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    Name:  1486583584123-932726756.jpg
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Size:  476.5 KB for part iv where does the root 3 come from?
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    I presume that root 3 is the area of the graph between y=1 and the x-axis, that you would need to find to determine R.
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    (Original post by Nikita_2313)
    I presume that root 3 is the area of the graph between y=1 and the x-axis, that you would need to find to determine R.
    When I work that out, I get 8-3root3/6, which is no where near root3
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    (Original post by gooner1010)
    When I work that out, I get 8-3root3/6, which is no where near root3
    How'd you get that? The line y=1 intersects the graph at \sin \theta = \frac{1}{2} so x = 1 \pm \frac{\sqrt{3}}{2}, since \cos \theta = \pm \sqrt{1 - \sin^2 \theta}

    This forms a rectangle with width 1 + \sqrt{3}/2 - 1 + \sqrt{3}/2 = \sqrt{3} and height 1. So area \sqrt{3}.

    Where did you go wrong?
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    (Original post by Zacken)
    How'd you get that? The line y=1 intersects the graph at \sin \theta = \frac{1}{2} so x = 1 \pm \frac{\sqrt{3}}{2}, since \cos \theta = \pm \sqrt{1 - \sin^2 \theta}

    This forms a rectangle with width 1 + \sqrt{3}/2 - 1 + \sqrt{3}/2 = \sqrt{3} and height 1. So area \sqrt{3}.

    Where did you go wrong?
    Why do you form a rectangle? I thought the area of R was the area of the whole semi circle minus the area of the semi circle below 1. Wouldn't some of the area of the semi circle below 1 be missing?
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    the two equations meet at 1+root3/2 and 1-root3/2. Basically, root 3 is the definite integral of the equation y=1 (when you put in the limits)
    integral of 1 is x. so you just subtract the lower x value from the upper one which gives (1 +root3/2) - (1 - root3/2) =root3/2 + root3/2 = root3
    hope i explained it alright.
    havent done integration in a while because I am on gap year so this was a refresher.
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    (Original post by gooner1010)
    Why do you form a rectangle? I thought the area of R was the area of the whole semi circle minus the area of the semi circle below 1. Wouldn't some of the area of the semi circle below 1 be missing?
    Eh? You're quite off track. When you integrate between x=1-sqrt(3)/2 and x =1 + sqrt(3)/2, you get the full area of the curve, as shaded here:



    To get the required area the question wants, you have to subtract the area of that rectangle I've shaded in. This rectangle has height 1 and width sqrt(3).

    Do you understand what it means to integrate between two limits...?
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    (Original post by k_popper)
    the two equations meet at 1+root3/2 and 1-root3/2. Basically, root 3 is the definite integral of the equation y=1 (when you put in the limits)
    integral of 1 is x. so you just subtract the lower x value from the upper one which gives (1 +root3/2) - (1 - root3/2) =root3/2 + root3/2 = root3
    hope i explained it alright.
    havent done integration in a while because I am on gap year so this was a refresher.
    Oh I get it it, I did it wrong before as I tried working out the area of the whole semi circle. Thank you and Zacken.
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    (Original post by Zacken)
    Eh? You're quite off track. When you integrate between x=1-sqrt(3)/2 and x =1 + sqrt(3)/2, you get the full area of the curve, as shaded here:



    To get the required area the question wants, you have to subtract the area of that rectangle I've shaded in. This rectangle has height 1 and width sqrt(3).

    Do you understand what it means to integrate between two limits...?
    Thanks, I get it now.
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    (Original post by gooner1010)
    Oh I get it it, I did it wrong before as I tried working out the area of the whole semi circle. Thank you and Zacken.
    my pleasure. I actually enjoyed solving this T_T i wish i could take a small maths course on the side at uni lmao
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    (Original post by gooner1010)
    Thanks, I get it now.
    No worries
 
 
 
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