A hole in the cylinder

One cross-section that is $x$ away from the center of the cylinder is as follows:

By symmetry $x = OF = OE$ , and since the radius is $6$ , $OA = OB = OC = OD = 6$ .

By trigonometry on $\triangle OFC$ , $FC = \sqrt{36 - x^2}$ and $\angle COF = \cos^{-1} (\frac{x}{6})$ , which means $\angle COD = 2 \cos^{-1} (\frac{x}{6})$

The area of sector $COD$ is then $A_{COD} = \frac{1}{2} \cdot 6^2 \cdot 2 \cos^{-1} (\frac{x}{6}) = 36 \cos^{-1} (\frac{x}{6})$

And the area of $\triangle BOC$ is then $A_{\triangle BOC} = \frac{1}{2} \cdot 2x \cdot \sqrt{36 - x^2} = x \sqrt{36 - x^2}$ .

The area of the pink cross-section is therefore $A_{\text{pink}} = 2A_{COD} + 2A_{\triangle BOC} = 72 \cos^{-1} (\frac{x}{6}) + 2x \sqrt{36 - x^2}$ .

The volume of the section drilled out of the cylinder is then $2 \int_{0}^{6} (72 \cos^{-1} (\frac{x}{6}) + 2x \sqrt{36 - x^2}) dx = 4[\frac{1}{3}(x^2 - 144)\sqrt{36 - x^2} + 36x \cos^{-1}(\frac{x}{6})]^6_0 = \boxed{1152}$ .

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.

Then the volume of the hole is,

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

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

$V = \displaystyle \int_{x=-R}^{R} 4 (R^2 - x^2) dx = 4 ( 2 R^3 - \frac{2}{3} R^3 ) = \dfrac{16}{3} R^3$

Plugging in $R = 6$ , we get $V = 1152$ .