Find volume under given contraints on the Cartesian plane.

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 $rho^2+z^2leq 64, rho^2leq 16, rho^2leq z^2, zgeq 0$. 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 $2pi$). 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 $rho;drho;dz;dphi$. So the first will be

$$int_0^{2pi}int_0^4int_0^z 1times ; rho;drho;dz;dphi$$$

Similarly the second will be $$int_0^{2pi}int_4^{sqrt{48}}int_0^4 1times ; rho;drho;dz;dphi$$

and the third will be $$int_0^{2pi}int_{sqrt{48}}^8int_0^{sqrt{64-z^2}} 1times ; rho;drho;dz;dphi$$

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)

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.