A hole in the cylinder - Part 3

Let us set up a reference frame at the center of the hole with its z-axis being along the axis of the hole cylinder, and its x-axis horiztonal and y-axis vertical upward. From a point $(x, y, 0)$ , the corresponding point on the surface of the drilled cylinder is $z = f(x,y) = \sqrt{R^2 - x^2}$ . Taking a point on the boundary of the hole, then $x = R \cos \phi$ and the area integral expression is

$A = \displaystyle \int_{\phi = 0 }^{\phi = 2 \pi } 2 R \sqrt{R^2 - R^2 \cos^2 \phi} d \phi$

Simplifying and integrating,

$A = 2 R^2 \displaystyle \int_{\phi = 0}^{\phi = 2 \pi} | \sin \phi | d \phi = 8 R^2$

Plugging in $R = 6$ , we get $A = 288$ .