Find the volume of a sphere with triple integral Consider this figure:

As shown in the figure. The center of the sphere is at $(0,0,-2)$ and the radius of the sphere is $4$. My question is how to calculate the volume of the sphere when $z\geq 0$ (upper half space). How can i apply triple integral on this problem?
My idea is shifting the sphere to above up to $2$. First i need to find the angle between the cone and $xy$ plane. That is:
$$\arcsin\left(\frac 12\right) = \frac{\pi}{6}$$
It means the complement of the angle is $\frac{\pi}{3}$ and calculating this integral:
$$\int_0^{\frac{\pi}{3}} \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \rho^2 \sin(\phi) \,\Bbb d\rho\,\Bbb d\theta\, \Bbb d\phi$$
Finally, the volume of the sphere when $z\geq 0$ is that integral minus the volume of the cone. Anyway, do you have the shortest way to solve this problem using triple integral that requires one step only? (Mine is 2 steps, calculate the integral and the cone)
 A: You can do it with cylindrical coordinates. Note that\begin{align}x^2+y^2+(z+2)^2\leqslant16&\iff\rho^2+z^2+4z\leqslant12\\&\iff\rho^2\leqslant12-4z-z^2.\end{align}So, compute$$\int_0^{2\pi}\int_0^2\int_0^\sqrt{12-4z-z^2}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta.$$You will get $\frac{40\pi}3$.
A: Can you integrate an interior of a circle with a double integral so that you calculate its area $S(r)$ for any given $r$...?
If so, you can present the problem by integrating 'horizontal' slices of a sphere by $dz$ above the $XY$ plane.
By a Pythagoras' theorem, the radius of a $z=\text{const.}$ section of a sphere at a chosen $z$ is $$r(z)=\sqrt{R^2-(z-z_A)^2}$$ where $R=8$ is a radius of a sphere and $z_A=-2$ is a height of the sphere's center.
Then the volume of the upper spherical cap is $$V_{z>0}=\int\limits_{z=0}^{R+z_A}S(r(z))\,dz$$
Just insert a double integral over the $XY$ circle with radius $r$ for $S(r)$ and you'll have a triple integral.
Which doesn't mean you must integrate all three at once. You can resolve them one by one from the innermost outwards.
