Thushan I need your brain.
There's a thread about this already but hopefully this thread has more traffic and will get your attention more:
Question 11
The region in the first quadrant enclosed by the curve y=sin(x), the line y=0 and the line x=pi/6 is rotated about the x-axis. Find the volume of the resulting solid of revolution.
Now most people I know rotated the area from 0 to pi/6 but the limit in the question isn't clear. Can't the region in the first quadrant also be from pi/6 to pi/2 since it didn't say it had to be bounded by the y-axis (x=0)? In this case, either we had to give two cases, or VCAA really screwed up.
What's your take on it Thushan?
We'll wait on Thushan's answer to this. However, few things I just want to point out...
1. pi/6 to pi/2 is NOT bound. There is no line at x = pi/2 as this was not stated. Hence, the very word bound implies this is not the case.
2. The y-axis (x = 0) has absolutely NO IMPACT on this question. Draw the graph. It intersects with the x-axis at x=0. Hence, it is bound without saying anything about the y-axis. Does that make sense? The y-axis has no relevance to the question and thus, was not stated. Think about it.... The graph will clear a lot of stuff up in your mind.
Thus, the only acceptable answer is from 0 to pi/6. There is no VCAA screw up and the question can really only be interpreted in one way.
Anyways, GOOD LUCK FOR EXAM 2 EVERYONE!! (and accounting for those who have that as well)