A hole in the sphere - Part 3

Let us attach a reference frame to the hole, with its origin at the midpoint of the two intersections of the axis of the hole cylinder and the sphere. Let's choose the $z$ -axis to be along the axis of the hole cylinder, and the $y$ -axis to point vertically up, passing through the center of the sphere at $y = -\frac{R}{2}$ , and this also determines the $x$ -axis.

To find the area of the hole (yellow area), we'll move on the boundary of the hole in the $xy$ plane, which is a circle of radius $\frac{R}{2}$ , and find the $z$ coordinate of the intersection of the hole with the sphere. This is given by

$z = \sqrt{ R^2 - (\frac{R}{2} + \frac{R}{2} \sin \phi )^2 - (\frac{R}{2} \cos \phi )^2 }$

where $0 \le \phi \lt 2 \pi$ . Simplifying the above expression, we obtain,

$z = \sqrt{ \frac{R^2}{2} (1 - \sin \phi) } = \dfrac{R}{\sqrt{2}} \sqrt{1 - \sin \phi}$

Let $\psi = \frac{\pi}{2} - \phi$ , then

$z = \dfrac{R}{\sqrt{2}} \sqrt{1 - \cos \psi}$

Now, we can use the fact that $\sin^2 x = \frac{1}{2} (1 - \cos(2x) )$ , to write the above as,

$z = R | \sin \dfrac{\psi}{2} |$

So that, the area integral is,

$A = \displaystyle \int_0^{2 \pi} 2 (\dfrac{R}{2}) z(\psi) \hspace{6pt} d\psi = R^2\int_0^{2 \pi} | \sin \dfrac{\psi}{2} | \hspace{6pt} d\psi$

Applying a change of variable, $u = \dfrac{\psi}{2}$ , results in,

$A = \displaystyle = 2 R^2\int_0^{\pi } | \sin u | \hspace{6pt} d u =2 R^2\int_0^{\pi } \sin u \hspace{6pt} d u = 4 R^2$

Substituting $R = 10$ , gives us, $A = \boxed{400}$ .