Intuition for spherical coordinates of a highly symmetric point Consider the sphere $\{(x,y,z)\in{\mathbb{R}^3}:x^2+y^2+z^2=1\}$ and the point on the sphere $(x,y,z)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Intuitively, I thought &quot;oh, that point lies on the &quot;dead center&quot; of the sphere as far the first octant is concerned&quot;, so I thought in spherical coordinates the same point should be $\left(r,\theta,\varphi)=(1,\frac{\pi}{4},\frac{\pi}{4}\right)$, but later when I was getting some inconsistent calculations I came back to this statement and when I applied the formal conversions:
$$x(r,\theta,\varphi)=r\sin(\theta)\cos(\varphi)$$
$$y(r,\theta,\varphi)=r\sin(\theta)\sin(\varphi)$$
$$z(r,\theta)=r\cos(\theta)$$
I got that the point $(r,\theta,\varphi)=\left(1,\frac{\pi}{4},\frac{\pi}{4}\right)$ should actually be $(x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$, and this has been driving me insane! Either there's some really silly arithmetic error I'm making or my geometric intuition is failing me. Which one is it?
Your intuition is failing you.  If you do the conversion, the point $(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$ you first need $\cos (\theta) =\frac 1{\sqrt 3}$, which says $\theta\approx 0.955$.  You do have $\phi=\frac \pi 4$
The sphere intersects the $x$, $y$, and $z$ axis at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. To see if a point is in the &quot;dead center&quot; you can find the distance to each of these points and see if they are the same. For $(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}})$, the distances are $1$, $1$, $\sqrt{2-\sqrt{2}}$, so it isn't in the &quot;dead center&quot;. However, for $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$, the distances are all $\sqrt{2-\frac{2}{\sqrt{3}}}$.