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Math question SteveB 10-07-2009
---> Re: Math question norminn@earthli...10-07-2009
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Posted by blueman on October 7, 2009, 5:36 pm
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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.
Posted by SteveB on October 7, 2009, 7:20 pm
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For my use, I took four widths, averaged them, then multiplied by the
length.
Close enough.
Steve
Posted by HeyBub on October 7, 2009, 4:04 pm
SteveB wrote:
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You can't. That's what Integral Calculus is for.
Posted by MikeB on October 7, 2009, 6:23 pm
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So what is the formula then, or how would one use integral calculus to
derive the area of the pool?
Posted by HeyBub on October 7, 2009, 7:46 pm
MikeB wrote:
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First you write the equation for the curve as a function of x: f(x) =
equation.
Area = the integral [from 0 to max x] f(x)dx. Turning the crank gives the
answer.
http://hyperphysics.phy-astr.gsu.edu/Hbase/integ.html#c3
An alternative is the Monte Carlo method.
Surround the curve with a box. Generate random points that will land inside
the box. Determine whether each generated point is inside the curve or
outside. If 62% of the random points lie within the curve, the area of the
curve is 62% of the area of the box. Obviously precision grows as a function
of the sheer number of points.
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