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the mother of all p2 volumes of rev. question watch

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    Find the volume when the following shaded region is rotated 360 degrees about the y-axis.


    Book says answer= pi.


    well i worked out the vol of shaded region on one side of the y axis- do i need to double this or is this the final answer?? because the shaded is symetrical about yaxis, and surely the same volume is achieved by not doubling??
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    (Original post by hihihihi)
    Find the volume when the following shaded region is rotated 360 degrees about the y-axis.


    Book says answer= pi.


    well i worked out the vol of shaded region on one side of the y axis- do i need to double this or is this the final answer?? because the shaded is symetrical about yaxis, and surely the same volume is achieved by not doubling??


    Integral [ 2-x^-2-x^2] between the point of intersection....
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    omfg!! i would guess that it should equal half of volume... but how do you calculate it in the first place.???? Could you give a step-by-step please...!!
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    (Original post by Mathemagician)
    omfg!! i would guess that it should equal half of volume... but how do you calculate it in the first place.???? Could you give a step-by-step please...!!
    try looking at it side ways

    get x in terms of y, so that you can integrate x with respect to y.
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    (Original post by hihihihi)
    Find the volume when the following shaded region is rotated 360 degrees about the y-axis.


    Book says answer= pi.


    well i worked out the vol of shaded region on one side of the y axis- do i need to double this or is this the final answer?? because the shaded is symetrical about yaxis, and surely the same volume is achieved by not doubling??
    volume=pi[INT(2-y) (between 1 and 2)+INTy (between 0 and 1)]
    =pi((2y-y^2/2)+(y^2/2))
    =pi(4-2-2+1/2+1/2)
    =pi!!!
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    im glad this didnt come up on AQA pure 2!!!
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    (Original post by lgs98jonee)
    volume=pi[INT(2-y) (between 1 and 2)+INTy (between 0 and 1)]
    =pi((2y-y^2/2)+(y^2/2))
    =pi(4-2-2+1/2+1/2)
    =pi!!!
    so you dont need to double? :eek:
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    (Original post by hihihihi)
    so you dont need to double? :eek:
    well no cos it is symmetrical and so the other side goes through the same volume as the right hand side so no doubling
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    I got it!!!!!!!! vol = pi*int(2-x^2)^2 = 19/3pi - 10/3pi = 3pi (limits 1 and -1)

    other volume = pi*int(x^2)^2 = pi (limits are 0 and -1). Do this twice as it is a reflection (limits 1 and 0). Add these together to get 2pi. (if done with limits 1 and -1, they will cancel each other out to give 0)

    so volume shaded = 3pi-2pi
    = pi
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    (Original post by Mathemagician)
    I got it!!!!!!!! vol = pi*int(2-x^2)^2 = 19/3pi - 10/3pi = 3pi (limits 1 and -1)

    other volume = pi*int(x^2)^2 = pi (limits are 0 and -1). Do this twice as it is a reflection (limits 1 and 0). Add these together to get 2pi. (if done with limits 1 and -1, they will cancel each other out to give 0)

    so volume shaded = 3pi-2pi
    = pi
    wouldn you get x^5/5 when """other volume = pi*int(x^2)^2 = pi (limits are 0 and -1). Do this twice as it is a reflection (limits 1 and 0)""" therefore 1/5+1/5=2/5pi ??
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    My book says its 16pi
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    (Original post by lgs98jonee)
    volume=pi[INT(2-y) (between 1 and 2)+INTy (between 0 and 1)]
    =pi((2y-y^2/2)+(y^2/2))
    =pi(4-2-2+1/2+1/2)
    =pi!!!
    shouldnt you square f(x) when finding the volume of revolution?
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    (Original post by lexazver203)
    shouldnt you square f(x) when finding the volume of revolution?
    he did, but didnt write all the steps
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    i get it to be 28/5pi but i know this is aint right
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    (Original post by hihihihi)
    he did, but didnt write all the steps
    doesnt look so to me
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    split it into 2 sections

    y= x²
    => x=√y
    Vol. = Pi * ∫ (x²)dy
    = pi * ∫ y dy
    =pi*[ y²/2] between 1 and 0
    etc.

    y=2-x²
    => x=√(2-y)
    Vol. = Pi * ∫ (x²)dy
    = pi * ∫ (2-y) dy
    = pi * [2y - y²/2] between 2 and 1.
    etc.

    i think that's right yeah?
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    y = x^2
    y = 2 - x^2
    x^2 = 2 - x^2
    2x^2 = 2
    x = +/- 1

    Area = (1/-1){ (2 - x^2) dx - (1/1){ x^2 dx
    = [2x - x^3/3](1/-1) - [x^3/3](1/-1)
    = {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
    = {[10/6] - [2/3]}
    = 1 square unit.

    Square and multiply by pi:

    pi.1^2 = pi cube units.
    Great question!
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    (Original post by mik1a)
    y = x^2
    y = 2 - x^2
    x^2 = 2 - x^2
    2x^2 = 2
    x = +/- 1

    Area = (1/-1){ (2 - x^2) dx - (1/1){ x^2 dx
    = [2x - x^3/3](1/-1) - [x^3/3](1/-1)
    = {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
    = {[10/6] - [2/3]}
    = 1 square unit.

    Square and multiply by pi:

    pi.1^2 = pi cube units.
    Great question!
    This is driving me crazy... I'm trying to work through yours to see where I've gone wrong, but this little bit is driving me insane...

    = {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
    = {[10/6] - [2/3]}

    How the hell do you get 10/6 ? I've tried it over and over again, I get 10/3!!!

    (2 - 1/3) - (-2 + 1/3) = 10/3
    aaarrgh! whyyyyyyy?
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    (Original post by kimoni)
    This is driving me crazy... I'm trying to work through yours to see where I've gone wrong, but this little bit is driving me insane...

    = {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
    = {[10/6] - [2/3]}

    How the hell do you get 10/6 ? I've tried it over and over again, I get 10/3!!!

    (2 - 1/3) - (-2 + 1/3) = 10/3
    aaarrgh! whyyyyyyy?
    :confused:
    I can't see myself... I guess it's a mistake.
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    (Original post by mik1a)
    :confused:
    I can't see myself... I guess it's a mistake.
    I don't think your method works.. I've even drawn up the graphs in a graphing program, which calculates the area under the curve. It tells me the area under x^2 is 2/3, and the area under the other is 10/3... so the area under the curve is 8/3

    I don't think you can just square and times by pi... dammit. back to square 1

    I don't see why i can't just do
    (1/-1) pi INT: (2-x^2) - (x^2) dx
    it doesnt work for me... argh
 
 
 
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