Question

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

Answer 1

You are correct.

Answer 2

its possible; you just get a different shape

Answer 3

no but you are revolving about itself twice

Answer 4

then subtract he part that gets doubled...

Answer 5

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

Answer 6

hmmmm, that doesnt make sense to me

Answer 7

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...

Answer 8

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

Answer 9

i would say the question is ill defined

Answer 10

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?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

that all depends onhow the curves and their boundaries interact

Answer 16

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)

Answer 17

in this case; you can safely ignore the left part

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

right, thats a good word. it annhilates :)

Answer 22

nulls them

Answer 23

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

Answer 24

whoa

Answer 25

how did you graph that?

Answer 26

i used paint in windows lol

Answer 27

nice :)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

haha, by displacement,

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

what method of integration are you doing, disc?

Answer 42

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

Answer 43

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?

Answer 44

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

Answer 45

volume is not defined for negative

Answer 46

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

Answer 47

you have a contradictory curve

Answer 48

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

Answer 49

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

Answer 50

no your curves are redundant

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

hey contrasat can u help me

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

i guess im not as flexible as you are

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

amistre, two people disagree with you

Answer 70

so far

Answer 71

lets ask watchmath

Answer 72

can u help me contrsat

Answer 73

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

Answer 74

yes it does

Answer 75

lol.... area is an absolute value

Answer 76

wait, waht do you mean?

Answer 77

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

Answer 78

whats the radius?

Answer 79

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

Answer 80

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

Answer 81

its not possible to do that, using disc method

Answer 82

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

Answer 83

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

Answer 84

so what's going on here? :)

Answer 85

lol..... well...

Answer 86

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

Answer 87

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.

Answer 88

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

Answer 89

exactly :)

Answer 90

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

Answer 91

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

Answer 92

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.