anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
You are correct.
amistre64
  • amistre64
its possible; you just get a different shape
anonymous
  • anonymous
no but you are revolving about itself twice

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amistre64
  • amistre64
then subtract he part that gets doubled...
anonymous
  • anonymous
the limits of itnegration are -1 to 2 and we are revolving about y axis (using shell method )
anonymous
  • anonymous
hmmmm, that doesnt make sense to me
amistre64
  • 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
  • anonymous
ive never seen a book say, subtract the region that gets doubled ,
anonymous
  • anonymous
i would say the question is ill defined
amistre64
  • 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
  • anonymous
because you have to ignore the negative part, or quadrant 3 and 4 of the curve
amistre64
  • amistre64
if that region is consumed by the rest, then yes, ignore it lol
amistre64
  • amistre64
it part of that region is left exposed; count the part that is exposed ;)
anonymous
  • anonymous
then one should just adjust the limits, start at x = 0 for lower limit
amistre64
  • amistre64
that all depends onhow the curves and their boundaries interact
anonymous
  • 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
  • amistre64
in this case; you can safely ignore the left part
anonymous
  • anonymous
ok , so then its ok to start at x = 0 ?
anonymous
  • 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
  • amistre64
it is for the first question becasue the rotation covers the left parts completely and nulls them
anonymous
  • anonymous
right, thats a good word. it annhilates :)
anonymous
  • anonymous
nulls them
amistre64
  • amistre64
when the rotation eats into itself, you adjust for the part that gets doubled...
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anonymous
  • anonymous
whoa
anonymous
  • anonymous
how did you graph that?
amistre64
  • amistre64
i used paint in windows lol
anonymous
  • anonymous
nice :)
amistre64
  • amistre64
i created some curves that would account for a condition where you spin and have some parts redundant
anonymous
  • anonymous
can you state the problem though, what are the regions, etc
amistre64
  • amistre64
if you really wanna find the volume; dunk it in a bathtub filled with water and measure the volume of overflow lol
anonymous
  • anonymous
haha, by displacement,
anonymous
  • anonymous
can you state the problem though, what are the regions, etc
amistre64
  • 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
  • amistre64
but then again, a function and its equation are not the same thing are they...
anonymous
  • anonymous
i mean can you set up a problem where your situation happens , like in this graph
amistre64
  • amistre64
sure, its some cubic coupled with a linear an bounded at their common points; then spin around the y axis..
amistre64
  • amistre64
the image as is just shows the accounting for spin by applying the mirror image apon the original positions
amistre64
  • amistre64
all the steps to accont for the red portion tho; might take longer than what I got for tonight lol
anonymous
  • anonymous
ok , i think you will end up in a contradiction though , i suspect it
amistre64
  • amistre64
nah; its goblet shaped in the end; so a goblet is just as feesible as any other shape created by rotation.. right?
anonymous
  • anonymous
what method of integration are you doing, disc?
anonymous
  • anonymous
also the shell method will produce a negative answer for your quadrant 2 curve
amistre64
  • 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
  • amistre64
negative answers are not a problem when accounting for volume; since volume doesnt care about direction...
anonymous
  • anonymous
volume is not defined for negative
amistre64
  • amistre64
Do you know how to account for the 'contradiction' of multiplying an inequaility by a negative?
anonymous
  • anonymous
you have a contradictory curve
anonymous
  • anonymous
the quadrant 1 and 4 overlaps will intersect quadrant 1 and 3 right?
amistre64
  • amistre64
the curve is perfectly representative of any other; its a line and a cubic;
anonymous
  • anonymous
no your curves are redundant
amistre64
  • amistre64
the volume produced by creating this shape on a potters wheel does not produce a contradiction does it?
anonymous
  • anonymous
the portion in Q 1 and 4 will revolve right over Q 2 and 3
anonymous
  • anonymous
your goblet is just the portion in Q 1 and 4 revolving
amistre64
  • amistre64
the mirror image is not a part of the original functions; they are simply there to show what happens during rotation
amistre64
  • amistre64
the curve itself is bounded from q1 into q2 and ends in q3; then spun around the y axis
amistre64
  • amistre64
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anonymous
  • anonymous
oh can you graph this without shading , that got me confused
amistre64
  • 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
  • toxicsugar22
hey contrasat can u help me
amistre64
  • amistre64
i gotta get to sleep; so yall have fun :)
anonymous
  • anonymous
amistre, ok , there are two porblems with this, if youre revolving about y axis, it should be in Q1 or Q4 onlye
anonymous
  • anonymous
i think thats a theorem, im going to make it a theorem :)
amistre64
  • amistre64
it is inwhat it is in; you can adjust it up and down for sure; but thats a given
anonymous
  • anonymous
i guess im not as flexible as you are
amistre64
  • amistre64
if mathis only good for a few solutions; it isnt good for any solutions right ;)
anonymous
  • anonymous
well math is idealized, so you have set up your problem to match it i guess
anonymous
  • anonymous
well, the problem here is that volume is always positive, and does not annul itself
amistre64
  • 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
  • amistre64
put the curves on the table in a way that they can be analysed; and then the rest is easy
anonymous
  • anonymous
amistre, two people disagree with you
anonymous
  • anonymous
so far
anonymous
  • anonymous
lets ask watchmath
toxicsugar22
  • toxicsugar22
can u help me contrsat
amistre64
  • amistre64
does the area of a circle disappear when its at the origin?
anonymous
  • anonymous
yes it does
amistre64
  • amistre64
lol.... area is an absolute value
anonymous
  • anonymous
wait, waht do you mean?
amistre64
  • amistre64
i mean, does the area of a circle vanish if its center is at the origin?
anonymous
  • anonymous
whats the radius?
amistre64
  • amistre64
radius doesnt matter; make it 10 if you want, or 500000
anonymous
  • anonymous
what do you mean vanish? do you mean if you revolve the region about the x axis?
anonymous
  • anonymous
its not possible to do that, using disc method
amistre64
  • amistre64
its possible; you jsut gotta use common sense and see that area or volume is an absolute value
amistre64
  • amistre64
does the area above the xaxis null the area below it? or do quadrants cancel each other out? or what?
watchmath
  • watchmath
so what's going on here? :)
amistre64
  • amistre64
lol..... well...
amistre64
  • amistre64
cantor says I cant determine the volume of my curves i think
watchmath
  • 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
  • watchmath
Also mathematically speaking we want to allow that kind of situation because we can have a more complex solid object.
amistre64
  • amistre64
exactly :)
amistre64
  • 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
  • amistre64
if it gets to complicated; we can break it up, examine parts and come to a conclusion by summing up our findings
watchmath
  • 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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