Point lies inside of the sphere Let $a,\alpha,\beta$ be any real numbers with $a>0$. Can someone show and explain why the point say $(x,y,z)=(a\sin{\alpha}\cos{\beta},a\sin{\alpha}\sin{\beta},\sqrt{3}a\sin{\alpha})$ lies inside, or on, the sphere with a radius of $2a$ which is centered at the origin. 
 A: The equation of such this sphere is $x^2+y^2+z^2=(2a)^2=4a^2$. Now set $$x=a\sin\alpha\cos\beta, y=a\sin\alpha\sin\beta, z=\sqrt{3}~a\sin\alpha$$ in the equation and check if $x^2+y^2+z^2\le4a^2$
A: The distance $\ d$ of the point $\ (x,y,z)$ from the origin is
$$\ d=\sqrt{x^2+y^2+z^2}=\sqrt{a^2(\sin^2\alpha \cos^2\beta+\sin^2\alpha\sin^2\beta+3\sin^2\alpha)}=$$
$$\ =a\sqrt{4\sin^2\alpha}=2a|\sin\alpha|$$
So the point lies inside the sphere if $\ \alpha≠(\frac{2k+1}{2})\pi$
Instead, if $\ \alpha=(\frac{2k+1}{2})\pi$ then the point lies on the surface of the sphere
A: The distance of $(x,y,z)$ to the origin is
$d = \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2} = \sqrt{x^2 + y^2 + z^2 }$
$= \sqrt{(a\sin(\alpha)\cos(\beta))^2 + (a\sin(\alpha)\sin(\beta))^2 + (\sqrt{3}a\sin(\alpha)^2}$
$= \sqrt{a^2\sin(\alpha)^2\cos(\beta)^2 + a^2\sin(\alpha)^2\sin(\beta)^2 + (\sqrt{3})^2a^2\sin(\alpha)^2}$
$= \sqrt{a^2\sin(\alpha)^2 (\cos(\beta)^2 + \sin(\beta)^2) + 3a^2\sin(\alpha)^2}$
$= \sqrt{a^2\sin(\alpha)^2 + 3a^2\sin(\alpha)^2}$
$= \sqrt{4a^2\sin(\alpha)^2}$
$= 2a|\sin(\alpha)|$
Since $|\sin(\theta)| \leq 1$ for all $\theta\in \mathbb{R}$, we have $d \leq 2a$. 
So $(x,y,z)$ lies in or on the sphere.
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By the way, in case you want to separate points on the sphere and inside the sphere: $(x,y,z)$ lies on the sphere if and only if $d = 2a$. Since $d = 2a\sin(\alpha)$, $(x,y,z)$ lies on the sphere if and only if $\sin(\alpha) = 1$, so $\alpha = k\pi + \frac{\pi}{2}$ for $k \in \mathbb{Z}$. 
Otherwise (so for all other $\alpha$), we have $d < 2a$ and $(x,y,z)$ is inside of (and not on) the sphere.
