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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have no idea how to do this :( .

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.3The 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can't use Pappus' theorem.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.3Why not? Specifically proscribed? Not yet presented? Or just not able?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First of all, I need to sketch this thing. I can't figure out how the thing looks like.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.3Think 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!

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.3Why 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But that changes the entire question.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But if you switch the x and y's that should change the entire shape of the graph.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359609779642:dw Something like that? REALLY bad sketch though.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I know. I felt lazy.

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