I'm trying to find the unit normal field n on S. Where S is the ellipsoid $2x^2+2y^2+z^2=2$ However i get stuck in trying to substitute f into the $|\nabla f(x,y,z)|$.
My steps:
(i) I calculate n with $n= \frac{(\nabla f(x,y,z)}{|\nabla f(x,y,z)|}$
(ii) Using partial differentiation i find $\nabla f=<4x,4y,2z>$
(iii) $|\nabla f(x,y,z)|= {\sqrt((f_x)^2+(f_y)^2+(f_z)^2)}={\sqrt(16x^2+16y^2+4z^2)} ={2\sqrt(4x^2+4y^2+z^2)}$
The next step is where i get stuck at. I think they substitute the $f=2x^2+2y^2+z^2=2$ into the formula but they don't match nicely and i'm not sure how to deal with it.
I believe that $n= \frac{<4x,4y,4z>}{6}$ should be the answer.
Thanks for any help!
$<4x,4y,2z>/2\sqrt{2x^2 + 2y^2 + 2}$
If you'd like, you could express 2z in terms of x and y using f = 2,