[tex]\bf \begin{cases}
y=sin\left( \frac{\pi x}{6} \right)\\
0\le x\le 6\\
\textit{about the x-axis, or }y=0
\end{cases}\\\\
-----------------------------\\\\
\textit{using the disc method}
\\\\
V=\int\limits_{0}^{6}\pi \left[ sin\left( \frac{\pi x}{6} \right) \right]^2\cdot dx\implies \pi\int\limits_{0}^{6} \left[ sin\left( \frac{\pi x}{6} \right) \right]^2\cdot dx
\\\\\\
\pi\int\limits_{0}^{6} sin^2\left( \frac{\pi x}{6} \right) \cdot dx\\\\
-----------------------------\\\\[/tex]
[tex]\bf \textit{now, let us check the double angle identities}
\\\\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\\
2cos^2(\theta)-1
\end{cases}
\\\\\\
thus\implies cos(2\theta)=1-2sin^2(\theta)\implies 2sin^2(\theta)=1-cos(2\theta)
\\\\\\
sin^2(\theta)=\cfrac{1-cos(2\theta)}{2}\qquad thus\\\\
-----------------------------[/tex]
[tex]\bf \pi\int\limits_{0}^{6} \left[ sin\left( \frac{\pi x}{6} \right) \right]^2\cdot dx\implies
\pi\int\limits_{0}^{6} \cfrac{1-cos\left(2\cdot \frac{\pi x}{6} \right)}{2}\cdot dx
\\\\\\
\pi\int\limits_{0}^{6}\cfrac{1}{2}dx-\pi \cdot \cfrac{1}{2}\pi\int\limits_{0}^{6}cos\left(\frac{\pi x}{3} \right)dx
\\\\\\
\left[\cfrac{\pi x}{2}-\cfrac{\pi }{2}\cdot \cfrac{sin\left(\frac{\pi x}{3} \right)}{\frac{\pi x}{3} } \right]\implies \left[ \cfrac{\pi }{2}x-\cfrac{3sin\left(\frac{\pi x}{3} \right)}{2x} \right]_0^6[/tex]
and surely, you'd know how to get the values for the bounds there
