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Core Maths 3: Integration to find volume of revolution ii

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    Hi there

    Please could you help me with this one:



    I have tried integrating both the line and the curve for the value of x = 8, and I am getting 2 pi for the area under the curve and 32 pi for the area under the line y = 2, the answer I have in my book for the whole thing is 18 pi.

    I don't really know how to work out the shaded area, any ideas??
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    32pi is wrong for your cylinder ... I think you have used 8 instead of 6
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    The curve is wrong too

    Not sure how you have got the 2pi
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    When you integrated you say you used 8

    What was your other limit
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    (Original post by jackie11)
    Hi there

    Please could you help me with this one:

    I have tried integrating both the line and the curve for the value of x = 8, and I am getting 2 pi for the area under the curve and 32 pi for the area under the line y = 2, the answer I have in my book for the whole thing is 18 pi.

    I don't really know how to work out the shaded area, any ideas??
    Required volume  \displaystyle = \pi\left(\int^8_2 2^2dx - \int^8_2 \left(\frac4{x}\right)^2dx \right)
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    (Original post by jackie11)
    Hi there

    Please could you help me with this one:

    I have tried integrating both the line and the curve for the value of x = 8, and I am getting 2 pi for the area under the curve and 32 pi for the area under the line y = 2, the answer I have in my book for the whole thing is 18 pi.

    I don't really know how to work out the shaded area, any ideas??
    Use the reverse bracket rule when integrating and you should end up with 18 -0 when you sub in the 8 and 2 hence 18pi.

    I just tried it out on a piece of paper and got the right answer.
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    (Original post by raheem94)
    Required volume  \displaystyle = \pi\left(\int^8_2 2^2dx - \int^8_2 \left(\frac4{x}\right)^2dx \right)
    I find it odd that both the OP and yourself would use integration to find the volume of a cylinder
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    (Original post by TenOfThem)
    32pi is wrong for your cylinder ... I think you have used 8 instead of 6

    (Original post by TenOfThem)
    The curve is wrong too

    Not sure how you have got the 2pi

    (Original post by TenOfThem)
    When you integrated you say you used 8

    What was your other limit
    I did use 8, I didn't use another limit as I assumed it was 0
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    (Original post by raheem94)
    Required volume  \displaystyle = \pi\left(\int^8_2 2^2dx - \int^8_2 \left(\frac4{x}\right)^2dx \right)
    aww thank you, this is great!!!
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    (Original post by jackie11)
    I did use 8, I didn't use another limit as I assumed it was 0
    Firstly ... just looking at the diagram you should have realised that the other limit was 2

    Secondly ... even when a limit is 0 you cannot just ignore it as the value may not always be 0
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    (Original post by TenOfThem)
    I find it odd that both the OP and yourself would use integration to find the volume of a cylinder
    Integration here takes the same time as using the area of cylinder takes, so i prefer to use integration rather than  \displaystyle \pi r^2h
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    (Original post by raheem94)
    Integration here takes the same time as using the area of cylinder takes [/latex]
    really?

    pi x 4 x 6 does not really take any time
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    (Original post by TenOfThem)
    really?

    pi x 4 x 6 does not really take any time
     \displaystyle \pi \int^8_2 4dx = \pi \left[4x\right]^8_2=\pi(32-8)=24\pi

    The above will also not take much time, may be just 5 seconds more.
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    (Original post by raheem94)
     \displaystyle \pi \int^8_2 4dx = \pi \left[4x\right]^8_2=\pi(32-8)=24\pi

    The above will also not take much time, may be just 5 seconds more.
    Much more scope for error though ... not saying that you are wrong ... just that I find it odd
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    (Original post by TenOfThem)
    Much more scope for error though ... not saying that you are wrong ... just that I find it odd
    Still the probability of error in such a question will be very less. This is basic integration.
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    (Original post by raheem94)
    Still the probability of error in such a question will be very less. This is basic integration.
    Given the errors made by the OP I beg to differ
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    (Original post by TenOfThem)
    Given the errors made by the OP I beg to differ
    Actually i was indicating my case not the OP's. I have seen several simple integration mistakes made by the OP so its best for the OP to avoid integration as much as possible.

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Updated: April 5, 2012
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