Alternate formulas for calculating area of a circle??

This thread here is to compile a list of alternate formulas or techniques to calculate area of a circle if you know any please post it thank you.

 

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A=2pi*r^2 - pi*r^2

 

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Gotta laugh a little everyday!

Thanks.

 

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A formula less obvious that doesn't involve pi or tau?

 

For the area of any finite area which is trapped between an upper curve and a lower curve, method I is the integral between the points where the curves meet of the upper curve minus the lower curve. For the area of any finite area which is bounded by a piece-wise smooth closed curve, the integral of \(\frac{x dy - y dx}{2}\) around the curve in a counter clockwise manner gives the area, this is method II.



I: \(\int_{-R}^{R}\; \sqrt{R^2 - u^2} - \left( - \sqrt{R^2 - u^2} \right) \; du \\ = 2 \int_{-R}^{R}\; \sqrt{R^2 - u^2} \; du \\ = 2 R^2 \int_{-1}^{1}\; \sqrt{1 - u^2} \; du \)
II: \(\oint_{x^2 + y^2 = R^2} \; \frac{x dy - y dx}{2} \\ = \oint_{x^2 + y^2 = R^2} \; \frac{-x^2 - y^2}{2 y} dx \\ = \int_{R}^{-R} \; \frac{-R^2}{2 \sqrt{R^2 - x^2} } \; dx \quad + \quad \int_{-R}^{R} \; \frac{-R^2}{-2 \sqrt{R^2 - x^2} } dx \\ = R^2 \int_{-1}^{1} \; \frac{1}{\sqrt{1 - x^2} } \; dx \)

Not surprisingly, \( \int_{-1}^{1}\; 2 \sqrt{1 - x^2} \; dx = \int_{-1}^{1} \; \frac{1}{\sqrt{1 - x^2} } dx \), as geometry and algebra require.

The observation that \(\frac{d}{dx} \left( x \sqrt{1 - x^2} \right) = 2 \sqrt{1 - x^2} - \frac{1}{\sqrt{1 - x^2} } \) confirms it.

Since \(\sin^2 x + \cos^2 x = 1\) and \( \frac{d}{dx} \sin x = \cos x\) we have \(\frac{d}{dx} \sin^{-1} x = \frac{1}{\cos \sin^{-1} x} = \frac{1}{\sqrt{1 - x^2}}\) and so the area of a circle divided by \(R^2\) is given two ways to be equal to \(\sin^{-1} 1 - \left( \sin^{-1} ( -1 ) \right) = 2 \sin^{-1} 1 = 4 \tan^{-1} 1 = 2i \ln \frac { i + 1}{ i - 1} = 2 i \ln (-i) = 2i \times \left( - \frac{\pi i}{2} \right) = \pi\)

It is left as an exercise for the reader to show that methods I and II work for general triangles and quadrilaterals.

 

0.78539816339*Di*4

 

apparently the area of a circle is related to:
sin(x),sin(x+90).
plotting the above coordinates onto a plane produces a circle.

 

If you like, I can compute the area of the circle from your degree-based sine function.

