# Thread: Area of a segment when r changes

1. ## 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

2. Originally Posted by Shizzle
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.

3. Hello Shizzle

Welcome to Math Help Forum!
Originally Posted by Shizzle
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
$\tfrac12\int_{\alpha}^{\beta}r^2d\theta$

4. 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?

5. Hello Shizzle
Originally Posted by Shizzle
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:
$=\frac{\pi}{96}\Big((1.95+1.92)^2 + (1.92+1.82)^2+...\Big)$