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Math Help - Area of a segment when r changes

  1. #1
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    Area of a segment when r changes

    Hi
    I want to find the area of a segment. I know that the area of a segment is given by \frac{1}{2}\theta r^2
    But how do I find the area when r isn't constant? e.g when r goes from 1.85 to 1.89
    Is it just a case of integrating the equation for area between the two rs?

    Thanks
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  2. #2
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    Quote Originally Posted by Shizzle View Post
    Hi
    I want to find the area of a segment. I know that the area of a segment is given by \frac{1}{2}\theta r^2
    But how do I find the area when r isn't constant? e.g when r goes from 1.85 to 1.89
    Is it just a case of integrating the equation for area between the two rs?

    Thanks
    You can't do that. Because both θ and r are changing.
    Last edited by sa-ri-ga-ma; April 20th 2010 at 07:25 PM.
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  3. #3
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    Hello Shizzle

    Welcome to Math Help Forum!
    Quote Originally Posted by Shizzle View Post
    Hi
    I want to find the area of a segment. I know that the area of a segment is given by \frac{1}{2}\theta r^2
    But how do I find the area when r isn't constant? e.g when r goes from 1.85 to 1.89
    Is it just a case of integrating the equation for area between the two rs?

    Thanks
    You are on the right lines. In polar coordinates, the position of a point P is fixed by its distance, r, from a fixed point O and the angle, \theta, that the line OP makes with a fixed line.

    If a curve is defined by giving r as a function of \theta, then the area enclosed between the curve and the rays \theta = \alpha and \theta = \beta is
    \tfrac12\int_{\alpha}^{\beta}r^2d\theta
    Grandad
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  4. #4
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    Thanks, I'm not sure if I made myself clear enough. The reason I'm asking this question is because I have some results from an experiment I need to analyse.

    From \theta=0 to 180 degrees every 15 degrees I measured r.

    Although r changes with \theta, the extent to which it changes depends on which 15 degree increment it is measured over.

    E.g from 0 to 15 degrees r changed from 1.95 to 1.92, but from 15 to 30 degrees r changed from 1.92 to 1.82.

    Don't I need to use these values of r to calculate the area instead of the angle?
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  5. #5
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    Hello Shizzle
    Quote Originally Posted by Shizzle View Post
    Thanks, I'm not sure if I made myself clear enough. The reason I'm asking this question is because I have some results from an experiment I need to analyse.

    From \theta=0 to 180 degrees every 15 degrees I measured r.

    Although r changes with \theta, the extent to which it changes depends on which 15 degree increment it is measured over.

    E.g from 0 to 15 degrees r changed from 1.95 to 1.92, but from 15 to 30 degrees r changed from 1.92 to 1.82.

    Don't I need to use these values of r to calculate the area instead of the angle?
    Since you're working with numerical data, you'll only get an approximation to the area. So I think the best and simplest option is probably to take the mean value of r in each 15^o ( = \pi/12 radians) range, and use \tfrac12r^2\theta with that value. Provided the change in r isn't too dramatic in any given range, that should give a good approximation to the area.

    So, with the values you mention, the areas of the first two sectors are (approximately):
    \frac12\left(\frac{1.95+1.92}{2}\right)^2\cdot\fra  c{\pi}{12} and \frac12\left(\frac{1.92+1.82}{2}\right)^2\cdot\fra  c{\pi}{12}
    You'll be able to simplify it a little by factorising, taking out \frac12\times\frac14\times\frac{\pi}{12}, to get:
    =\frac{\pi}{96}\Big((1.95+1.92)^2 + (1.92+1.82)^2+...\Big)
    Grandad
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