How do you find a normal vector of this set $T=\{(x,y,z)\mid x^2+y^2 \leq 1, z=0 \}$?

Question

How do you find a normal vector of this set $T=\{(x,y,z)\mid x^2+y^2 \leq 1, z=0 \}$?

The solution that my teacher has given is $(0, 0, -1)$, but I don't know how to find it. The vector must go downwards.

Answer 1

Try to visualize the surface. Sketch it. In this case you have a disk on the xy plane. What is perpendicular to this disk?

If you are unable to visualize your surface, can you paramerize it?

Suppose you have $x = u(s,t), y = v(s,t), z = w(s,t)$

The vector $(\frac {\partial x}{\partial s},\frac {\partial y}{\partial s}, \frac {\partial z}{\partial s})$ will be a vector parallel to the surface. As will $(\frac {\partial x}{\partial t},\frac {\partial y}{\partial t}, \frac {\partial z}{\partial t})$

Their cross product will be normal to the surface.

More tools:

If you have $z = f(x,y)$ then you can use $x,y$ as your parameters.

$(\frac {\partial x}{\partial x},\frac {\partial y}{\partial x}, \frac {\partial z}{\partial x}) = (1,0,\frac {\partial z}{\partial x})\\ (\frac {\partial x}{\partial y},\frac {\partial y}{\partial y}, \frac {\partial z}{\partial y}) = (0,1,\frac {\partial z}{\partial x})\\ (1,0,\frac {\partial z}{\partial x})\times(0,1,\frac {\partial z}{\partial y}) = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1)$

Finally, if you have $F(x,y,z) = k, \nabla F$ will be normal to your surface.

Answer 2

The surface $T$ is contained in the $xy-$plane, right? And the vectors $v=(1,0,0)\,w=(0,1,0)$ belong to the surface $T$, right? Hence a vector $n$ normal to $T$ is orthogonal to both vectors. $n$ is equal to $w\times v$. Notice that also $v\times w$ is normal.