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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?

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  • $\begingroup$ Which, if either, method gives $4\pi a^2$? $\endgroup$ – Lord Shark the Unknown Jun 8 '17 at 5:30
  • $\begingroup$ The second one does (which should be correct) because the original question was to derive the surface area of a sphere. $\endgroup$ – Twenty-six colours Jun 8 '17 at 5:31
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The second method is correct. It should be clear that operations that preserve normalness don't necessarily preserve the integral... a simpler example than the one you cite is just multiplying by a constant.

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 $\phi_0,\theta_0$.

The surface element is computed by method 2 above. The fact that it's correct has nothing to do with the fact that the cross product of the tangent vectors points normal to the surface and everything to do with the fact that its length is the area of the paralellogram formed by the tangent vectors.

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  • $\begingroup$ RIght!! I forgot the way that it was originally derived (the integral). It's because we take infinitesimally small parallelograms (which happen to be the modulus of the two tangent vectors crossed)? So it's not because we require a normal, it's just that the area of the parallelogram happens to be the length of a specific normal? $\endgroup$ – Twenty-six colours Jun 8 '17 at 5:38

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