Finding a unit normal to a surface

[dyn]

Hi.
I'm trying to self-study vector calculus and I have to admit I struggle with it. As regards finding normals to a surface I know 2 ways - one involves writing the surface equation as f(x.y.z) = 0 and taking the gradient. The other method involves writing the surface in terms of 2 parameters and taking the vector product of the 2 partial derivatives. Finally, my question. If I have a circle such as x² + y² = a² I know the unit normal is k , but how do I show this ? My 2 methods don't work !
Thanks

 

[Charles Link]

The circle equation is the boundary of the surface. ($x^2+y^2=a^2$ is the equation for a cylinder.) The actual equation of the plane in which the circle $x^2+y^2=a^2$ resides is simply $z=0$. (I'm presuming you are wanting your surface to be a flat circle with center at the origin in the x-y plane). What you have for a surface is the endface of the cylinder. Take the gradient of $z=0$ and you do get $\hat{n}=\hat{k}$.

 

[dyn]

Yes i am taking the surface as a flat circle in the x-y plane. To take the gradient I need the form f(x,y,z)=0. What do I take f as ?

 

[Charles Link]

Yes i am taking the surface as a flat circle in the x-y plane. To take the gradient I need the form f(x,y,z)=0. What do I take f as ?

$f(x,y,z)=z$. Comes from $z=0$ is your equation of $f(x,y,z)=0$. $\nabla z=\hat{k}$. $\\$ If the plane was instead $z=5$, $f(x,y,z)=z-5$, and again $\nabla (z-5)=\hat{k}$.

 

[dyn]

Thanks for your help. On a similar topic ; is it possible to find the unit normal to the curved surface of a cylinder orientated along the z-axis in this way. I can't find a way to write this surface as f(x,y,z) = 0 but I can get an answer using my other method using the vector product

 

[Charles Link]

Thanks for your help. On a similar topic ; is it possible to find the unit normal to the curved surface of a cylinder orientated along the z-axis in this way. I can't find a way to write this surface as f(x,y,z) = 0 but I can get an answer using my other method using the vector product

The equation of the cylinder is simply $x^2+y^2=a^2$. Thereby $f(x,y,z)=0$ is $x^2+y^2-a^2=0$.

 

[dyn]

Thanks again. Much appreciated.