Given $G = \{(x,y,z) | x^2 + y^2 + z^2 \le 1, x \ge 0, y \ge 0, z \ge 0\}$ find angles Given $G = \{(x,y,z) | x^2 + y^2 + z^2 \le 1, x \ge 0, y \ge 0, z \ge 0\}$ find angles and find radius
Well, In the solution it says that it is eigth of a ball, and angles are from $0$ to $90$ degrees, both angles (teta and $\phi$). and radius is from $0$ to $1$ which is true because $r = x^2 + y^2 + z^2 \le 1$.
But, I don't understand how they calculated 90 degrees. If it was 4th a ball, then I could understand that it is 90. but for me it seems like $180/8 = 22.5 $degrees.
What am I missing here?
Edit: Maybe it is because $sin$, $cos$ are positive ($\ge 0$) from $0$ to $90$ degrees, that why this is the range?
 A: Set $x=\rho\cos\theta\sin\phi$ , $x=\rho\sin\theta\sin\phi$ and $z=\rho\sin\theta\cos\phi$. In spherical coordinate we have
$$\rho\ge 0$$ 
$$0\le \theta\le 2\pi$$ 
$$0\le \phi \le \pi$$ 

If $x,y\ge 0$ then $0\le \theta\le \frac \pi 2$
If $z\ge 0$ then $0\le \phi\le \frac \pi 2$
A: Basically the comment in your edit is correct:
Given the usual spherical coordinates
\begin{align*}x& =r \, \sin\theta \, \cos\varphi\\
y& =r \, \sin\theta \, \sin\varphi \qquad r\in[0,R],\theta\in[0,\pi),\varphi\in[0,2\pi)\\ z&=r \, \cos\theta,
\end{align*}
you immediately see that in order to get a positive $z$ coordinate you need $\theta$ to lie in $[0,\pi/2],$ but then $\sin(\theta)$ is also positive there, and so you need to have $\varphi$ also in the first quadrant $\left([0,\pi/2]\right)$ to have both $\cos\varphi,\sin\varphi$ simultaneously positive.
A: A geometric argument like the one Behrouz Maleki gives is the easiest way to solve this,but I think it's worth the effort to seeing how a calculational argument would look. 
The simplest way to do this is to convert the rectangular coordinates of the sphere to spherical coordinates. 
(1) $x^2 + y^2+ z^2 \leq 1$ 
(2)$\rho^2\leq 1$
(3) x =$\rho\sin(\theta)\cos(\phi)\geq 0$ 
(4) y=$\rho\sin(\theta)\sin(\phi)\geq 0$
(5) z=$\rho\cos(\phi)\geq 0$
We need $0\leq \rho \leq 1$ and we're given that $x,y,z \geq 0$.Using these bounds in equation (5) yields  $0\leq\cos(\phi)\leq 1$ or $0\leq\phi\leq \frac{\pi}{2}$.Now square (3) and (4),add them and let z=0.This gives: 
(6)  $x^2 + y^2 =\rho^2\sin^2(\theta)\cos^2(\phi)+ \rho^2\sin^2(\theta)\sin^2(\phi)$ $\rightarrow $ 
$\rho^2\sin^2(\theta)(\cos^2(\phi)+ sin^2(\phi))= \rho^2\sin^2(\theta)$  
Substituting the value ranges of $\rho$ and $\psi$ give: 
  $0\leq \sin^2(\theta)\leq 1$ $\rightarrow $ $|sin(\theta)|\leq 1$ 
And this implies $0\leq \theta\leq \frac{\pi}{2}$.  
