A classical mechanics problem by Rocco Dalto

The points $O: (0,0,\dfrac{h}{2})$ and $A: (x,0,\dfrac{h}{2})$ are two points on the line $z = \dfrac{h}{2}$ in the $xz$ plane. $$

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

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

$\vec{OB} = x \vec{i} + y \vec{j} + (z - \dfrac{h}{2}) \vec{k}$ $$

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

$\vec{OB} \: X \: \vec{OA} = 0 \vec{i} + x(z - \dfrac{h}{2}) \vec{j} + xy \vec{k}$

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

$\vec{r}(\theta,z) = z cos\theta \vec{i} + z sin\theta \vec{j} + z \vec{k}$ $$

$\dfrac{\partial \vec{r}}{\partial \theta} = -z sin\theta \vec{i} + z cos\theta \vec{j} + 0 \vec{k}$ $$

$\dfrac{\partial \vec{r}}{\partial z} = cos\theta \vec{i} + sin\theta \vec{j} + \vec{k}$ $$

$N = \dfrac{\partial \vec{r}}{\partial \theta} \: X \: \dfrac{\partial \vec{r}}{\partial z} = z cos\theta \vec{i} + z sin\theta \vec{j} - z \vec{k} \implies |N| = \sqrt{2} z$ $$

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

$\sqrt{2} \int_{0}^{2\pi} \int_{0}^{h} (z^3 - h z^2 + \dfrac{h^2}{4} z + z^3 sin^2(\theta) ) \: dz \: d\theta =$ $$

$\sqrt{2} \int_{0}^{2\pi} (\dfrac{z^4}{4} - \dfrac{h z^3}{3} + \dfrac{h^2 z^2}{8} + \dfrac{z^4 sin^2(\theta)}{4})|_{0}^{h} \: d\theta =$ $$

$\sqrt{2} \int_{0}^{2\pi} (\dfrac{h^4}{24} + \dfrac{h^4}{8} (1 - cos(2\theta))) \: d\theta =$ $$

$\sqrt{2} \: (\dfrac{h^4}{24} \theta + \dfrac{h^4}{8} (\theta - \dfrac{1}{2} sin(2\theta)))|_{0}^{2\pi} =$ $$

$2 \sqrt{2} \pi (\dfrac{h^4}{24} + \dfrac{h^4}{8})$ =
$\sqrt{2} \pi (\dfrac{h^4}{3}) = 2 \pi (\dfrac{h^4}{\sqrt{2} * 3}) = a \pi (\dfrac{h^4}{\sqrt{a} * b})$ $$

$2a + b = \boxed{7}$ .