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    i've just done this beast of a P2 volumes of revolution question.

    try it, it's good practice.


    The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

    Find the volume when R is rotated 360 degrees around:
    a) the x axis
    b) y axis





    answers:
    a)12pi b)206pi/15
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    (Original post by hihihihi)
    i've just done this beast of a P2 volumes of revolution question.

    try it, it's good practice.


    The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

    Find the volume when R is rotated 360 degrees around:
    a) the x axis
    b) y axis




    answers:
    a)12pi b)206pi/15
    a. volume= 20pi-pi*int(x-1)[5 and 1]
    =20pi-pi(x^2/2-x)
    =20pi-pi(12.5-5-0.5+1)
    =20pi-8pi
    =12pi
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    0 and 1 i think
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    (Original post by lgs98jonee)
    wot r the limits when u integrate around the xaxis??
    well you know y = 2...

    cmon dude, dont slip now!
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    oh yeh lol i fell in the same trap i didnt see y=2! deary me
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    (Original post by lgs98jonee)
    a. area= 6-pi*int(x-1)[3 and 1]
    =6-pi(x^2/2-x)
    =6-pi(4.5-0.5-3+1)
    =6-2pi
    area?? it's volumes!!
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    yeh i just noticed the y=2! oops
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    (Original post by hihihihi)
    you also need to find another limit..clue: when y=2, x=...
    oh thnx for that clue.. :mad: .give us a chance to do the question....seems like a really patronisng tone..ok then children wot if y=1+x and x=1, y=...
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    r u sure that it is not 12.5pi?
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    I got for a) (25pi-5pi)-(1pi-1pi) = 20pi and for b) 404/15pi...
    When y = 2... 4 = x-1 so x=5... You seem to have integrated the upper value as being 4... Why?
    For a) y was squared already so I just integrated x^2 - x between 5 and 1...

    Have I done something horrible wrong?
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    why do u do 20pi-what u get? where are u minusing it from?
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    I am such a retard... I forgot to find the intersection points... I will fail this bloody exam!
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    (Original post by hihihihi)
    i've just done this beast of a P2 volumes of revolution question.

    try it, it's good practice.


    The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

    Find the volume when R is rotated 360 degrees around:
    a) the x axis
    b) y axis





    answers:
    a)12pi b)206pi/15

    b. volume=piINT(y^2+1)^2 dx
    =pi INT(y^4+2y^2+1)
    =pi(y^5/5+2y^3/3+x) [between 2 and 0]
    =pi(32/5+16/3+2)
    =96/15+80/15+30/15
    =206pi/15 :cool: :cool:
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    (Original post by Sapphira)
    why do u do 20pi-what u get? where are u minusing it from?
    well i did the big cylinder (pi *2^2*5)-the area under the curve between x=0 and x=5
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    (Original post by lgs98jonee)
    a. volume= 20pi-pi*int(x-1)[5 and 0]
    =20pi-pi(x^2/2-x)
    =20pi-pi(12.5-5)
    =20pi-7.5pi
    =12.5pi
    you need to do between 5 and 1??
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    (Original post by lgs98jonee)
    b. volume=piINT(y^2+1)^2 dx
    =pi INT(y^4+2y^2+1)
    =pi(y^5/5+2y^3/3+x) [between 2 and 0]
    =pi(32/5+16/3+2)
    =96/15+80/15+30/15
    =206pi/15 :cool: :cool:
    high five!!
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    (Original post by hihihihi)
    high five!!
    they werent easy
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    The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

    Find the volume when R is rotated 360 degrees around:
    a) the x axis
    b) y axis

    a)

    2^2 = x-1
    x = 5

    Vol. = (5/0){ pi(y^2) dx
    = (5/0){ pi x - pi) dx
    = [pi.x^2/2 - pi.x] (5/0)
    = [pi.25/2 - 5pi] - 0
    = 12.5pi

    b)
    y^2=x-1
    x = (y^2 + 1)
    x^2 = y^4 + 2y^2 + 1

    Vol. = (2/0){ pi(x^2) dy
    = (2/0){ (pi.y^4 + 2pi.y^2 + pi) dy
    = [pi.y^5/5 + 2pi.y^3/3 + pi.y](2/0)
    = [(32/5)pi + (16/3)pi + (2)pi]
    = (206/15)pi

    Nice question, had to think hard about the (b) part!
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    (Original post by mik1a)

    Vol. = (5/0){ pi(y^2) dx
    = (5/0){ pi x - pi) dx
    = [pi.x^2/2 - pi.x] (5/0)
    = [pi.25/2 - 5pi] - 0
    = 12.5pi
    i got that but apparently it was 12pi :-(
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    I don't get why between 5 and 1. :confused:
 
 
 
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