Set up an intergal for the colume of the solid obtained by ratating the region bounded by the parabolas x=8y-2y^2, x= 4y-y^2 about the x-axis.

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Set up an intergal for the colume of the solid obtained by ratating the region bounded by the parabolas x=8y-2y^2, x= 4y-y^2 about the x-axis.

Mathematics
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I have no idea how to do this :( .
The maximum x-value for each is at y = 2. The intersections of the two are at (0,0) and (0,4). Would you like to chop it up into 3 pieces or simply apply Pappus' Theorem?
I can't use Pappus' theorem.

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Why not? Specifically proscribed? Not yet presented? Or just not able?
We won't learn it.
@saifoo.khan
First of all, I need to sketch this thing. I can't figure out how the thing looks like.
Think it up yourself and surprise everyone! 3 pieces then. 1) Biggest Piece x from 4 to 8 and y from \(2 - \sqrt{\dfrac{8-x}{2}}\) to \(2 + \sqrt{\dfrac{8-x}{2}}\) 2) Top Piece x from 0 to 4 and y from \(2 + \sqrt{4-x}\) to \(2 + \sqrt{\dfrac{8-x}{2}}\) 3) Bottom Piece x from 0 to 4 and y from \(2 - \sqrt{\dfrac{8-x}{2}}\) to \(2 + \sqrt{4-x}\) Go!
Why not flip it around to where you are more familiar and solve that problem. y = 8x-2x^2 and y= 4x-x^2 and wrap it around the y-axis!
But that changes the entire question.
No. The question is to calculate the volume. Unique answers do not care how you find them. It MUST be considered a valid methodology.
Anyway, you can flip it around just to get a look at it, if you can't relate to the present condition.
But if you switch the x and y's that should change the entire shape of the graph.
That's silly. Why would you think that? Absolutely not the case. It is the EXACT SAME problem.
|dw:1359609779642:dw| Something like that? REALLY bad sketch though.
|dw:1359609887178:dw| A little better. You crossed the x-axis. That's no good.
yeah I know. I felt lazy.
Well, go ahead and do it, then. I gave you all the limits up above. Good luck. gtg
Thanks!

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