Question

Find the exact area of the surface obtained by rotating the curve about the x-axis.

Answer 1

$\bf \begin{cases} y=sin\left( \frac{\pi x}{6} \right)\\ 0\le x\le 6\\ \textit{about the x-axis, or }y=0 \end{cases}\\\\ -----------------------------\\\\ \textit{using the disc method} \\\\ V=\int\limits_{0}^{6}\pi \left[ sin\left( \frac{\pi x}{6} \right) \right]^2\cdot dx\implies \pi\int\limits_{0}^{6} \left[ sin\left( \frac{\pi x}{6} \right) \right]^2\cdot dx \\\\\\ \pi\int\limits_{0}^{6} sin^2\left( \frac{\pi x}{6} \right) \cdot dx\\\\ -----------------------------$

$\bf \textit{now, let us check the double angle identities} \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ \boxed{1-2sin^2(\theta)}\\ 2cos^2(\theta)-1 \end{cases} \\\\\\ thus\implies cos(2\theta)=1-2sin^2(\theta)\implies 2sin^2(\theta)=1-cos(2\theta) \\\\\\ sin^2(\theta)=\cfrac{1-cos(2\theta)}{2}\qquad thus\\\\ -----------------------------$

$\bf \pi\int\limits_{0}^{6} \left[ sin\left( \frac{\pi x}{6} \right) \right]^2\cdot dx\implies \pi\int\limits_{0}^{6} \cfrac{1-cos\left(2\cdot \frac{\pi x}{6} \right)}{2}\cdot dx \\\\\\ \pi\int\limits_{0}^{6}\cfrac{1}{2}dx-\pi \cdot \cfrac{1}{2}\pi\int\limits_{0}^{6}cos\left(\frac{\pi x}{3} \right)dx \\\\\\ \left[\cfrac{\pi x}{2}-\cfrac{\pi }{2}\cdot \cfrac{sin\left(\frac{\pi x}{3} \right)}{\frac{\pi x}{3} } \right]\implies \left[ \cfrac{\pi }{2}x-\cfrac{3sin\left(\frac{\pi x}{3} \right)}{2x} \right]_0^6$

and surely, you'd know how to get the values for the bounds there