Understanding reasoning behind certain bounds in double integration. So I am looking at the following question

Find the volume of the ice-cream cone shape given by the region bounded between the upper half of the sphere $x^{2}+y^{2}+z^{2}= 16$ and the cone $z=\frac {1}{\sqrt{3}}\sqrt{x^{2}+y^{2}}$.

I can easily understand that we need to convert this into spherical coordinates, and can see that I will have $0 < p < 4$ and $0 < \theta < 2\pi$. But how do I know the bounds of integration for $\phi$?
In my notes, it is simply noted that the intersection when squaring the cone equation and replacing $z^{2}$ with the subsequent value leads to $x^{2} + y^{2} = 12$, which occurs in the plane $z=2$. What does this mean? Why does it occur in this plane? 
Further, it then notes that the cone equation forms an angle of $\frac{\pi}{3}$ with the z-axis, and thus we have $0 < \phi < \frac{\pi}{3}$. I don't want to have to draw out or plot my cone to see this, I would much rather know how to solve for my bounds on $z$ and subsequently $\phi$, but I'm struggling to understand where these values are coming from. 
 A: @muaddib has answered one part of your question so I'll answer the other part.  Actually, you've practically answered it yourself.
At z = 1 and y = 0 what is the value of x on the cone?  This will allow you to draw a triangle from which you can determine $\phi$ using basic trig.  You'll find that it is the $\pi/3$ value that your notes tell you it should be.
Now to integrate over the whole volume of the shape you need to integrate $\phi$ from the z-axis out to the surface of the cone.  Given the above, do you see what this range is?
A: We can solve for $z$ as follows.  On one hand you have
$$x^2 + y^2 + z^2 = 16$$
You can square both sides of your cone equation and then multiply by 3 to get
$$3z^2 = x^2 + y^2$$
So substituting the second equation into the first we obtain
$$3z^2 + z^2 = 16$$
This has two possible solutions $z = -2, 2$.  However, looking at the origin equation for the cone, we see that $z$ has to be positive so $z = 2$.  Now substituting back $z = 2$ into the first equation we obtain
$$x^2 + y^2 + 4 = 16$$
