Finding unknown bounds for the cylindrical and spherical cases:
Cylindrical coordinates:
Notice that how far we must go out in the "$r$" direction (in $(r,\theta, z)$ space) is dependent on the value of $z$. If we're at the origin, $r$ has a maximum of $0$ because the vertex is a mere point, but at $z = 1/\sqrt{3}$, $r$ can go all the way out to the edge on the flat top of the cone. Imagine drawing a right triangle with height $z$ and hypotenuse along the outer edge of the cone. Knowing one of the angles of this triangle is $\pi/3$ allows us to "solve the triangle", giving us $\displaystyle r = \frac{\sqrt{3}}{z}$, meaning $r$ is going to run between $0$ and this $z$-dependent value.
Spherical coordinates:
We can take an approach similar to the above to find the bounds for $\rho$. Notice that the upper bound on $\rho$ for given values of $\theta$ and $\phi$ is dependent only on the value of $\phi$. For example, when $\phi = 0$, the maximum value $\rho$ can attain is at its shortest of them all ($0 \leq \rho \leq 1/\sqrt{3}$), but when $\phi = \pi/3$, we have $\rho$ able to attain its longest possible value for the whole cone. Again, draw a triangle as we did above. This time, the triangle will have a height of $1/\sqrt{3}$, and a known angle, allowing you to solve for the hypotenuse, which is $\rho$.
A tip for both of the above:
We are allowed, in this scenario, to rearrange the order of integration. As an example, $\displaystyle \iiint \text{ stuff } \ dr \ dz \ d\theta$ will be the same as $\displaystyle \iiint \text{ stuff } \ dz \ d \theta \ dr$. Let's take advantage of this to the fullest extent possible. In particular, in both cases above, theta will run between $0$ and $2 \pi$, and this is entirely independent of whatever the other variables are doing. So if you think about it, you'll notice that ordering the integrals so that the $d \theta$ is outermost, we'll have:
$$\displaystyle \iiint \text{ stuff }\ dz \ dr \ d\theta = 2 \pi \iint \text{ same stuff } dz \ dr$$
And just like that we have only $2$ integrals to worry about instead of $3$.