I'm not sure how to approach the problem, though I know it needs to be broken up into two integrals before it can be evaluated based on the way the answer input is set up. I do not know how to start visualizing what is going on correctly though, so any help on that front would be greatly appreciated.
Problem: $$\int_0^2 \int_0^\sqrt{4 - x^2} \int_0^\sqrt{16 - x^2 - y^2} \sqrt{x^2 + y^2} \, dz \, dy \, dx$$
On
Consider an integral of this form: $$ \int_0^4 \int_0^{\sqrt{16-x^2}} \int_0^{\sqrt{16-x^2-y^2}} f(x,y,z) \, dz\, dy\, dx. \tag1 $$ The region of integration of $f(x,y,z)$ in this integral is the portion of a sphere of radius $4$ around the origin for which $x>0$, $y>0$, and $z>0$. \tag2
Next, consider an integral of the form $$ \int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^4 f(x,y,z) \, dz\, dy\, dx. $$ The region of integration of $f(x,y,z)$ in this integral is the portion of a cylinder of radius $2$ and height $4$ around the $z$-axis for which $x>0$ and $y>0$.
Compare your region of integration. It is smaller than the sphere, because it eliminates all the points where $x>2$, and when $x\leq 2$ it eliminates all the points where $y > \sqrt{4-x^2}$. That is, it eliminates everything outside a cylinder of radius $2$ around the $z$-axis. But your region of integration is smaller than the cylinder of radius $2$ and height $4$, because when $x>0$ or $y>0$, the upper bound of $z$ is $\sqrt{16-x^2-y^2} < 4$.
What you actually have is the intersection of the two regions of integration in formulas $(1)$ and $(2)$. If you take a ray straight out from the origin, depending on the direction in which you go, you're either going to hit the surface of the sphere first, or you're going to hit the surface of the cylinder first, and when either of those happens the ray passes outside the region of integration, so that point is at your maximum value of $\rho$ in that direction.
The reason you need two integrals is to have one for all the directions where $\rho$ is limited by the surface of the sphere, and another for the directions where $\rho$ is limited by the surface of the cylinder.
Hint: If $x \in [0, 2]$, then: $$ y \leq \sqrt{4 - x^2} \iff y^2 \leq 4 - x^2 \iff x^2 + y^2 \leq 4 \iff r^2 \leq 4 \iff r \leq 2 $$ Note that in the $xy$-plane, $\{(x, y) \mid x \in [0, 2] \text{ and } y \in [0, \sqrt{4 - x^2}\}$ represents the Quadrant I section of a circle centred at the origin with radius $2$ (that is, it is the top-right quarter circle). So what should $\theta$ range between?
When switching from Cylindrical to Spherical, imagine taking a cross-section by slicing the region at an arbitrary angle using the half-plane $\theta$ to obtain an $rz$-plane. In this cross-section, we want: $$ S = \{(r, z) \mid r \in [0, 2] \text{ and } z \in [0, \sqrt{16 - r^2}]\} $$ That is, we are taking a quarter-circle of radius $4$ and making a vertical slice at $r = 2$ and keeping the left piece. Now because $(0, 4), (2, 0) \in S$, we are forced to let $\phi \in [0, \pi/2]$. As for $\rho$, it's a bit tricker as we transition from the arc $z = \sqrt{16 - r^2}$ to the line $r = 2$, which intersect at the threshold $(2, \sqrt{12})$. Notice that at this threshold, we have $\rho = 4$, so: $$ 2 = r = \rho\sin\phi = 4\sin\phi \implies \phi = \frac{\pi}{6} $$ Hence, for $\phi \in [0, \pi/6]$, the outermost boundary for $\rho$ is: $$ z = \sqrt{16 - r^2} \iff r^2 + z^2 = 4^2 \iff \rho = 4 $$ while for $\phi \in [\pi/6, \pi/2]$, the outermost boundary for $\rho$ is: $$ r = 2 \iff \rho\sin\phi = 2 \iff \rho = 2\csc\phi $$
Putting it all together, we obtain: \begin{align*} &\int_0^2 \int_0^\sqrt{4 - x^2} \int_0^\sqrt{16 - x^2 - y^2} \sqrt{x^2 + y^2} \, dz \, dy \, dx \\ &= \int_0^{\pi/2} \int_0^2 \int_0^\sqrt{16 - r^2} r (r \, dz \, dr \, d\theta) \\ &= \int_0^{\pi/2} \int_0^{\pi/2} \int_0^{\min\{4, 2\csc\phi\}} (\rho \sin \phi)(\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta) \\ &= \int_0^{\pi/2} \left[ \int_0^{\pi/6} \int_0^{4} \rho^3 \sin^2 \phi \, d\rho \, d\phi + \int_{\pi/6}^{\pi/2} \int_0^{2\csc\phi} \rho^3 \sin^2 \phi \, d\rho \, d\phi \right] d\theta \end{align*}