# Finding the normal to a sphere at any point with spherical coordinates

For a sphere with radius $a$ centred at the origin, why are these two methods not valid?

So for any point on the sphere, can be parametrized in spherical coordinates as so:
$${\textbf{x}}= \begin{pmatrix}a \cos \theta \sin \phi \\ a \sin \theta \sin \phi \\ a \cos \phi\end{pmatrix}$$

By intuition, this is also the normal vector to the surface of the sphere at the point.
Now this will have length equal to $a$.

For the surface integral $$\iint_{S} dS = \iint_{S} ||{\textbf{N}}||\mathrm{d}\theta \mathrm{d}\phi$$
I would substitute $a$ for the integrand.

However, using a different method (taking the partial derivatives of the parametric vector and finding the cross product, another normal vector is
$${\textbf{N}} = \begin{pmatrix} a^2 \cos \theta \sin^2\phi \\ a^2\sin\theta\sin^2\phi\\-a^2\sin\phi\cos\phi\end{pmatrix}$$ with length equal to $a^2 \sin \phi$.
Now substituting this into the integrand, I'd get a different answer.

My question is: I know that the two normal vectors I get are different, but they are still normal vectors to the sphere. By the surface integral, shouldn't they both be allowed to be substituted into the integral?

• Which, if either, method gives $4\pi a^2$? – Lord Shark the Unknown Jun 8 '17 at 5:30
• The second one does (which should be correct) because the original question was to derive the surface area of a sphere. – Twenty-six colours Jun 8 '17 at 5:31

So you don't want any normal vector, you want a particular one. Actually, since you're taking the length at the end of the day, you want a particular function of $\theta$ and $\phi$... the direction doesn't actually matter.
More specifically you want what's called the surface element, which (vaguely) tells you what you need to multiply $d\theta d\phi$ by to get the area of the small patch of surface $\theta_0< \theta < \theta_0 + d\theta$, $\phi_0< \phi < \phi_0 + d\phi$ near the point $\phi_0,\theta_0$.