A hole in the cylinder - Part 2

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. Taking point $(x, y, 0)$ , then the corresponding point on the surface of the drilled out surface is $z = f(x,y) = \sqrt{R^2 - x^2}$

The surface area hole is given by,

$A = \displaystyle \int_{x=-R}^{R} \int_{y=-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}} 2 \sqrt{1 + f_x^2 + f_y^2} dy dx$

We have $f_x = \dfrac{-x}{\sqrt{R^2 - x^2} }$ and $f_y = 0$ . Plugging these in the above expression and simplifying, we get,

$A = \displaystyle \int_{x=-R}^{R} \int_{y=-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}} 2 \dfrac{R}{\sqrt{R^2 - x^2}} dy dx$

Integrating with respect to $y$ first, we obtain,

$A = \displaystyle \int_{x=-R}^{R} 4 R dx = 8 R^2$

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