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    Hey

    I don't understand what numbers to use where in this question.

    The region R is bounded by the curve y = x^2 + 2 the line x = 1 and the x and y axis. Calculate the volume of the solid generated when R is rotated through 360 degrees about the y-axis.

    Thanks if you can help me.
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    If you try and sketch the region, you'll find, from the information you've given, that this is not a valid region; unless I missed something (always possible).
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    It was supposed to be the line x = 1, sorry :p:
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    The region specified is invalid, otherwise it would be infinite.Did you perhaps mean x=1?

    EDIT: Oops, reread. It's about the y-axis. So rearrange to get sqrt(y-2)=x. Then pi * integral of sqrt(y-2) from 0 to 1.
    Or more simply, just [(y^2)/2-2y] from 0 to 1. Then 1/2-2=3/2pi
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    volume of revolution is the integral of (pi)(y^2)

    And because pi is a number you just integrate y^2

    So square y, then integrate the answer = pi * integral(x^2 +1)^2 plus c
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    (Original post by Afrikaans Boytjie)
    The region specified is invalid, otherwise it would be infinite.Did you perhaps mean x=1?

    EDIT: Oops, reread. It's about the y-axis. So rearrange to get sqrt(y-2)=x. Then pi * integral of sqrt(y-2) from 0 to 1.
    Or more simply, just [(y^2)/2-2y] from 0 to 1. Then 1/2-2=3/2pi
    That's exactly what I got.. but the answers say it's \frac{5\pi}{2}
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    (Original post by MikeL230)
    volume of revolution is the integral of (pi)(y^2)

    And because pi is a number you just integrate y^2

    So square y, then integrate the answer = pi * integral(x^2 +1)^2 plus c
    It's about the y-axis.
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    (Original post by Adam92)
    That's exactly what I got.. but the answers say it's \frac{5\pi}{2}
    I get \frac{5\pi}{2}

    I can't think how to explain it to you, so I tried drawing a sketch.



    If this was a "normal" volumes of revoultion about the y-axis question, you would integrate between 0 and 3 with respect to y and be happy with that. Can you see from the sketch why this doesn't work in this case? The curve "crosses" the region that you are integrating, which messes things up, so can you see on the sketch a different region that you could find the volume of revolution for in the "normal" way?

    Spoiler:
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    could you take the volume \displaystyle \pi \int_2^3 y-2 \, \mathrm{d}y away from something to find the volume of R?


    Sorry if that made no sense, I'm terrible at explaining these things! :o:
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    Why are you using limits of 0 and 1 for the y value?

    Edit: I see sonodot's just beaten me to it.
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    (Original post by sonofdot)
    I get \frac{5\pi}{2}

    I can't think how to explain it to you, so I tried drawing a sketch.



    If this was a "normal" volumes of revoultion about the y-axis question, you would integrate between 0 and 3 with respect to y and be happy with that. Can you see from the sketch why this doesn't work in this case? The curve "crosses" the region that you are integrating, which messes things up, so can you see on the sketch a different region that you could find the volume of revolution for in the "normal" way?

    Spoiler:
    Show
    could you take the volume \displaystyle \pi \int2^3 1 \, \mathrm{d}y away from something to find the volume of R?


    Sorry if that made no sense, I'm terrible at explaining these things! :o:
    That made absolute sense. So I do \displaystyle \pi \int_0^3 1 \, \mathrm{d}y  -  \displaystyle\int_2^3 y - 2 \, \mathrm{d}y to get \frac{5\pi}{2}! yay thanks !
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    (Original post by ghostwalker)
    Why are you using limits of 0 and 1 for the y value?

    Edit: I see sonodot's just beaten me to it.
    The problem was I didn't know which numbers to use..
 
 
 
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