Triple integrals, change of coordinates - how understand new area of integration

Question

$K=\{(x,y,z)\in \mathbb{R}^{3}|1\leq x^{2}+y^{2}+z^{2}\leq 2,x+y\geq0,\sqrt3x-y\leq0,z\geq0\} \rightarrow \\ \rightarrow\{spherical\quad coordinates\}\rightarrow K:1\leq r\leq \sqrt2,\quad0\leq \theta\leq \pi/2,\quad\pi/3\leq\phi\leq3\pi/4.$

My question is how we arrive at these inequalities/intervals for the new variables? I can understand the $r$ part, just plug all new expressions into the first inequality and we get the answer. As for the others, I don't see it. Some kind combinations of the inequalities?

Edit:

Spherical coordinates $ :\\ x=r\sin{\theta}\cos{\phi}\\ y=r\sin{\theta}\sin{\phi}\\ z=r\cos{\theta}$

Answer 1

Since $z \ge 0$. $\theta$, the angle from the vertical axis, covers from the peak of the sphere to the equator. Hence $$0 \le \theta \le \frac{\pi}2.$$

Now to adress $\phi$,the angle on the $x-y$ plane that measures in a counter clockwise manner from the positive $x$-direction, a plot helps. We plot the region $\{(x,y): x+y \ge 0, \sqrt3x - y \le 0\}$

$$\tan^{-1}\left(\sqrt{3} \right) \le \phi\le\tan^{-1}\left( -1\right)$$

$$\frac{\pi}3 \le \phi \le \frac{3\pi}4$$

Answer 2

Hints: $$ x = -y\implies\cos\phi = -\sin\phi\implies\tan\phi = -1\implies\phi = \cdots $$ $$ \sqrt3 x = y\implies\sqrt3\cos\phi = \sin\phi\implies\tan\phi = \sqrt3\implies\phi = \cdots $$ $$\cdots$$