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Math Help - Calculating area from a graph.

  1. #1
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    Calculating area from a graph.

    I have to calculate the area of the shaded graph (in the link). Seeming as no equations are given, how do I calculate it?

    Untitled | Flickr - Photo Sharing!

    It's hard to read the graph but it's just 0.1π, 0.2π... 2.0π.
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  2. #2
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    To me it seems like it is a sine and cosine graph.

    You would need to develop an equation to be able to calculate the area accurately:

    y = a*sin(bx)
    y = a * cos(bx)

    Where a and b are constant values.
    To me it seems its maximum is 4.5 and minimum is -4.5. Thus constant a must be 4.5.
    One full cycle seems to be 2.0π. If you are working in radians, constant b must be 2\pi/2. If working in degrees, b = 360/2.


    y = 4.5*sin(\pi {x})
    y = 4.5 * cos(\pi {x})
    (For working in radians)

    There once you have your 2 equations, you can integrate between 0 and 2 to find the area.
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  3. #3
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    Of course! Thankyou so much.
    Can I just ask, how do you know to integrate between 0 and 2?
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  4. #4
    Senior Member Educated's Avatar
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    Oh sorry I misinterpereted the figures on your graph.

    Now I take a closer look, 2.0π actually meant 2\pi. I thought π was the units...

    Forget what I said about b being a constant. It stretches out like a normal graph - from 0 to 2\pi.

    Thus:
    y = 4.5*sin(x)
    y = 4.5 * cos (x)

    To make finding the area of the shaded part easier, move the graph up by 4.5 so that no part of it is in the negative part, below the x-axis.


    y = 4.5*sin(x)+4.5
    y = 4.5 * cos (x)+4.5

    To find the area:
    Integrate between 0 to \frac{1}{4}\pi, and do the cosine graph subtracted the sine graph.
    Then integrate between \frac{1}{4}\pi to \frac{5}{4}\pi, and do the sine graph subtracted the cosine graph this time.
    Finally integrate between \frac{5}{4}\pi to \frac{8}{4}\pi, and do the cosine graph subtracted the sine graph.
    Add all of these together and you will find your shaded area.
    Last edited by Educated; August 21st 2010 at 10:28 PM. Reason: Didn't see the complexity of the problem, I wasn't thinking before.
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  5. #5
    Senior Member Educated's Avatar
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    Also the sine and cosine graphs go onto infinity - I'm only guessing the shaded parts are only between 0 to 2\pi as that is as far as the graph goes.
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