Originally Posted by
HallsofIvy The line y= 1/2 crosses the curve y= sin x when sin x= 1/2 which is at $\displaystyle x= \pi/6$, $\displaystyle x= \pi- \pi/6= 6\pi/6$, $\displaystyle \pi+ \pi/6= 7\pi/6$ and $\displaystyle 2\pi- \pi/6= 11\pi/6$. Also, perhaps by looking at a graph, you should be able to see that some of those areas are exactly the same and that the total area is 4 times the area from x= 0 to $\displaystyle \pi/6$ plus 2 times the area from $\displaystyle \pi/6$ to $\displaystyle \5\pi/6$. Those two have to be done separately because between x= 0 and $\displaystyle \pi/6$ 1/2 is lower than sin x and between $\displaystyle x= \pi/6$ and $\displaystyle 5\pi/6$, 1/2 is higher than sin x.
Now, since you put this in "geometry" rather than "calculus", what do you know how to do? If you have learned how to "integrate" then you should know this area is $\displaystyle 4\int_0^{\pi/6} sin(x)- 1/2 dx+ 2\int_{\pi/6}^{5\pi/6} 1/2- sin(x) dx$.
If you do not know how to integrate then I don't think it will be possible for you to do this problem.