Answer
$$\frac{\pi^2}{2}$$
Work Step by Step
y=sin(x)
so one arc of the curve y=sin(x) is half one period which can be from x=0 to x=$\pi$
volume generated by revolving = $\pi\int y^2 dx$
$y^2= sin^2(x)$
$\pi\int\sin^2(x)dx$
$\pi\int\frac{1}{2}-\frac{cos(2x)}{2}$
$\pi(\frac{x}{2}-\frac{sin(2x)}{4})+c$
then substitute as it is definit intgral his boundry are 0 and $\pi$
$\pi\times(\frac{\pi}{2}-\frac{sin2\pi}{4})- \pi\times(\frac{0}{2}-\frac{sin(0)}{4}) = \frac{\pi^2}{2}-0$
