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Dido525
Group Title
Set up an intergal for the colume of the solid obtained by ratating the region bounded by the parabolas
x=8y2y^2, x= 4yy^2
about the xaxis.
 one year ago
 one year ago
Dido525 Group Title
Set up an intergal for the colume of the solid obtained by ratating the region bounded by the parabolas x=8y2y^2, x= 4yy^2 about the xaxis.
 one year ago
 one year ago

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Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I have no idea how to do this :( .
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
The maximum xvalue 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?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I can't use Pappus' theorem.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
Why not? Specifically proscribed? Not yet presented? Or just not able?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
We won't learn it.
 one year ago

soty2013 Group TitleBest ResponseYou've already chosen the best response.0
@saifoo.khan
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
First of all, I need to sketch this thing. I can't figure out how the thing looks like.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
Think it up yourself and surprise everyone! 3 pieces then. 1) Biggest Piece x from 4 to 8 and y from \(2  \sqrt{\dfrac{8x}{2}}\) to \(2 + \sqrt{\dfrac{8x}{2}}\) 2) Top Piece x from 0 to 4 and y from \(2 + \sqrt{4x}\) to \(2 + \sqrt{\dfrac{8x}{2}}\) 3) Bottom Piece x from 0 to 4 and y from \(2  \sqrt{\dfrac{8x}{2}}\) to \(2 + \sqrt{4x}\) Go!
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
Why not flip it around to where you are more familiar and solve that problem. y = 8x2x^2 and y= 4xx^2 and wrap it around the yaxis!
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
But that changes the entire question.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
No. The question is to calculate the volume. Unique answers do not care how you find them. It MUST be considered a valid methodology.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
Anyway, you can flip it around just to get a look at it, if you can't relate to the present condition.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
But if you switch the x and y's that should change the entire shape of the graph.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
That's silly. Why would you think that? Absolutely not the case. It is the EXACT SAME problem.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1359609779642:dw Something like that? REALLY bad sketch though.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
dw:1359609887178:dw A little better. You crossed the xaxis. That's no good.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
yeah I know. I felt lazy.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.3
Well, go ahead and do it, then. I gave you all the limits up above. Good luck. gtg
 one year ago
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