First establish a relation between degree-based SIN and the usual radian-based sine function of mathematics, and use this to compute the derivatives:
\( i = e^{\frac{i \pi}{2}} \\ \frac{ d \quad }{dx} e^{ax} = a e^{ax} \\ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \\ \textrm{SIN}(x) = \sin \frac{\pi x}{180} = \frac{e^{\frac{ i \pi x}{180}} - e^{-\frac{ i \pi x}{180}}}{2i} \\ \textrm{SIN}(x + 90) = \sin \frac{\pi (x + 90)}{180} = \sin \left(\frac{\pi x}{180} + \frac{\pi}{2} \right) = \frac{e^{\frac{ i \pi x}{180}}e^{\frac{i \pi}{2}} - e^{-\frac{ i \pi x}{180}} e^{-\frac{i \pi}{2}}}{2i} = \frac{ i e^{\frac{ i \pi x}{180}} + i e^{-\frac{ i \pi x}{180}} }{2i} = \frac{ e^{\frac{ i \pi x}{180}} + e^{-\frac{ i \pi x}{180}} }{2} \\ \left( \textrm{SIN}(x + 90) \right)^2 + \left( \textrm{SIN}(x) \right)^2 = \frac{ e^{\frac{ i \pi x}{90}} + 2 + e^{-\frac{ i \pi x}{90}} }{4} + \frac{e^{\frac{ i \pi x}{90}} - 2 + e^{-\frac{ i \pi x}{90}}}{-4} = 1 \\ \frac{ d \quad }{dx} \textrm{SIN}(x) = \frac{\left( \frac{ i \pi}{180} \right) e^{\frac{ i \pi x}{180}} - \left( \frac{ -i \pi}{180} \right) e^{-\frac{ i \pi x}{180}}}{2i} = \left( \frac{\pi}{180} \right) \frac{e^{\frac{ i \pi x}{180}} + e^{-\frac{ i \pi x}{180}}}{2} = \frac{\pi}{180} \, \textrm{SIN}(x + 90) \\ \frac{ d \quad }{dx} \textrm{SIN}(x + 90) = \frac{ \left( \frac{ i \pi}{180} \right) e^{\frac{ i \pi x}{180}} + \left( - \frac{ i \pi}{180} \right) e^{-\frac{ i \pi x}{180}} }{2} = i \left( \frac{ i \pi}{180} \right) \frac{ e^{\frac{ i \pi x}{180}} - e^{-\frac{ i \pi x}{180}} }{2 i} = - \frac{\pi}{180} \textrm{SIN}(x) \)

Since \(x = \textrm{SIN}(\theta), y = \textrm{SIN}(\theta + 90)\) describes a circle of radius 1 in the clockwise direction, we compute its area as:

\(x = \textrm{SIN}(\theta), y = \textrm{SIN}(\theta + 90), dx = \frac{\pi}{180} \textrm{SIN}(\theta + 90) \, d\theta, dy = - \frac{\pi}{180} \textrm{SIN}(\theta) \, d \theta\)
\(dA = - \frac{x dy - y dx}{2} = \frac{\pi}{360} \left( \left( \textrm{SIN}(\theta + 90) \right)^2 + \left( \textrm{SIN}(\theta) \right)^2 \right) d \theta = \frac{\pi}{360} d \theta\)
\( \oint_{\theta \in \left[0, 360\right) } - \frac{x dy - y dx}{2} = \oint_{\theta \in \left[0, 360\right) } dA = \frac{\pi}{360} \int_{0}^{360} d\theta = \pi\)

 

actually, what i posted was an off shoot of a BASIC program i was working on.
the formulas i actually used were:
sin(d/57.29583)*170+175 and sin((d+90)/57.29583)*170+175
due to monitor resolution, the x variable had to have an aspect multiplier of 1.48 and an extra 60 pixels added to get center screen.

i do have a question though, is what i posted in 7 above a 2 unit circle?

edit:
the aspect ratio was found by trial and error.
i used an ordinary school ruler to measure it.
not exact, but good enough.

 

I am a bit curious about "57.29583" since 180/π ≈ 57.295779513....
If you were using IEEE doubles, a common choice with general purpose computing, the best approximation of \(\frac{180}{\pi}\) is \( \frac{1007958012753983}{2^{44}} \approx 57.29577951308232 \), while if you have lower-precision IEEE floats, the best you can do is \(\frac{15019745}{2^{18}} \approx 57.295780\).

