Math question

Alone, you don't.

_MINIMUM_ area is that of circle of same circumference, how much greater depends on the eccentricity.

Example of magnitude difference depending on shape, multiplier is pi for a circle, 4 for a square of the same "radius" so square would bound 4/pi

--> ~33% greater area.

How critical is the measurement? I would sketch out the length and width, cut off triangles for the belly of the kidney and outside the curves....area of the rectangle less the areas (roughly triangular) outside of the curves should give a fairly close measurement.

You can't. That's what Integral Calculus is for.

You can estimate the area by overlaying the circumference of a couple of circles, figuring the area of each, then adding those areas together. Take the remaining area not covered by your circles, and estimate that area, adding it to the previous area to obtain your final rough estimate.

Jon

If accuracy is important, I'd use the Simpson's Rule formula, where you take measurements across the pool at interals and plug those distances into the formula. You also have to plug the interval distance into the formula.

Why do you ask?

Found something. See problem #6 in the following link. It shows an example without too much math jargon

On 10/7/2009 1:18 PM mike spake thus:

Ackshooly, that's called "Simpson's approximation", but yes, it does work as you described. It's a weighted-average method of approximating the area under a curve.

Wikipedia says it's Simpson's Rule:

MAXIMUM not minimum is bounded by a circle

NO. Circle has area: pi * r^2 and perimeter 2*pi*r so multiplier is r/2

And a square with equivalent perimeter has area: pi^2 * r^2/4 So square is pi/4 as large - or about 22% SMALLER than a circle with equivalent perimeter

The OP asked for AREA not volume in gallons. Also your formula at best is some vague type of approximation since there is no standard kidney-shape and certainly irregular-shaped is even less well-defined. Although since the site doesn't define what A and B are, the formula will by definition be true for some values of A and B ;)

This is probably the best simple way if an approximation is OK. You can get as precise as you want by making the sketch more precise and projecting it on a fine grid and counting the "squares" and fractions of "squares" covered by the pool.

blueman wrote: ...> NO.

Brain fart... :(

circle has minimum perimeter for given area, and i turned it around w/o thinking....

--

And, as delta-x approaches zero, you get the integral.

So what is the formula then, or how would one use integral calculus to derive the area of the pool?

Use SketchUp. It's probably the easiest way.

Actually it's a geometry question...

For my use, I took four widths, averaged them, then multiplied by the length.

Close enough.

Steve

And the answer is ...................?