# Volume of off-center cylindrical hole in Sphere

$S$ is a sphere of radius $2$, centered at origin. A cylindrical hole of radius $1$, centered at $(1,0)$ is drilled through $S$.

How much materials from $S$ was removed?

Attempt

$z = \sqrt{4 - x^2 - y^2}$ is the equation of the top half of the sphere.

Let $R$ be the region bounded by the equation of the intersection between the cylindrical hole and x-y plane.

Then, integrate the top half of the sphere over this region:

$$\int \int_R \sqrt{4 - x^2 - y^2} dx dy$$

Find the volume of that hole cut from the sphere, by finding one fourth of that volume first. As you noted, the area is as follows:

The correct volume is $$V=4\int_0^{\pi/2}\int_0^{r=2\cos(\theta)} \sqrt{4-r^2}rdrd\theta$$= 16/9 (3 π - 4) ≈ 9.64405

• Thanks! Especially the diagram! – Legendre Oct 11 '12 at 22:49
• The diagram is wonderful! How'd you make it? – Namaste Mar 27 '13 at 3:26

This is a complicated question. The "cap" volume is what is difficult to evaluate. As Jack mentioned, it leads to an elliptical integral. However, since Sphere Radius = 2*eccentricity = 2*hole radius, you fall in to a special case where all of the elliptical integrals disappear. This is also known as "Viviani's Curve". This link has the formulas for the volume of the remaianing sphere: http://diynovice.wordpress.com/2012/12/02/hole-in-sphere/

By translating the domain of integration and switching to polar coordinates I get:

$$V=4\int_{0}^{1}\int_{0}^{\pi}\rho\,\sqrt{3-\rho^2-2\rho\cos\theta}\,d\theta\,d\rho,$$

that leads to an elliptic integral.