Magnitude of the differential arc segment in spherical coordinates This has always bugged me.  What is the magnitude of a vector in spherical coordinates?
In Cartesian this is simple, $\vert \mathbf x\vert =\sqrt{x^2 + y^2 + z^2}$
Solutions I have read online say that the magnitude of
$$r\,{\rm d}r (\hat r) + r  \,{\rm d}\theta (\hat\theta)  + r \sin(\theta) \,{\rm d}\phi (\hat\phi)$$
is simply the square root of the dot product
$$\vert\mathbf x\vert=\sqrt{r^2 \,{\rm d}r^2 + (r\,{\rm d}\theta)^2 + (r \sin(\theta) \,{\rm d}\phi)^2}$$
I don't know why but I have a very difficult time believing this. Cartesian its true by Pythagorean theorem but this just doesn't seem correct to me.
By linear algebra it would seem to be accurate as well since a vector's dot product with itself is just the magnitude squared.  I guess a proof of this would be nice; can anyone shed light on this?
 A: Your intuition did not fail you, the formula in your question was incorrect indeed. The correct formula for a 3D metric in polar coordinates is:
$${\rm d}s=\sqrt{{\rm d}r^2 + (r\,{\rm d}\theta)^2 + (r \sin(\theta) \,{\rm d}\phi)^2}$$
In case of an arc on the surface of a sphere, $\,{\rm d}r=0\,$ and you have:
$${\rm d}s=r\sqrt{{\rm d}\theta^2 + (sin(\theta) \,{\rm d}\phi)^2}$$
Both still are the same Pythagorian theorem, if you look at them closely.
A: The magnitude of a vector in spherical coordinates is just $r$, by definition.
A: Not sure what is confusing you here. Based on the construction of spherical coordinates, , 
you see immediately that the magnitude (=length) of the vector is equal to $r$. You could say that this is part of the definition of spherical coordinates.
Alternatively, if you accept that the length in cartesian coordinates is, $l=\sqrt{x^2+y^2+z^2}$, you can just insert the transformations:
$$x=r\sin\theta\cos\phi\\
y=r\sin\theta\sin\phi
z=r\cos\theta$$
and obtain (using the identity $\sin^2\alpha+\cos^2\alpha = 1$):
$$l=\sqrt{r^2(\sin^2\theta\cos^2\phi+\sin^2\theta\sin^2\phi+\cos^2\theta)}\\
=r\sqrt{\sin^2\theta (\cos^2\phi+\sin^2\phi)+\cos^2\theta}\\
=r\sqrt{\sin^2\theta +\cos^2\theta}\\
=r$$
But this is not so much of a "proof" but rather confirms that the transformation is done properly.
