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anonymous
 5 years ago
PLEASE SOLVE :)
let R be the region bounded by the graph of y=x^2 1 and the graph of x=y^2 .... find the volume of the solid generated when r is rotated about the vertical line x=2, and then write an integral expression that can be used to find the volume of the solid generated when R is rotated about the line y=1
anonymous
 5 years ago
PLEASE SOLVE :) let R be the region bounded by the graph of y=x^2 1 and the graph of x=y^2 .... find the volume of the solid generated when r is rotated about the vertical line x=2, and then write an integral expression that can be used to find the volume of the solid generated when R is rotated about the line y=1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can get you started, but I may have to leave soon. If I help you set up the volume element, can you integrate from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I'll start anyway...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to send through an image of the plots you need to look at.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is hard without a calculator. I got the integral would be \[\pi*\int\limits_{?}^{?}(2y^2)^2(2\sqrt{y+1})^2dy\] and then i would need a calulator to get the limits where the two function are equal...is it supposed to have such nasty algebra or did I mess up soehwhere?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Assuming your equations are correct, the shaded region is what you're looking at. I would plan on using cylindrical shells, and doing the integral in two parts: one where\[y=\sqrt{x}\]and one where\[y=\sqrt{x}\]and then add the volumes up.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0plotting it was easy :) how we find the solutions? or is there another way?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, put cyl. shells on hold. When I first looked at your problem, I took the yaxis as a natural bound, but that's not the case since it's not in your question. When considering the region shaded, discs may be better. I want to work it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm deriving it  it's easy to do on paper, but hard to explain. online

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0deriving sounds interesting.... would you see where the slopes are perp to each other?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0been trying to solve it like system of equation....but no luck; then again, i could be missing something glaringly obvious :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the integral came out \[\pi*[y^5/54y^3/3y^2/2+8/3*y^(3/2)y]...\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0jpick: you just subtract the equations from each other?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but to get bounds for y i have to solve the values where y^4y1=0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah.... been trying that one for hours :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0just thinking outloud, but would newtons approximation work for us? or is that just to clumsy?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, it seems whatever method you use here (cylindrical shells or discs) the algebra is going to be crappy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Given that, I'm going to persist with cylindrical shells.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe, some numerical approximation (or algebra solver program) will be needed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0finding the bounds is like pulling teeth, but without the lollipop at the end :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is it possible to set your own bounds and still get an answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure what you mean? the bounds of integration are set by the intersection, or do you mean the approximation getting close enough will be our "own bounds"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0either way... I was wondering if going further than the bounds would have a canceling out effect, but most likely that is just wishful thinking

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, the basic construction for the cylindrical shell is\[\delta V = 2\pi(radius)(height)(elemental.thickness)\]\[=2\pi (x2)(\sqrt{x}(x^21)) \delta x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this element is for the section to the right of the line in the following image.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A similar situation will arise with the section to the left, where you have\[\delta V=2\pi (2x)(\sqrt{x}(\sqrt{x})) \delta x=2\pi(2x)(2\sqrt{x}) \delta x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a mistake in the first element: x2 should read as 2x, the radius.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to communicate with me. Is this working for you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your limits of integration for the element I just derived above (the one with 2sqrt{x} for height) will be from x=0 to x=t where \[\sqrt{t}=t^21\] (i.e. get a number out for t).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that will work, the only thing we need are the actual bounds, either in terms of x or y. I do have one question for you lokisan about the earlier differential equation thread. I got a different answer than you, even though we did it the same way given it was an exact equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limits of the initial section I derived will be from x = the t you just found to x=s where \[\sqrt{s}=s^21\](i.e. get an approximation for s).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The one where he wanted substitution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bethanymichalski, you there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, I got x^2/2+y^/2+xyx=c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's what I got using exact differential equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, i thought it said log(x+y)...my bad nm :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, he wanted to use a substitution method in the end. I didn't bother to check if the answers were the same, ;p
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