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    Im getting 4160, exactly double the answer in the book.

    I found the area of the rectangle around it (12 x 9)
    Then minused the definite integral from 6 to -6 of 1/4x^2
    Then minused the definite integral of 2 to -2 of 1-(1/4)x^2 which is the bottom white part of the cross section

    Then multiplied this by 60, the length.
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    (Original post by Feynboy)
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    Im getting 4160, exactly double the answer in the book.

    I found the area of the rectangle around it (12 x 9)
    Then minused the definite integral from 6 to -6 of 1/4x^2
    Then minused the definite integral of 2 to -2 of 1-(1/4)x^2 which is the bottom white part of the cross section

    Then multiplied this by 60, the length.
    Me too :s

    By the way, an easier way to do this is to write it in terms of x, or in other words invert the axes, so then it's just a basic curve with a single integral to get the answer.

    I got:

    y = \frac{1}{4}x^2 \longrightarrow x = 2 \sqrt{y}

    Then the volume is

    60 * 2 * \displaystyle\int^9_1 2\sqrt{y}\ dy = 4160
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    Also getting the same answer here.
    If I recognise the book correctly as one of the old AQA Textbooks then there are usually a few incorrect answers in them, so yours is probably correct
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    (Original post by Pessimisterious)
    Me too :s

    By the way, an easier way to do this is to write it in terms of x, or in other words invert the axes, so then it's just a basic curve with a single integral to get the answer.

    I got:

    y = \frac{1}{4}x^2 \longrightarrow x = 2 \sqrt{y}

    Then the volume is

    60 * 2 * \displaystyle\int^9_1 2\sqrt{y}\ dy = 4160
    Is the *2 because the integral only calculates the area to one side of the axis?
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    (Original post by Feynboy)
    Is the *2 because the integral only calculates the area to one side of the axis?
    Precisely
 
 
 

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