Spesh Exam

[funkyducky]

Luffy is 100% correct. draw the sin graph and the line x=pi/6 - the only enclosed region within the domain xE[0,pi/2] is the one b/w 0 and pi/6.

[jane1234]

^ Concur with this even though I said it on the other thread. x=0 to x=pi/6 was the only option which fit the criteria given in the question...

[acinod]

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)

Oh I get it now! I thought x=pi/2 was an x-intercept. Look so stupid now. Thank you for clarifying luffy

[illuminati]

okay i haven't read the posts above
but interpreting the question it never said specifically it was x in the first quadrant, rather the region itself
so what about the area bound by x = pi/6 and x = pi and the x axis rotated about the x axis

[thushan]

Hmm; pi/6 to pi/2 is not a valid bound, as the others have already established.
However, I had another peek at the exam paper and yes, pi/6 to pi is an equally valid bound! Well, looks like VCAA has some issues here...

[illuminati]

surely they'll just accept both answers
unless they did it intentionally and require both answers.
in which case GG

[thushan]

that would be soooo tight!

[funkyducky]

Hmm; pi/6 to pi/2 is not a valid bound, as the others have already established.
However, I had another peek at the exam paper and yes, pi/6 to pi is an equally valid bound! Well, looks like VCAA has some issues here...

It says first quadrant, implying x is a first quadrant angle in terms of the unit circle

[illuminati]

Hmm; pi/6 to pi/2 is not a valid bound, as the others have already established.
However, I had another peek at the exam paper and yes, pi/6 to pi is an equally valid bound! Well, looks like VCAA has some issues here...

It says first quadrant, implying x is a first quadrant angle in terms of the unit circle

it just says the region is in the first quadrant. not the angle. so it can be interpreted.

[acinod]

Hmm; pi/6 to pi/2 is not a valid bound, as the others have already established.
However, I had another peek at the exam paper and yes, pi/6 to pi is an equally valid bound! Well, looks like VCAA has some issues here...

It says first quadrant, implying x is a first quadrant angle in terms of the unit circle

Could also mean the area is in the first quadrant of the cartesian plane.

[thushan]

Hmm, I took "first quadrant" to be a reference to the actual area being in the first quadrant rather than the angle, but now I think of it, that would have been redundant...if they meant to be a reference to the angle (x-value) then that's horribly sloppy wording!

[funkyducky]

Hmm, I took "first quadrant" to be a reference to the actual area being in the first quadrant rather than the angle, but now I think of it, that would have been redundant...if they meant to be a reference to the angle (x-value) then that's horribly sloppy wording!

Pretty much, it means nothing to say "the region in the first quadrant bound by x=pi/6 and y=sin(x)" instead of "the region bound by x=pi/6 and y=sin(x)", so they must have meant it in reference to the x-value (why else would they bother?). It is fairly ambiguous phrasing, though, VCAA could have done better.

[BubbleWrapMan]

I suppose they thought if they mentioned x = 0 some might have thought it was the area above the curve. Can't win 'em all I guess.

They will most likely allow both regions.

[jane1234]

Hmm; pi/6 to pi/2 is not a valid bound, as the others have already established.
However, I had another peek at the exam paper and yes, pi/6 to pi is an equally valid bound! Well, looks like VCAA has some issues here...

But you could argue that from pi/6 onwards, so long as the area was above the x-axis, it could be a 'valid bound'. As in from pi/6 to pi and then 2pi to 3pi and then 4pi to 5pi etc etc etc... so long as the area was in the "first quadrant". I don't think it would be a valid intepretation because you would end up with an infinite volume, technically speaking.

[thushan]

Here we go into semantics...