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

Although r changes with $\displaystyle \theta$, 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 $\displaystyle r$ in each $\displaystyle 15^o$ ($\displaystyle = \pi/12$ radians) range, and use $\displaystyle \tfrac12r^2\theta$ with that value. Provided the change in $\displaystyle r$ 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):$\displaystyle \frac12\left(\frac{1.95+1.92}{2}\right)^2\cdot\fra c{\pi}{12}$ and $\displaystyle \frac12\left(\frac{1.92+1.82}{2}\right)^2\cdot\fra c{\pi}{12}$

You'll be able to simplify it a little by factorising, taking out $\displaystyle \frac12\times\frac14\times\frac{\pi}{12}$, to get:$\displaystyle =\frac{\pi}{96}\Big((1.95+1.92)^2 + (1.92+1.82)^2+...\Big)$

Grandad