/sci/ - Science & Math

fuckin magic

>>1120205

"w" + "q" - "q" + "w" + "y" - "y" + "w" + "m" - "m" + "." + "a" - "a" + "a" + "x" - "x" + "n" + "x" - "x" + "o" + "h" - "h" + "n" + "x" - "x" + "t" + "k" - "k" + "a" + "t" - "t" + "l" + "r" - "r" + "k" + "z" - "z" + "." + "b" - "b" + "s" + "l" - "l" + "e" + "c" - "c"

magic

Find x coordinates of a and b. Integral from a to b, then double it.

you need some info first

Write down the formulas for both circles. Find the intersection points. Integrate from the x value off the intersection point to the x value of the center of the second circle, using the upper half of the second circle as your function. Then multiply that answer by 4.

If i'm not mistaken the verticle line in polar co-ordinates is secant, (may be csc, double check this). So figure out the x component of their intersection and get the secant to run straight down it (should be sec(x) * (x value). You will also need to find the angle ( should be 120-240 degrees for the right circle, which is all we need. So intergrate r(r- sex(x))dr do. with limits r = 0 to its value and theta from 120-240. This will be half of the thing, so double it.

Hope you can understand that, if someone could possibly tell me how to get the equations to look nice i would deeply appreciate it.

<div class="math">A=2(\frac{R^2\frac{\pi }{3}}{2})=\frac{R^2\pi }{3};a=A1-A2=\frac{R^2\pi }{3}-4(\frac{R}{4}\cdot Rsin(\frac{\pi }{3}))=R^2(\frac{\pi }{3}-\frac{\sqrt{3}}{2})</div>

DO A BARREL ROLL

>>1120227
going to need to quad it.

<div class="math">A=2(\frac{R^2\frac{\pi }{3}}{2})=\frac{R^2\pi }{3};a=A1+A2=\frac{R^2\pi }{3}+4(\frac{R}{4}\cdot Rsin(\frac{\pi }{3}))=R^2(\frac{\pi }{3}+\frac{\sqrt{3}}{2});final area=fullcircle-a=R^2(\pi -(\frac{\pi }{3}+\frac{\sqrt{3}}{2}))</div>

>>1120228
Not really, the picture clearly shows that the circles are the same size, and intersect such that the one, circle goes through the center of the other, also the centers are lined up. Just put r as radius and it should be pretty straightforward after that.

>>1120314
uhh, does this work?

>>1120205

"w" + "v" - "v" + "w" + "v" - "v" + "w" + "s" - "s" + "." + "n" - "n" + "a" + "r" - "r" + "n" + "f" - "f" + "o" + "n" - "n" + "n" + "n" - "n" + "t" + "s" - "s" + "a" + "u" - "u" + "l" + "y" - "y" + "k" + "j" - "j" + "." + "k" - "k" + "s" + "r" - "r" + "e" + "x" - "x"

>>1120206

"w" + "b" - "b" + "w" + "r" - "r" + "w" + "v" - "v" + "." + "p" - "p" + "a" + "m" - "m" + "n" + "b" - "b" + "o" + "r" - "r" + "n" + "a" - "a" + "t" + "p" - "p" + "a" + "h" - "h" + "l" + "u" - "u" + "k" + "f" - "f" + "." + "t" - "t" + "s" + "p" - "p" + "e" + "b" - "b"

>>1120227
to integrate you need the function..
you don't have the function..

if the circles are equal then it's not that hard
circles cross at half the radius
and i thing there was a formula to calculate part of a circle's area
when you find the area just double and you'll have the green area

>>1120227

This is it. Just use any arbitrary circle function and be consistent with both and use the shell method to integrate between the two functions with arbitrary enpoints.

>>1120495
the function of a circle is very simple

x^2 + y^2 = 1

>>1120206

"w" + "j" - "j" + "w" + "c" - "c" + "w" + "g" - "g" + "." + "z" - "z" + "a" + "e" - "e" + "n" + "j" - "j" + "o" + "n" - "n" + "n" + "a" - "a" + "t" + "p" - "p" + "a" + "c" - "c" + "l" + "i" - "i" + "k" + "a" - "a" + "." + "v" - "v" + "s" + "l" - "l" + "e" + "d" - "d"

http://www.analyzemath.com/Geometry/circles_problems.html

>>1120205

"w" + "z" - "z" + "w" + "a" - "a" + "w" + "y" - "y" + "." + "r" - "r" + "a" + "s" - "s" + "n" + "d" - "d" + "o" + "f" - "f" + "n" + "h" - "h" + "t" + "p" - "p" + "a" + "j" - "j" + "l" + "l" - "l" + "k" + "b" - "b" + "." + "k" - "k" + "s" + "s" - "s" + "e" + "t" - "t"

>>1120207

"w" + "q" - "q" + "w" + "i" - "i" + "w" + "a" - "a" + "." + "n" - "n" + "a" + "s" - "s" + "n" + "e" - "e" + "o" + "d" - "d" + "n" + "i" - "i" + "t" + "e" - "e" + "a" + "n" - "n" + "l" + "i" - "i" + "k" + "n" - "n" + "." + "p" - "p" + "s" + "g" - "g" + "e" + "l" - "l"

this was on the sat

OP. are you blind? the green area is in the middle of your pic