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anonymous
 5 years ago
can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis. this is not possible, since part of the region is in quadrant 2
anonymous
 5 years ago
can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis. this is not possible, since part of the region is in quadrant 2

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its possible; you just get a different shape

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no but you are revolving about itself twice

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then subtract he part that gets doubled...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the limits of itnegration are 1 to 2 and we are revolving about y axis (using shell method )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmmm, that doesnt make sense to me

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if I have a region in the form of a rectangle that overlaps the y axis and its domain is from 1 to 5 with a height of 3; when I rotate it about the yaxis; I just have to account for the part that get counted twice...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ive never seen a book say, subtract the region that gets doubled ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would say the question is ill defined

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you dont need a book to tell you that the volume of a solid isnt counted twice and to adjust for the computational errors :) do you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because you have to ignore the negative part, or quadrant 3 and 4 of the curve

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if that region is consumed by the rest, then yes, ignore it lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it part of that region is left exposed; count the part that is exposed ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then one should just adjust the limits, start at x = 0 for lower limit

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that all depends onhow the curves and their boundaries interact

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you said to ignore the region counted twice. we can do this by starting at x = 0 i have a different question, revolve region bounded by y = sqrt x, y = 0, x = 4, x = 6 , about the line x = 6 (using shell method i think)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0in this case; you can safely ignore the left part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , so then its ok to start at x = 0 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when would i not, i think im going to state a general theorem soon, but can you help me with that other shell problem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it is for the first question becasue the rotation covers the left parts completely and nulls them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, thats a good word. it annhilates :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when the rotation eats into itself, you adjust for the part that gets doubled...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you graph that?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i used paint in windows lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i created some curves that would account for a condition where you spin and have some parts redundant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you state the problem though, what are the regions, etc

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if you really wanna find the volume; dunk it in a bathtub filled with water and measure the volume of overflow lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, by displacement,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you state the problem though, what are the regions, etc

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i wouldnt be able to state that graph with any accuracy becasue I didnt use any defined functions to create it; I used artisitic liscense lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but then again, a function and its equation are not the same thing are they...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean can you set up a problem where your situation happens , like in this graph

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sure, its some cubic coupled with a linear an bounded at their common points; then spin around the y axis..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the image as is just shows the accounting for spin by applying the mirror image apon the original positions

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0all the steps to accont for the red portion tho; might take longer than what I got for tonight lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , i think you will end up in a contradiction though , i suspect it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0nah; its goblet shaped in the end; so a goblet is just as feesible as any other shape created by rotation.. right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what method of integration are you doing, disc?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also the shell method will produce a negative answer for your quadrant 2 curve

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you dont have to confine yourself to one method; use the tools available to you to determine the easiest way to arrive at the solution; break it up, and glue it back when your done right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0negative answers are not a problem when accounting for volume; since volume doesnt care about direction...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0volume is not defined for negative

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know how to account for the 'contradiction' of multiplying an inequaility by a negative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have a contradictory curve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the quadrant 1 and 4 overlaps will intersect quadrant 1 and 3 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the curve is perfectly representative of any other; its a line and a cubic;

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no your curves are redundant

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the volume produced by creating this shape on a potters wheel does not produce a contradiction does it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the portion in Q 1 and 4 will revolve right over Q 2 and 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your goblet is just the portion in Q 1 and 4 revolving

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the mirror image is not a part of the original functions; they are simply there to show what happens during rotation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the curve itself is bounded from q1 into q2 and ends in q3; then spun around the y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh can you graph this without shading , that got me confused

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats the original concept for the curves to be used in the rotation and the grey are is the area that creates the volume of rotation

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0hey contrasat can u help me

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i gotta get to sleep; so yall have fun :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre, ok , there are two porblems with this, if youre revolving about y axis, it should be in Q1 or Q4 onlye

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think thats a theorem, im going to make it a theorem :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it is inwhat it is in; you can adjust it up and down for sure; but thats a given

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess im not as flexible as you are

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if mathis only good for a few solutions; it isnt good for any solutions right ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well math is idealized, so you have set up your problem to match it i guess

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, the problem here is that volume is always positive, and does not annul itself

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i have a cubic and a line that can create a bounded region that canbe solved by integration thru rotation about the y axis... math can do it :) just have to adjust the graphs for common sense and solve

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0put the curves on the table in a way that they can be analysed; and then the rest is easy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre, two people disagree with you

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0can u help me contrsat

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0does the area of a circle disappear when its at the origin?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol.... area is an absolute value

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, waht do you mean?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i mean, does the area of a circle vanish if its center is at the origin?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0radius doesnt matter; make it 10 if you want, or 500000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean vanish? do you mean if you revolve the region about the x axis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not possible to do that, using disc method

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its possible; you jsut gotta use common sense and see that area or volume is an absolute value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0does the area above the xaxis null the area below it? or do quadrants cancel each other out? or what?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0so what's going on here? :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cantor says I cant determine the volume of my curves i think

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I think so far I agree with you Amistre. The important think is that do we get a solid object is some of the region is on the left and some on the right of the axis of rotation? yes we do? But how to compute the volume of that object maybe not as simple as applying the standard method of computing volume.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Also mathematically speaking we want to allow that kind of situation because we can have a more complex solid object.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0just as when we have an off centered rotation; we adjust the function to get them so that we can examine them from a reasonable stage

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if it gets to complicated; we can break it up, examine parts and come to a conclusion by summing up our findings

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0One way to find the volume is to reflect the one of the region about the axis of the symmetry so that now everything is on one side only. In the first cantorset example when we reflect the left part to the right, we don't get any new region on the right. So to compute volume we just integrate from x=0 to x=2.
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