Is this for real? What the fuuuu?

OP here. Found this in /b/

>(pi * 20^2)/4 - 10^2/2 cm^2

Can anyone verify?

oops...

>pi * 10^2

green part is 5.375 cm^2, blue part is half the circle, red part is half area of the angle they're equivalent in area so who cares)

[math]S_{square}=10^2\\
S_{inscribed}=\pi5^2\\
S_{big}=\frac{\pi10^2}{4}\\
S_{rest}=S_{square}-S_{big}-S_{inscribed}\\
S_{shaded}=2(S_{square}-S_{circle}-S_{rest})[/math]
It can't possible be that simple, right?
>t. yuropoor brainlet highschool junior

[math]2(S_{square}-S_{circle}-S_{rest})[/math] should be [math]2(S_{square}-S_{big}-S_{rest})[/math] obviously, sorry

inscribed and big are equivalent, 4 is 2^2, you end up with pi5^2 anyway

Just need to know the area of a square and area of a circle, then add and remove pieces until you manage to get the grey area. Could be done in elementary school if you illustrate it with paper cut-outs and such

I'm surprised that you actually need integrals to find this

You don't. There was a thread about this on >/pol/, and I managed to solve it with some basic knowledge about geometry and some diddly do with the circle's equations. Maybe it would be a lot easier with integrals but I'm too bad at them to be able to tell. Also I don't know math terms that well in English so sorry in advance of this sounding retarded.
Just calculate the dark area on the top and multiply that by two. Add the two relevant circles to an xy-axis, figure out the intersection points using the circles' equations, draw lines from the origins of the circles to those intersection points, draw relevant triangles, figure out some relevant lengths in said triangles and then the angles of the edges that form in the circles' origins. After that you should be pretty much set to go as far as the information that you need to collect. Just calculate the areas like in this picture.

*blocks your geometry

I had a similar method, something close to 25.12cm^2 eventually

assume that the side of the square is 1, then
c+d=pi/8
a+b=1/8-pi/32
2a+c=pi/4-1/2
3a+3b+c+d=1/2

You do not need calculus you utter brainlet

blue area minus orange area x2?

Sci has discussed this problem many times. Leaving aside the "double", the real point of the problem is to compare the sector cut off between a circle and a smaller circle having half the radius, oriented in a "square" position with respect to the larger circle.

It is possible to solve the problem without using calculus-as-such. To do this, you have to look up formulas for circular sectors/segments, identify 4-6 critical points, and grind some trig. It is also possible to re-orient the problem so as to make a solution in calculus straightforward. Careful comparison of both methods shows that they give the same answer, and so are consistent.

Exact forms of the solution are given using fairly ugly trig forms; even simplifying, you still have like 4-8 trig forms IIRC. Using the above first approach for the first time I came up with something like 16 trig forms, ugly as fuck, but correct.

Still, it's a fairly fun problem in the sense that it requires a bit of grinding, but not a STUPID amount of grinding. Any adult who has had a year of calc/precalc and is willing to look up formulas and apply themselves can find the correct answer by applying themselves.

>he thinks he made the problem harder
If you could find the answer to the original problem all you'd have to do is multiply by 4

Integrate between these two points with this graph

integral-calculator.com/#expr=2*(-(x^2+25)^.5-5+((x^2+100)^.5+5*(1+2^.5)))&lbound=-5/2*(7/2)^.5&ubound=5/2*(7/2)^.5&simplify=1&simpallroots=1

Found a solution, check my work

Rotate it 45degrees like I did, and you only need to integrate 1 section instead of 3

Also, if you guys aren't feeling the calc, this way is more geometrical but I'm too lazy to plug in all the numbers

Is that someone's asshole? Chinks are known to be perverted sexually so this is that it could be some mental image of

Just think of it as a graph. What areas can you easily compute, what can those lead to, and so on.

Actually never mind, it's a vagina. It's even referred to as "阴影“

makes the problem easier actually because you can see actually see how to compute the arc area now

It's a Pussy
>elementary school...
A Loli Pussy

What is the physical significance of finding the shaded area? What does it represent?

Wait nevermind its a way to make little kids subconsciously interested in sex. Chinks are sick, wtf!

>Chinks are known to be perverted sexually

They no different from white people in that regard.

>23
>couldn't solve
legitimately think i should just kill myself at this point

almost got a similar solution but then i gave up
looking at yours i realize my system of equations was literally staring me in the face but i didn't realize

god i'm a brainlet

That's actually simplifies the problem user, now you can see the arc is a quarter of a circle with radius 10

reminds me of a 3D variant of this problem featured in a chinese cartoon about a monstrous assassin working as a teacher

>Mathematical achievement is not related to culture in any way, it is purely genetic

You're both useless, these aren't linearly independent equations so you wont get a unique solution.

Meanwhile in brainlet land

ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/final-exam/MIT18_01SCF10_final.pdf

ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/final-exam/MIT18_06SCF11_final_ex.pdf

How can us cucks even compete.
HAAHAHAHAHAHHAHAHAHA

Find determinant of 2*2 matrics

Ahahahahahaha

What a joke!

we are number 1 in the world

Ahahahahahahahaha

Wtf this is really lame shit, are we brainlets?

aren't h2 and r2 are both radii of the larger circle. shouldn't they both just be 10?

Ahem:

That is because math is math, not sex ed.

The fuck is this bs? I had to prove the cayley-hamilton theorem on my first linear algebra final

docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxtYXRoMTE1YWhzcHJpbmcyMDE1fGd4Ojc0OGI4NzhiODBkMDc2NGM

(It starts out babby shit but gets hard to do towards the end, I'm thankful tho bc there was no curve)

Dude, this is not hard unless you are brainlet.
Problem that OP posted was not even collage matertial and its harder than some of the problems from your test.

>some of the problems
No shit sherlock, thats what I said. The issue is that 60% babby shit won't save you from the harder problems when there's no curve and 12ish questions.

No h2 is not the radius of the larger circle. h2 starts from the origin of the larger circle and ends at point C.
From what we got the correct answer was around 29,3.
Oh ok that's pretty interesting.

ayyyyy 29.276 ...
lost some accuracy along the way the first time

i just solved for the intersection then shove that shit into wolfram alpha

holy FUCK i'm a retard
i had to read the thread and it took me way longer than it should, but at least i got it

Lazy but inefficient solution
rotate 45°, find intersection, integrate