Graph of $y = a \sin \dfrac{\pi x}{L}$
Differential area
$dA = y \, dx$
$dA = a \sin \dfrac{\pi x}{L} \, dx$
Area by integration
$\displaystyle A = a \int_0^L \sin \dfrac{\pi x}{L} \, dx$
$\displaystyle A = \dfrac{aL}{\pi} \int_0^L \left( \sin \dfrac{\pi x}{L} \right)\left( \dfrac{\pi}{L} \, dx \right)$
$A = \dfrac{aL}{\pi} \left[ -\cos \dfrac{\pi x}{L} \right]_0^L$
$A = -\dfrac{aL}{\pi} \left[ \cos \pi - \cos 0 \right]$
$A = -\dfrac{aL}{\pi} \left[ -2 \right]$
$A = \dfrac{2aL}{\pi}$
Location of centroid
$\displaystyle A \, \bar{y} = \int_a^b y_c \, dA$
$\displaystyle \dfrac{2aL}{\pi} \, \bar{y} = \int_0^L (\frac{1}{2}y)(y \, dx)$
$\displaystyle \dfrac{2aL}{\pi} \, \bar{y} = \frac{1}{2}\int_0^L y^2 \, dx$
$\displaystyle \dfrac{2aL}{\pi} \, \bar{y} = \frac{1}{2}\int_0^L \left( a \sin \dfrac{\pi x}{L} \right)^2 \, dx$
$\displaystyle \dfrac{2aL}{\pi} \, \bar{y} = \dfrac{a^2}{2}\int_0^L \sin^2 \dfrac{\pi x}{L} \, dx$
From
$\cos 2\theta = 1 - 2\sin^2 \theta$
$\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)$
Assign $\theta = \dfrac{\pi x}{L}$
Thus,
$\displaystyle \dfrac{2aL}{\pi} \, \bar{y} = \dfrac{a^2}{2}\int_0^L \frac{1}{2} \left( 1 - \cos \dfrac{2\pi x}{L} \right) \, dx$
$\dfrac{2aL}{\pi} \, \bar{y} = \dfrac{a^2}{4}\left[ x - \dfrac{L}{2\pi}\sin \dfrac{2\pi x}{L} \right]_0^L$
$\dfrac{2aL}{\pi} \, \bar{y} = \dfrac{a^2}{4}\left[ \left( L - \dfrac{L}{2\pi}\sin 2\pi \right) - \left( 0 - \dfrac{L}{2\pi}\sin 0 \right) \right]$
$\dfrac{2aL}{\pi} \, \bar{y} = \dfrac{a^2}{4}\left[ L \right]$
$\dfrac{2aL}{\pi} \, \bar{y} = \dfrac{a^2L}{4}$
$\bar{y} = \dfrac{\pi a}{8}$ answer
