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

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- anonymous

You are correct.

- amistre64

its possible; you just get a different shape

- anonymous

no but you are revolving about itself twice

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## More answers

- amistre64

then subtract he part that gets doubled...

- anonymous

the limits of itnegration are -1 to 2
and we are revolving about y axis (using shell method )

- anonymous

hmmmm, that doesnt make sense to me

- amistre64

if 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

ive never seen a book say, subtract the region that gets doubled ,

- anonymous

i would say the question is ill defined

- amistre64

you 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

because you have to ignore the negative part, or quadrant 3 and 4 of the curve

- amistre64

if that region is consumed by the rest, then yes, ignore it lol

- amistre64

it part of that region is left exposed; count the part that is exposed ;)

- anonymous

then one should just adjust the limits, start at x = 0 for lower limit

- amistre64

that all depends onhow the curves and their boundaries interact

- anonymous

but 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

in this case; you can safely ignore the left part

- anonymous

ok , so then its ok to start at x = 0 ?

- anonymous

when would i not, i think im going to state a general theorem soon, but can you help me with that other shell problem

- amistre64

it is for the first question becasue the rotation covers the left parts completely and nulls them

- anonymous

right, thats a good word. it annhilates :)

- anonymous

nulls them

- amistre64

when the rotation eats into itself, you adjust for the part that gets doubled...

##### 1 Attachment

- anonymous

whoa

- anonymous

how did you graph that?

- amistre64

i used paint in windows lol

- anonymous

nice :)

- amistre64

i created some curves that would account for a condition where you spin and have some parts redundant

- anonymous

can you state the problem though, what are the regions, etc

- amistre64

if you really wanna find the volume; dunk it in a bathtub filled with water and measure the volume of overflow lol

- anonymous

haha, by displacement,

- anonymous

can you state the problem though, what are the regions, etc

- amistre64

i 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

but then again, a function and its equation are not the same thing are they...

- anonymous

i mean can you set up a problem where your situation happens , like in this graph

- amistre64

sure, its some cubic coupled with a linear an bounded at their common points; then spin around the y axis..

- amistre64

the image as is just shows the accounting for spin by applying the mirror image apon the original positions

- amistre64

all the steps to accont for the red portion tho; might take longer than what I got for tonight lol

- anonymous

ok , i think you will end up in a contradiction though , i suspect it

- amistre64

nah; its goblet shaped in the end; so a goblet is just as feesible as any other shape created by rotation.. right?

- anonymous

what method of integration are you doing, disc?

- anonymous

also the shell method will produce a negative answer for your quadrant 2 curve

- amistre64

you 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

negative answers are not a problem when accounting for volume; since volume doesnt care about direction...

- anonymous

volume is not defined for negative

- amistre64

Do you know how to account for the 'contradiction' of multiplying an inequaility by a negative?

- anonymous

you have a contradictory curve

- anonymous

the quadrant 1 and 4 overlaps will intersect quadrant 1 and 3
right?

- amistre64

the curve is perfectly representative of any other; its a line and a cubic;

- anonymous

no your curves are redundant

- amistre64

the volume produced by creating this shape on a potters wheel does not produce a contradiction does it?

- anonymous

the portion in Q 1 and 4 will revolve right over Q 2 and 3

- anonymous

your goblet is just the portion in Q 1 and 4
revolving

- amistre64

the mirror image is not a part of the original functions; they are simply there to show what happens during rotation

- amistre64

the curve itself is bounded from q1 into q2 and ends in q3; then spun around the y axis

- amistre64

##### 1 Attachment

- anonymous

oh can you graph this without shading , that got me confused

- amistre64

thats 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

hey contrasat can u help me

- amistre64

i gotta get to sleep; so yall have fun :)

- anonymous

amistre, ok , there are two porblems with this, if youre revolving about y axis, it should be in Q1 or Q4 onlye

- anonymous

i think thats a theorem, im going to make it a theorem :)

- amistre64

it is inwhat it is in; you can adjust it up and down for sure; but thats a given

- anonymous

i guess im not as flexible as you are

- amistre64

if mathis only good for a few solutions; it isnt good for any solutions right ;)

- anonymous

well math is idealized, so you have set up your problem to match it i guess

- anonymous

well, the problem here is that volume is always positive, and does not annul itself

- amistre64

i 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

put the curves on the table in a way that they can be analysed; and then the rest is easy

- anonymous

amistre, two people disagree with you

- anonymous

so far

- anonymous

lets ask watchmath

- toxicsugar22

can u help me contrsat

- amistre64

does the area of a circle disappear when its at the origin?

- anonymous

yes it does

- amistre64

lol.... area is an absolute value

- anonymous

wait, waht do you mean?

- amistre64

i mean, does the area of a circle vanish if its center is at the origin?

- anonymous

whats the radius?

- amistre64

radius doesnt matter; make it 10 if you want, or 500000

- anonymous

what do you mean vanish? do you mean if you revolve the region about the x axis?

- anonymous

its not possible to do that, using disc method

- amistre64

its possible; you jsut gotta use common sense and see that area or volume is an absolute value

- amistre64

does the area above the xaxis null the area below it? or do quadrants cancel each other out? or what?

- watchmath

so what's going on here? :)

- amistre64

lol..... well...

- amistre64

cantor says I cant determine the volume of my curves i think

- watchmath

I 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

Also mathematically speaking we want to allow that kind of situation because we can have a more complex solid object.

- amistre64

exactly :)

- amistre64

just 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

if it gets to complicated; we can break it up, examine parts and come to a conclusion by summing up our findings

- watchmath

One 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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