You should be able to find the red area. Brainlets need not

>>8886315
not possible

>>8886316
It should absolutely be possible

>>8886315
Nice homework thread

>>8886326
I can tell you that I've already solved this. That's why I know it's solvable using freshman math.

Equation of the large circle
[eqn]x^2 + y^2 = R^2 [/eqn]
Equation of the small circle
[eqn] \left(x - \frac{R}{2} \right)^2 + \left(y + \frac{R}{2} \right)^2 = \frac{R^2}{4} [/eqn]
Set them equal
[eqn]x = \left( \frac{5}{8} \pm \frac{1}{8}\sqrt{7} \right) R [/eqn]

So the area is equal to
[eqn] \int_{\left( \frac{5}{8} - \frac{1}{8}\sqrt{7} \right) R }^{\left( \frac{5}{8} + \frac{1}{8}\sqrt{7} \right) R } \left(-\sqrt{R^2 - x^2} + \frac{R}{2} + \sqrt{R x - x^2} \right) dx + 2 \int_{\left( \frac{5}{8} + \frac{1}{8}\sqrt{7} \right) R} ^{R} \sqrt{Rx - x^2} dx [/eqn]

Or something similiar.

>>8886315
>Hint: This problem is entirely solvable using freshman math.
That's a lie

>>8886315
>You should be able to find the red area.

Good, then I'm not going to bother.

File: solvetheredarea2.png (10KB, 550x550px) Image search: [Google]

10KB, 550x550px

>>8886346
You have the right idea, but your integral is equal to this area, not the entire red area.

https://en.wikipedia.org/wiki/Lune_of_Hippocrates

>>8886315

Method 1, which does not use calculus but which relies upon received formulae for circular segments ("cap" segments of a circle cut off by a chord):

-look up the formulae for circular segments at wiki: https://en.wikipedia.org/wiki/Circular_segment
-WLOG equate the large circle with the unit circle, and suppose that the small circle has a radius of 1/2 and is situated at the appropriate spot in the first quadrant, touching both axes. Both circles have equations in the cartesian plane, find these equations (feel free to look up the equations for a circle to do this, if necessary), and find and show that they share exactly two points in common. Identify those points, and find the distance between those points. Subtract one circular segment from another in order to produce your answer. Hint: the result has a fairly messy expression using trig terms.

Method 2, using calculus and building upon Method 1:

-Given the above picture of things, it is easy to slide the figure "45 degrees counter-clockwise", as it were, and to translate down so that the common points between the two circles rest on the x-axis. From another point of view, two functions of half-circles (the lower halves are now superfluous) are involved. Use an elementary integration technique which should be obvious to you at this point, to set up and compute the integral itself, which is a relatively straightforward process. Check that this new answer, which may be expressed in a somewhat different form, is nevertheless equal to the solution obtained by Method 1 (it is, or else we have a more serious problem). Hint: a diagonal associated with the first picture contains about 4-6 special points which are worth knowing and finding.

>>8886425
Ding ding ding

>>8886315
Computational methods: aka, Monte Carlo.

Solved in less than 45 seconds in Python.

http://www.wolframalpha.com/input/?i=int+(1%2F2)*(sqrt(r-4x%5E2)%2Bsqrt(2)*r)-sqrt(r%5E2-x%5E2)+dx,+x%3D-((sqrt(-4r%5E2%2B12r-1))%2F(4*sqrt(2)))..((sqrt(-4r%5E2%2B12r-1))%2F(4*sqrt(2)))

Here's the solution of the calculus method. Sadly enough WolframAlpha won't give it enough computation time to evaluate. Anyone here with WolframAlpha Pro or Mathematica?

>>8886425

I used polar coordinates and avoided most of that....

>>8886315
The correct answer is to use numerical methods.

>>8886454
You could start by replacing R with 1. The area is proportional to R^2 anyway.

