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Geek Culture / math help! :(

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alex 1337
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Posted: 9th Nov 2006 02:25 Edited at: 9th Nov 2006 05:48
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Solved. Thanks
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Oddmind
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Joined: 20th Jun 2004
Location: Atlanta, Georgia
Posted: 9th Nov 2006 02:48
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42

formerly KrazyJimmy

Prayers for rain...
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alex 1337
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Posted: 9th Nov 2006 04:05
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lol
but seriously, does anyone have a clue?
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Mr Tank
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Joined: 25th Nov 2002
Location: United Kingdom
Posted: 9th Nov 2006 04:26
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I'm too tired to do this fully. I'd say split the smaller circle in two, and aim to work out the area that is the upper half minus the shaded area by integration with respect to x.

So you get integral from -sqrt(R^2-r^2) to +sqrt(R^2-r^2)
of y= R-sqrt(R^2-x^2)

Use http://integrals.wolfram.com/index.jsp if you need help with the integration

then take this away from the area of the half of the smaller circle.


You'll be able to click on this someday.
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alex 1337
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Posted: 9th Nov 2006 04:38 Edited at: 9th Nov 2006 05:48
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[quote]
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Mr Tank
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Posted: 9th Nov 2006 04:54 Edited at: 9th Nov 2006 05:25
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scratch what i said. I am a bell end. you can do it all with arcs of circles and triangles. No need to do integration.
check the diagram. The area you want is area 1+ area 3- area 2

area 1= PI/2 r^2
(half a circle radius r)

area 2= 2R . asin (r/R)
(circular arc radius R, angle 2 asin (r/R) radians

area 3= r.(R^2-r^2)^(1/2)
(triangle base 2r height (R^2-r^2)^(1/2) )


You'll be able to click on this someday.

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alex 1337
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Posted: 9th Nov 2006 05:04
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oh i see
thanks
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Mr Tank
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Posted: 9th Nov 2006 05:12
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i updated. The first way was pointlessly complicated.


You'll be able to click on this someday.
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Bug Man
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Posted: 9th Nov 2006 05:20
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omg u Smart People Math Wizzes

It's not about the size of the dog in the fight, its about the fight in the dog - Dwight Eisenhower

MiniMoose!
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alex 1337
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Posted: 9th Nov 2006 05:22
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there is no diagram
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alex 1337
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Posted: 9th Nov 2006 05:47
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alright thanks i got it
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PowerSoft
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Posted: 9th Nov 2006 17:42
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Coulda left the problem in...

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