The lower cone begins at $\phi=\dfrac{3\pi}{4}$ and runs all the way down to $\phi=\pi$. If the starting point for $\phi$ isn't obvious, then you can find it as follows:
Let $x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi, z=r\cos\theta$. Substituting this into the cone equation, squaring, and rearranging, you obtain:
$$r^2\cos^2\theta = r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi\iff \tan^2\phi=1.$$
Since we're in the lower octants, this means $\tan\phi=-1$, i.e. $\phi=\dfrac{3\pi}{4}$.
Obviously, $0\le \theta\le 2\pi$.
$\rho$ will fun from the origin to the hemisphere. $0 \le \rho\le 2$.
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 $3$.
Best Answer
If the integral is
$\displaystyle \int_{-7}^7\int_{-\sqrt{49-x^2}}^\sqrt{49-x^2}\int_{x^2+y^2}^{49} x \, dz \, dy \, dx$
We are basically integrating over the region bound between paraboloid $z = x^2 + y^2$ and plane $z = 49$.
It is a bit complicated in spherical coordinates and you have to split it into two regions.
$x = \rho \cos \theta \sin \phi$
$y = \rho \sin \theta \sin \phi$
$z = \rho \cos \phi$
In spherical coordinates, $\phi$ is the angle made with $z$ axis so at $z = 49$, $\phi$ varies from $0$ to $\cot \phi = \frac{49}{7} = 7 \implies \cot^{-1}(7)$ with no change in value of $z$.
Also as $z$ is constant, $\rho$ as a function of $\phi$ will be $0 \leq \rho \leq 49 \sec \phi$ (from equation of $z$).
Now for $\cot^{-1}(7) \leq \phi \leq \frac{\pi}{2}$, we are traversing till the paraboloid boundary
So, $z = x^2 + y^2 \implies \rho \cos \phi = \rho^2 \sin^2 \phi \implies \rho = \cot \phi \csc \phi$
So, $I = \iiint \rho \cos \theta \sin \phi.\rho^2 \sin \phi \, dr d\phi d \theta$
with two integrations in two parts with limits of integration -
i) $0 \leq \rho \leq 49 \sec \phi, 0 \leq \phi \leq \cot^{-1}(7), 0 \leq \theta \leq 2\pi$.
ii) $0 \leq \rho \leq \cot \phi \csc \phi, \cot^{-1}(7) \leq \phi \leq \frac{\pi}{2}, 0 \leq \theta \leq 2\pi$.