At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
You are correct.
its possible; you just get a different shape
no but you are revolving about itself twice
then subtract he part that gets doubled...
the limits of itnegration are -1 to 2 and we are revolving about y axis (using shell method )
hmmmm, that doesnt make sense to me
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...
ive never seen a book say, subtract the region that gets doubled ,
i would say the question is ill defined
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?
because you have to ignore the negative part, or quadrant 3 and 4 of the curve
if that region is consumed by the rest, then yes, ignore it lol
it part of that region is left exposed; count the part that is exposed ;)
then one should just adjust the limits, start at x = 0 for lower limit
that all depends onhow the curves and their boundaries interact
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)
in this case; you can safely ignore the left part
ok , so then its ok to start at x = 0 ?
when would i not, i think im going to state a general theorem soon, but can you help me with that other shell problem
it is for the first question becasue the rotation covers the left parts completely and nulls them
right, thats a good word. it annhilates :)
how did you graph that?
i used paint in windows lol
i created some curves that would account for a condition where you spin and have some parts redundant
can you state the problem though, what are the regions, etc
if you really wanna find the volume; dunk it in a bathtub filled with water and measure the volume of overflow lol
haha, by displacement,
can you state the problem though, what are the regions, etc
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
but then again, a function and its equation are not the same thing are they...
i mean can you set up a problem where your situation happens , like in this graph
sure, its some cubic coupled with a linear an bounded at their common points; then spin around the y axis..
the image as is just shows the accounting for spin by applying the mirror image apon the original positions
all the steps to accont for the red portion tho; might take longer than what I got for tonight lol
ok , i think you will end up in a contradiction though , i suspect it
nah; its goblet shaped in the end; so a goblet is just as feesible as any other shape created by rotation.. right?
what method of integration are you doing, disc?
also the shell method will produce a negative answer for your quadrant 2 curve
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?
negative answers are not a problem when accounting for volume; since volume doesnt care about direction...
volume is not defined for negative
Do you know how to account for the 'contradiction' of multiplying an inequaility by a negative?
you have a contradictory curve
the quadrant 1 and 4 overlaps will intersect quadrant 1 and 3 right?
the curve is perfectly representative of any other; its a line and a cubic;
no your curves are redundant
the volume produced by creating this shape on a potters wheel does not produce a contradiction does it?
the portion in Q 1 and 4 will revolve right over Q 2 and 3
your goblet is just the portion in Q 1 and 4 revolving
the mirror image is not a part of the original functions; they are simply there to show what happens during rotation
the curve itself is bounded from q1 into q2 and ends in q3; then spun around the y axis
oh can you graph this without shading , that got me confused
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
hey contrasat can u help me
i gotta get to sleep; so yall have fun :)
amistre, ok , there are two porblems with this, if youre revolving about y axis, it should be in Q1 or Q4 onlye
i think thats a theorem, im going to make it a theorem :)
it is inwhat it is in; you can adjust it up and down for sure; but thats a given
i guess im not as flexible as you are
if mathis only good for a few solutions; it isnt good for any solutions right ;)
well math is idealized, so you have set up your problem to match it i guess
well, the problem here is that volume is always positive, and does not annul itself
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
put the curves on the table in a way that they can be analysed; and then the rest is easy
amistre, two people disagree with you
lets ask watchmath
can u help me contrsat
does the area of a circle disappear when its at the origin?
yes it does
lol.... area is an absolute value
wait, waht do you mean?
i mean, does the area of a circle vanish if its center is at the origin?
whats the radius?
radius doesnt matter; make it 10 if you want, or 500000
what do you mean vanish? do you mean if you revolve the region about the x axis?
its not possible to do that, using disc method
its possible; you jsut gotta use common sense and see that area or volume is an absolute value
does the area above the xaxis null the area below it? or do quadrants cancel each other out? or what?
so what's going on here? :)
cantor says I cant determine the volume of my curves i think
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.
Also mathematically speaking we want to allow that kind of situation because we can have a more complex solid object.
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
if it gets to complicated; we can break it up, examine parts and come to a conclusion by summing up our findings
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.