Problem: A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue.  Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\circ$ .  The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half.  The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime.  Find $a+b+c$
[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3);  picture lpic, rpic;  size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8));  size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W);  add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]

Solution: This problem can be calculus-bashed (for those like me who never noticed the surface was merely a stretch of its projection). Label points as in the first paragraph of Solution 1 ( $A$ and $B$ as given, $M$ the midpoint of $AB$ $O$ the center of the cylinder, $T$ the center of the bottom face of the cylinder). Because of the 120 degrees and right triangle calculations, we find $MT$ = 3, $OT$ = 4, $OM$ = 5). We will be integrating with respect to the y-coordinate which we define as distance downwards from $O$ (in this system, the $y$ -coordinate of the bottom face would be 4).
We note that by similar triangles, we have that the length from $O$ to the point on the unpainted surface of coordinate $y$ is $\ell = \frac{5}{4} y$ , and therefore $d\ell = \frac{5}{4} dy$ . Define the segment $A'B'$ to be the intersection of the painted surface with the circular cross section of the cylinder of coordinate $y$ , with endpoints $A'$ and $B'$ and midpoint $M'$ , with $T'$ the center of this circular cross section. Then, by similar triangles, $T'M' = \frac{3}{4} y$ and thus \[A'B' = 2A' M' = 2 \sqrt{ 6^2 - \left( \frac{3}{4}y\right)^2 } = \frac{3}{2} \sqrt{ 64 - y^2 }\] . We know that $A'B'$ is perpendicular to $OM$
Now we can set up our integral: we will integrate $y$ from 0 to 4 and multiply by two because the total height is 8. \[A = 2\int_0^4 \left(\frac{3}{2}\sqrt{ 64 - y^2 }\right) \left( \frac{5}{4} dy\right)\] \[A = \frac{15}{4} \int_0^4 \sqrt{ 64 - y^2 }dy\]
Then we substitute $y = 8\sin \theta$ with $dy = 8 \cos \theta d \theta$ , changing the bounds to 0 to $\frac{\pi}{6}$ as appropriate. \[A = \frac{15}{4} \int_0^\frac{\pi}{6} \sqrt{ 64 - 64\sin^2 \theta }\cdot 8\cos\theta d\theta\] \[A = 240 \int_0^\frac{\pi}{6} \cos^2 \theta d\theta\] \[A = 240 \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^\frac{\pi}{6}  = 240 \left[ \frac{\pi}{12} + \frac{\sqrt{3}}{8} \right] = 20{\pi} + 30\sqrt{3}\]
Therefore, $a + b + c = 20 + 30 + 3 = \boxed{053}$

Answer:  053