Moment of Inertia

Two points on line ${\bf z = \frac{h}{2} }$ in the ${\bf xz \: plane }$ are:

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${\bf A:(0,0,\frac{h}{2}) \: and \: B:(1,0,\frac{h}{2}) \implies }$

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Line ${\bf AB \: is: x = t, y = 0, z = \frac{h}{2} }$ , where the direction numbers of line ${\bf AB \:\ are \: \vec{u} = \vec{i} + 0\vec{j} + 0\vec{k} }$

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Let ${\bf P:(x,y,z) }$ be a point not on line ${\bf AB }$ and ${\bf \vec{v} = x\vec{i} + y\vec{j} + (z - \frac{h}{2})\vec{k} }$ be vector from point ${\bf A \: to \: P. }$

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The distance ${\bf d }$ from point ${\bf P }$ to line ${\bf AB }$ is: ${\bf d = \frac{|\vec{u} X \vec{v}|}{|\vec{u}|} }$

${\bf \vec{u} X \vec{v} = 0\vec{i} + -(z - \frac{h}{2})\vec{j} + y\vec{k} \: and \: |\vec{u}|^2 = 1 \implies}$

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${\bf d^2 = (z - \frac{h}{2})^2 + y^2 }$

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for ${\bf x^2 + y^2 = 1, (0 <= z <= h) \implies \vec{r}(u,z) = cos(u)\vec{i} + sin(u)\vec{j} + z\vec{k} }$ where ${\bf (0 <= u <= 2\pi) }$ and ${\bf (0 <= z <= h) \implies }$

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${\bf \vec{r_{u}} = -sin(u)\vec{i} + cos(u)\vec{j} + 0\vec{k}, }$ and ${\bf \vec{r_{z}} = 0\vec{i} + 0\vec{j} + \vec{k} \implies }$

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${\bf \vec{r_{u}} X \vec{r_{z}} = cos(u)\vec{i} + sin(u)\vec{j} + 0\vec{k} \implies |\vec{r_{u}} X \vec{r_{z}}| = 1}$

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${\bf \implies dA = |\vec{r_{u}} X \vec{r_{z}}| du \: dz = du \: dz }$ $$

and, ${\bf d^2 = (z - \frac{h}{2})^2 + y^2 = z^2 - hz + \frac{h^2}{4} + sin^2(u) \implies }$

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${\bf \int \int_{S} d^2 dA = \int_{0}^{h} \int_{0}^{2\pi} (z^2 - hz + \frac{h^2}{4} + sin^2(u)) du \: dz = }$

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${\bf \int_{0}^{h} ( (z^2 - hz + \frac{h^2}{4}) * u + \frac{1}{2} * (u - \frac{1}{2} sin(2u)) )|_{0}^{2\pi} dz= }$

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${\bf 2\pi * \int_{0}^{h} z^2 - hz + \frac{h^2}{4} + \frac{1}{2} \: dz = 2\pi * (\frac{z^3}{3} - \frac{h * z^2}{2} + \frac{h^2 * z}{4} + \frac{z}{2} )|_{0}^{h} = 2\pi (\frac{h^3}{12} + \frac{h}{2} ) }$

${\bf = \frac{\pi * h}{6} * (h^2 + 6) = \pi * h (\frac{h^2}{6} + 1) = \pi * h (\frac{h^2}{a} + b ) \implies a + b = 7. }$