# Second moment of area under arbitrary arc

\begin{align} I_{XX} = \frac{1}{96}R( & - (32a^{3} + 72aR^{2}) (\cos \theta_{2} - \cos \theta_{1} ) \\ & + (48a^{2}R + 12R^{3}) (\theta_{2} - \theta_{1}) \\ & - (24a^{2}R + 8R^{3}) (\sin 2\theta_{2} - \sin 2\theta_{1} ) \\ & + 8aR^{2} (\cos 3\theta_{2} - \cos 3\theta_{1} ) \\ & + R^{3} (\sin 4\theta_{2} - \sin 4\theta_{1} )) \end{align}

### Derivation

Second moment of area of that subtends angle $$d\theta$$ about $$X$$ axis

$dI_{XX} = - \frac{y^{3}}{3} dX\label{A}\tag{Eq.1}$

$$X$$ co-ordinate of arc

\begin{align} X & = R \cos \theta \\ \frac{dX}{d\theta} & = - R \sin \theta \\ - dX & = R \sin \theta \, d\theta\label{B}\tag{Eq.2} \\ \end{align}

$$Y$$ co-ordinate of arc

$y = a + R \sin \theta\label{C}\tag{Eq.3}$

Substitute $$\ref{B}$$ and $$\ref{C}$$ into $$\ref{A}$$

$dI_{XX} = \frac{1}{3}(a + R \sin \theta)^{3}(R \sin \theta)\, d\theta$

which integrates between $$\theta_{1}$$ and $$\theta_{2}$$ to:

\begin{align} I_{XX} = \frac{1}{96}R( & - (32a^{3} + 72aR^{2}) (\cos \theta_{2} - \cos \theta_{1} ) \\ & + (48a^{2}R + 12R^{3}) (\theta_{2} - \theta_{1}) \\ & - (24a^{2}R + 8R^{3}) (\sin 2\theta_{2} - \sin 2\theta_{1} ) \\ & + 8aR^{2} (\cos 3\theta_{2} - \cos 3\theta_{1} ) \\ & + R^{3} (\sin 4\theta_{2} - \sin 4\theta_{1} )) \end{align}