There are any number of rational numbers between any two distinct reals. So there are rational numbers in abundance, and like all real numbers they may be ordered in magnitude:

5729577/10^5 < 469367/2^13 < 1007958012753982/2^44 < 1204537366/21023143 < 180/π < 2061911869/35987151 < 1007958012753983/2^44 < 5729578/10^5 < 147995/2583 < 15019745/2^18 < 143927/2512 < 17819/311 < 18×10^6/314159 < 1007958864140413 / 2^44 < 7509879/2^17 < 5729583/10^5 < 13751/240 < 15019759/2^18

Notice, that 57.29583 = 5729583/10^5 is anomalously high for a computer representation and is much closer to 180/3.14159 than 180/π. So I strongly suspect it is actually obtained by using the decimal approximation of the binary approximation of the ratio of 180 with the decimal approximation 3.14159.

Why is that bad? Because it takes 360 degrees = 2 π radians to close a circle, and if you divide 360 by a number larger than 180/π then you get a number smaller than 2 π and your circle doesn't close (even by computer IEEE floating point approximations). How much does it not close by? At least 2π - ( 360 / (18×10^6/314159 )) ≳ 1/188424 radians or about
0.00008% of a circle. By contrast, dividing by 57.29577 causes the circle to overlap itself it by a smaller amount.

While there might be legitimate reasons for using a formula for a circle that does not close completely in computer graphics, there is no reason to use approximate arithmetic if one is trying to answer a geometric question about the area of a circle with known radius.

 

Unit has multiple meanings:

For "the first and least natural number" = 1, the circle described in post #7 is centered at the origin and has radius 1, and such a circle is referred to as "the unit circle."
For "a determinate quantity adopted as a standard of measurement", one could say the radius is 1 (unit), and the circle has a diameter of 2 (units) , a circumference of 2 π (units) and an area of π (units)².

But the phrase "2 unit circle" doesn't signify anything in English. It is better to say "a circle of radius 2 units" if that's what you wanted to mean.

 

well, i was taught a few things in trig class that i still remember.
first, there are 2 pi radians in a circle.
second, 5 places is usually sufficient.

the formula i presented is used in a graphics program and isn't meant to be instructive or informative.
it's accurate enough for the resolutions i have so i have no need to get any closer.

 

i've heard the phrase "unit circle" used and i believe the area of this circle is pi.
if i'm not mistaken, the diameter is one unit.
by extension, if the above is true then what i posted is a "2 unit" circle, although i've never heard of such a thing.
you gotta remember, i'm not exactly a math genius.

 

Correct.

Incorrect.

Incorrect

Incorrect.

Correct.

Correct.

Sources:
http://www.mathsisfun.com/geometry/unit-circle.html
http://www.mathmistakes.info/facts/TrigFacts/learn/uc/uc.html
http://www.themathpage.com/atrig/unit-circle.htm
http://en.wikipedia.org/wiki/Unit_circle
http://mathworld.wolfram.com/UnitCircle.html

 

you are sadly mistaken if you think the diameter isn't the unit in a unit circle.
it MUST be if the area is pi.

 

Pi*r^2 = area

So r = 1 and D = 2

 

Or assuming leopold doubts that Area = π r² = (π/4) D² applies to a circle (even though this was demonstrated in posts #5 and #8 above), we can prove that his circle of area π does not have a diameter of 1:

A disc of diameter x is a proper subset of a square with side x with the same center point. Therefore the area of the disc must be no larger than the square.

By definition, a square with unit side has an area of 1 square unit. Therefore a circle with diameter 1 unit has an area no larger than 1 square unit.

Yet the unit circle, as the sources and leopold agree upon, has area of π square units > 1 square unit, therefore it's diameter is necessarily larger than 1 unit.​

 

i don't know what i was thinking . . . apparently i wasn't.

 

(H*r)^2*2=A

H = sqrt{pi/2}

 

That fails to measure up to the topic of the thread as expressed in the OP. The formula is algebraically identical to the formulas derived in post #5 and extrapolated from post #8 but there is no general principle demonstrated. In addition, this post is repetitive lunacy cross-posted from another thread, e.g. http://www.sciforums.com/threads/h-pi-sqrt-pi-2.143482/ , in violation of forum rules.

You would have been better served by studying the general methods displayed in post #5 and attempt the simple exercises in the last sentence of that post.