Expert: Ahmed Salami - 12/12/2005

Question
I need to find the volume of the resulting solid:
Region bounded by y=sin(x), x=pi/6, x=pi/3, and y=0, which is revolved about the x-axis. Would my integral be from pi/6 to pi/3 of pi*(sinx)^2? If it is, I'm having problems solving that. I could have it setup incorrectly though too. Thanks for your help! I appreciate it!

Answer
Hi Emily,
Sorry for the time taken.
The integral is of course $pi.(sinx)^2 dx
To solve this, we make use of the trigonometric identity
cos2x = 1 - 2(sinx)^2 to give
(sinx)^2 = 1/2(1 - cos2x)
The integral therefore becomes
pi.$1/2(1 - cos2x) dx
= pi/2 [x - sin2x/2]pi/6 to pi/3
= pi/2 [(pi/3 - sqrt3 /4) - (pi/6 - sqrt3 /4)]
= pi/2 (pi/6)
= (pi)^2 / 12
I hope it is clear, you can always get back to me.
Regards.