A classical mechanics problem by Rocco Dalto

The points $O: (0,0,0)$ and $A: (1,1,0)$ are two points on the line $y = x, z = 0$ $$

Let point $B: (x,y,z)$ be any point not on the line above. $$

The distance from point $B$ to the line above is $d = \dfrac{|\vec{OB} \: X \: \vec{OA}|}{|\vec{OA}|}$
$$

$\vec{OB} = x \vec{i} + y \vec{j} + z \vec{k}$ $$

$\vec{OA} = \vec{i} + \vec{j} + 0 \vec{k}$ $$

$\vec{OB} \: X \: \vec{OA} = -z \vec{i} + z \vec{j} + (x - y) \vec{k}$

$\implies d = \sqrt{\dfrac{{2 z^2 + (x - y)^2}}{2}} = \sqrt{z^2 + \dfrac{1}{2} (x - y)^2}$ $$

$\vec{r}(\theta,z) = cos(\theta)\vec{i} + sin(\theta)\vec{j} + z\vec{k}$ where $(0 <= \theta <= 2\pi)$ and $(0 <= z <= h) \implies$

$$

$\dfrac{\partial \vec{r}}{\partial \theta} = -sin(\theta)\vec{i} + cos(\theta)\vec{j} + 0\vec{k},$ and $\dfrac{\partial \vec{r}}{\partial z} = 0\vec{i} + 0\vec{j} + \vec{k} \implies$

$$

$\vec{N} = \dfrac{\partial \vec{r}}{\partial \theta} \: X \: \dfrac{\partial \vec{r}}{\partial z} = cos(\theta)\vec{i} + sin(\theta)\vec{j} + 0\vec{k} \implies |\vec{N}| =1$ $$

$I = \int_{0}^{2\pi} \int_{0}^{h} (z^2 + \dfrac{1}{2} (cos\theta - sin\theta)^2)) \: dz \: d\theta =$ $$

$\int_{0}^{2\pi} \int_{0}^{h} (z^2 + \dfrac{1}{2} (1 - sin(2\theta))) \: dz \: d\theta =$ $$

$\int_{0}^{2\pi} (\dfrac{z^3}{3} + \dfrac{1}{2} (1 - sin(2\theta)) z)|_{0}^{h} \: d\theta =$ $$

$\int_{0}^{2\pi} (\dfrac{h^3}{3} + \dfrac{1}{2} (1 - sin(2\theta)) h) \: d\theta =$ $$

$\dfrac{h^3}{3} \theta + \dfrac{1}{2} (\theta + \dfrac{1}{2} cos(2\theta)) h|_{0}^{2\pi} =$ $$

$\dfrac{2 \pi}{3} h^3 + \pi h = \pi h(\dfrac{2}{3} h^2 + 1) = \pi h (\dfrac{a}{b} h^2 + c) \implies$ $$

$a + b + c = \boxed{6}$ .