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    Find the volume of the solid of revolution formed by rotating the area enclosed by the curve y= 2sin3x, the x-axis and the coordinates x=0 and x= Pi/9.

    I did this and got 7pi/6 units cubed and was wondering if anyone would be kind enough to attempt this question and help me out with it.

    Thanks

    Michelle
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    That isn't correct. Show some working?
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    Ok so i got

    pi ∫ (2sin3x)^2 and my limits were 0 to pi/9

    => pi ∫ 4sin^23x and my limits were 0 to pi/9

    => 4pi ∫ (sin3x)^2 and my limits were 0 to pi/9

    4pi [(-cos3x)^3/3] and the limits were 0 to pi/9

    so the i pulled in pi/9 and 0 and got

    4pi (-1/24 - (-1/3))

    = 4pi x 7/24

    = 28pi/24

    = 7pi/6 units cubed
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    (Original post by Michelle678)
    ...
    Use a trig identity at this line:

    4pi ∫ (sin3x)^2 and my limits were 0 to pi/9
    to allow the integral to be evaluated.

    Spoiler:
    Show
     \displaystyle \int sin^2(3x) dx = \int \dfrac{1 - cos(6x)}{2} dx
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    (Original post by Michelle678)
    Ok so i got

    pi ∫ (2sin3x)^2 and my limits were 0 to pi/9

    => pi ∫ 4sin^23x and my limits were 0 to pi/9

    => 4pi ∫ (sin3x)^2 and my limits were 0 to pi/9

    4pi [(-cos3x)^3/3] and the limits were 0 to pi/9

    so the i pulled in pi/9 and 0 and got

    4pi (-1/24 - (-1/3))

    = 4pi x 7/24

    = 28pi/24

    = 7pi/6 units cubed
    You have integrated wrongly.
    To integrate this, it might be worth noting that:
    \cos (6x) = 1 - 2\sin ^2(3x).
    Try to manipulate that to make \sin ^2(3x) the subject and then replace the \sin ^2(3x) in the integral with this.
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    As prime suspect says there is a standard trick you must use when integrating \sin^2 x or \cos^2 x and that is to express them in terms of \cos 2x

    Look through your notes, you will have been taught this. Learn it well. It is highly likely to be examined.
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    Mr M is completely right - they like to use it a lot in Edexcel C4 papers
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    First of all use the identity sin² A = ½(1-cos 2A) to see that sin² 3x = ½ - ½ cos 6x.

    This makes it 2╥ ∫(1-cos 6x) dx (with your limits). Which I believe ends up as (╥ (4╥ - 3√3))/18 units³
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    (Original post by wookymon)
    First of all use the identity sin² A = ½(1-cos 2A) to see that sin² 3x = ½ - ½ cos 6x.

    This makes it 2╥ ∫(1-cos 6x) dx (with your limits). Which I believe ends up as (╥ (4╥ - 3√3))/18 units³

    Thank you very much!! x
 
 
 
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