# Why is the normal vector different in cartesian coordinates vs. spherical coordinates?

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.

• Thanks for the reply! But if I was to calculate a surface integral, would both vectors give the same answer? Also, if the two vectors are different, then where did the vector $\mathbf{x}_u \times \mathbf{x}_v$ (in cartesian coordinates) get mapped to in sphericals? I guess I assumed that the mapping would preserve the same vector. – user523384 Apr 20 at 10:03
• But the vector obtained from a cross product in one coordinate system need not be mapped to the cross product of the same vectors in the new coordinate system? (Just want to make sure I haven't misunderstood something) i.e. If $a\times b = c$ then $T(a\times b)=?= T(c)$? – user523384 Apr 20 at 10:14
• Hmm but the jacobian over here would be $\rho^2 \sin(\phi)$ I'm not sure why that explains a difference (since in cartesian it's not just (x,y,z) but 1/z (x,y,z)) – user523384 Apr 20 at 10:30