The constrains are given as
$$x^2+y^2+z^2 \leqslant 64,\,x^2+y^2\leqslant 16,\,x^2+y^2\leqslant z^2,\,z\geqslant 0\,.$$
With the goal of finding the Volume. Personally, I have trouble interpreting the constrains in terms of integrals to find the Volume. However, logically z can be maximum of 8 and x or y no larger than 4.
On
Split the volume into three parts: cone, cylinder, and spherical cap:
The formulas for these three-dimensional figures are well known:
$$V_{cone} = \frac{1}{3} \pi r^2 h$$
$$V_{cylinder} = \pi r^2 h$$
$$V_{cap} = \frac{1}{3} \pi a^2 (3 R - a)$$
where $R$ is the radius of the sphere and $a$ is the distance between the plane "cutting" the cap and the center of the sphere.
All these variables are found easily through simple high-school algebra.
The problem is simpler if you convert to cylindrical coordinates. In this coordinate system, we have $\rho=\sqrt{x^2+y^2}$, $z=z$, and $\phi=\textrm{arctan}\left(\frac{y}{x}\right)$ (sort of - see the wikipedia article for a better explanation of how $\phi$ relates to $x$ and $y$). An important point is that $\rho$ is taken to be positive, while $\phi$ dictates direction completely.
Under this system, we have the bounds $\rho^2+z^2\leq 64, \rho^2\leq 16, \rho^2\leq z^2, z\geq 0$. From these inequalities, we have that $z$ may range anywhere from $0$ to $8$, and there is no restriction on $\phi$ (so it ranges from $0$ to $2\pi$). Combining the inequalities involving $\rho$, we get that as $z$ ranges from $0$ to $4$, $\rho$ ranges from $0$ to $z$. Then, as $z$ ranges from $4$ to $\sqrt{48}$, $\rho$ ranges from $0$ to $4$. Finally, as $z$ ranges from $\sqrt{48}$ to $8$, $\rho$ ranges from $0$ to $\sqrt{64-z^2}$.
We can write each of these three volumes as triple integrals. The volume element in cylindrical coordinates, $dV$, is $\rho\;d\rho\;dz\;d\phi$. So the first will be
$$\int_0^{2\pi}\int_0^4\int_0^z 1\times \; \rho\;d\rho\;dz\;d\phi$$$
Similarly the second will be $$\int_0^{2\pi}\int_4^{\sqrt{48}}\int_0^4 1\times \; \rho\;d\rho\;dz\;d\phi$$
and the third will be $$\int_0^{2\pi}\int_{\sqrt{48}}^8\int_0^{\sqrt{64-z^2}} 1\times \; \rho\;d\rho\;dz\;d\phi$$
Solve each triple integral from the inside out, add the three together, and you'll have your answer.
(feel free to comment or edit for any correction or suggestion)