>>8886454
>"Computation took too much time, upgrade to ProPrime?"
Or something like that, I have pro not upgrading again. I tried ; ;

File: 20170505_175041.jpg (1MB, 2560x1920px) Image search: [Google]

1MB, 2560x1920px

Can someone post what the numerical answer should be? This is what I did just using basic geometry, but there's a good chance I made some computational error along the way. Just wanted to check it

https://en.wikipedia.org/wiki/Lune_of_Hippocrates

>>8886390
You're a little nigger dude

File: png.png (12KB, 550x550px) Image search: [Google]

12KB, 550x550px

>>8886883

This is of course a function of the radii of the circles in the OP's figure, and linearly (or quadratically maybe, I should be careful here) depends upon either such radius. So WLOG let's take one example.

Let the radius of the large circle be unity. Then the numerical answer is something like 0.146, or 0.160, or something along those lines. Now let me check myself.

This isn't even hard to approximate so that you know that your numerical answer makes sense. Consider the following: In such an arrangement, then red + yellow + green = 1 - π/4, and green = 1/4 - π/16, so that their difference, red + yellow, is (12 - 3 π)/16 ~ 0.161 (notice how this is close to the upper suggestion I made before checking). Obviously R+G is a fairly good, strictly larger upper bound on R, and simply looking at this simple picture is enough to say, by sheer reasonability and size/number sense alone, that the final answer must be /on the order of/ 0.14 to about 0.16, as I'd as much as said earlier. And when you actually do the problem that's what it turns out to be, as I recall.

File: Sketch55182110.png (169KB, 720x1280px) Image search: [Google]

169KB, 720x1280px

You only need basic knowledge of geometry and basic linear algebra.

File: 1493909397087.jpg (470KB, 1920x1280px) Image search: [Google]

470KB, 1920x1280px

>>8886315
>You should be able to find the red area.

It is in the right bottom corner of that image.

You're welcome.

>>8886883
I'm not absolutely 100% certain but I think the exact answer in a similar form as yours should be

[math]\frac{r^2}{8}(\sqrt{7} + 2 tan^{-1}(\sqrt{7}) - 8 tan^{-1}(\frac{\sqrt{7}}{5}))[/math]

>>8886945
What happened to latex on /sci/ anyway? I guess it's readable anyway.

>>8886945
>>8886883

The arctan soup is familiar, you're both on the right track.

There is a rather messy pic which gets posted (surprised it hasn't appeared yet), and it uses a different approach from mine, but I remember concurring with its numerical approximation after careful review.

>>8886883
never mind this it was wrong, found mistakes

File: Screen Shot 2017-05-06 at 01.41.17.png (10KB, 189x66px) Image search: [Google]

10KB, 189x66px

You really only need to know the surface of this three corner shape, and for that you only need to know the x-value at which the circles cross (and then you integrate the surface under the two circles)

File: 1429466022664.png (7KB, 373x334px) Image search: [Google]

7KB, 373x334px

>>8886951
A similar problem has been posted on /sci/ before. http://imgur.com/a/4HkJc has the solution for it and at least for r = 1 >>8886945
seems to be correct:

http://www.wolframalpha.com/input/?i=1%2F8*(sqrt(7)+%2B+2+tan%5E(-1)(sqrt(7))+-+8+tan%5E(-1)(sqrt(7)%2F5))

File: 1429466807005.png (140KB, 1218x972px) Image search: [Google]

140KB, 1218x972px

>>8886951
This is probably the picture you're talking about

>>8887017

Yes it is. I've always kinda disliked it for multiple reasons: the shit text formatting, the too-cute non-sequitors, and I didn't even "like" the approach, despite the approach being valid.

Nevertheless, the numerical approximation given is correct; the author simply used a different method, as I'd already said.

>>8886883
Ah, I forgot the factor of 2 on both of the arctan(...) terms when finding the area of a sector, since we have two sectors with angle arctan(...).
Now it gives the correct (0.146...)*R^2

Just fucking move the smaller circle, so long as its center rsqrt(1/2) from the origin the area outside the large circle will remain the same.

Just move the small circle to be centered at (0,.7071r) and now it's a much easier integration.

>>8887773
https://www.desmos.com/calculator/0wibphscvi