Consider the sphere $x^2+y^2+z^2=1$. Let $\mathbf x(u,v)$ be a parameterization for the sphere.
Say I was trying to find specifically the normal vector given by $$ \frac{\partial \bf x}{\partial u} \times \frac{\partial \bf x}{\partial v} $$ (to for example, evaluate a surface integral)
Question. Why is the value of this normal vector different when I evaluate it in cartesian coordinates vs spherical coordinates?
1. Cartesian coordinates
$$\mathbf{x}(u,v)= \langle u, v, \pm\sqrt{1-u^2-v^2} \rangle$$
First consider the positive case.
Then, the partial derivatives $$ \mathbf{x}_u(u,v)= (1,0,\frac {-x}{\sqrt{1-x^2-y^2}}\mathstrut ) $$
$$\mathbf x_v(u,v)=(0,1,\frac {-y}{\sqrt{1-x^2-y^2}}\mathstrut ) $$
$$\mathbf{x}_u\times\mathbf{x}_v=\frac{1}{z}\langle x,y,z \rangle$$
2. Spherical coordinates
The sphere is given by $\rho = 1$ where $\phi \in [0,\pi]$ and $\theta \in [0,2\pi]$.
Computing the cross product in spherical coordinates gives
$$\mathbf x_u \times \mathbf x_v=\Biggl\vert \begin{array} 1\mathbf e_\rho & \mathbf e_\phi & \mathbf e_\theta \\ 0 & h_\phi & 0 \\ 0 & 0 & h_\theta \end{array}\Biggr\vert = \rho^2\sin(\theta) \mathbf{e}_\rho = z\langle x,y,z \rangle $$
Where $\mathbf{e}_\rho$ etc are the unit basis vectors for spherical coordinates, and $h_\rho$ etc are their respective scale factors.
Best Answer
The unit normal vector is the same. Different parametrizations of the same surface give different non-unit vectors.