Volumes when curve crosses the axis...

[jane1234]

Okay I just did VCAA 2008 Spesh exam 1. And question 9 bothers me...

Attached the graph below... the question says to find the volume of the solid of revolution formed if the graph is rotated about the y-axis from x=-1 to x=1.

Now, VCAA says that even though the graph is crossing the y-axis during the interval, you can just integrate as normal (like pi*int(cos^2(y)dy from 0 to pi) and you will get an answer of pi^2/2.

However, if you try and visualise this it doesn't seem to work. What I see is a cylinder of pi^2 (radius 1 and height pi) units with a bit cut out of the top. This "bit" is equivalent to the area from x=0 to x=1 rotated about the y axis which gives a volume of pi^2/4. So this cylinder with a bit cut out the top should be 3pi^2/4 cubic units.

Basically, the way I see it you can disregard the area from x=0 to x=1 because it is less than the area on the other side, and if you only visualise the left hand side area rotating you'll see what I mean.

I presume VCAA are right, but I don't see how the shape would be half the cylinder because if you look at it the part that would be cut out of the top, it less than half of the volume of the original cylinder.

http://en.wikipedia.org/wiki/File:Shell_integration.svg << that kind of shows what I mean about the cylinder with a bit cut out the top.

Anyway, needless to say I am thoroughly confused. I'm assuming I'm visualising this wrong somehow...

Please Help!

[xZero]

The volume would be a hour glass, think your visualising it wrong

[jane1234]

? That would only be if the area from y = pi/2 to y= pi was shaded on the left hand side adjacent to the y-axis... I think...

Because if you just rotate the area just on the LHS of the y axis you get a GREATER volume than if you rotate both the area on the LHS and the RHS... which doesn't make sense that you are rotating a greater area to yield a smaller volume...

[xZero]

Not sure if i understand you, the volume is between the graph and y-axis so LHS and RHS covers the same volume?

[jane1234]

Hmm... just rereading the question...

It says "find the exact volume if the graph shown is rotated about the y-axis"

Ah, I think I get it now... part (a) asks you to find a particular area (bound by the x-axis, x=-1 and the curve) and naturally that's the one i was assuming would be rotated.... silly VCAA, they should specify which area they wanted rotated rather than just "the graph".

Thanks anyway.