{n} | &= frac{1}{sqrt{1-x^2-y^2}} \ widehat{mathbf{n}} &= frac{mathbf{n}}{|mathbf{n}|} = mathbf{n} sqrt{1-x^2-y^2} = left(x, y, sqrt{1-x^2-y^2}right) = (x, y, f(x,y)) end{align*} $$

This makes sense: at any point on a sphere, the vector to that point from the origin is perpendicular to the sphere. And this time, it's clear to see that the limit works out as expected: If $x_0^2+y_0^2=1$, then

$$ lim_{substack{(x,y) to (x_0,y_0) \ x^2+y^2<1}} widehat{mathbf{n}} = (x_0, y_0, 0) $$