Give me your hardest "find the area of the shaded region" problems

Then tell me what the area would be when a = 1 and b = 2

3/4pi

area of A ≈ 0.79
area of B ≈ 3.14
area of C ≈ 7.07

(7.07/2) - (3.14/2) + (0.79/2) = 2.36

Very well, but you lose points for not having the solution in terms of a and b.

>Give me your hardest "find the area of the shaded region" problems.
provides an elementary school level problem
A(a) = (1/2)*pi*(((a+b)^2)+(a^2)-(b^2)) = pi*((a^2)+ab)
A(b) = (1/2)*pi*(((a+b)^2)-(a^2)+(b^2)) = pi*(ab+(b^2))

Look at it as 4 half circles and its easy.

Kek

>>Give me your hardest "find the area of the shaded region" problems.
>provides an elementary school level problem
THEN POST HARDER ONES.

(3/4pi)*(a/a)*(b/b)

Alright Veeky Forums let's get serious.

a*(a+b)*pi/4

(πab)/4

1/2 * pi * (a/2)^2 + ((a+b)/2)^2 * pi * 1/2 - (b/2) ^2 *pi * 1/2

1/2 * pi * [ (a/2)^2 + ((a+b)/2)^2 - (b/2) ^2 ]

1/2 * pi * [ a^2 * 1/4 + (a^2 + 2ab + b^2) * 1/4 - b^2 * 1/4 ]

1/8 * pi * [ a^2 + a^2 + 2ab + b^2 - b^2 ]

1/8 * pi * [2a^2 + 2ab ]

pi/4 * [ a^2 + ab ]

pi/4 * a * (a+b)

If Veeky Forums can't solve this introductory problem you are all nothing but popsci memers who should never discuss QM.

Mathfag here and there are about 20 terms that I do not know the definition for.

This is why math threads are way better when it is an actual complex problem that can be described with elementary terms and not your physics technobabble that could be solved by a toddler with a dictionary.

Take your pedophile cartoons back to .

Then grab a dictionary and let us see your "solution" you petulant brainlet.

Not all mathematical terms are elementary, either. Are you in high school?

>QM

Fuck off. This is a math thread.

Solved it in my head in like 4 minutes. Would've been quicker but I had a phone call from my terminally ill Gran and couldn't stop thinking about the Hodge conjecture.

something like this?

[math]L[/math] is the lattice [math]\left(a\mathbb{Z}/\Lambda\mathbb{Z}\right)^n[/math] where [math]a[/math] is the lattice spacing and [math]\Lambda[/math] is the lattice size. Compact support means the interaction term is zero on all but a finite subset of the lattice as [math]\Lambda \rightarrow \infty[/math].
There are no indices missing. Good job assuming the existence of a number operator you dunce.

(pi/8) * (2a^4 - b^4 + a^2 + b^2 + 2ab)

(Pi/8)(2a^2+2ab)

Right.
This simplifies to the right answer.
You fucked up user.

It takes me some time...

Fuck, I forgot to square it.

typical engineer
>how can i integrate this...

>two dimensional phase space
I thought physicists liked hard problems.

[math]x[/math] and [math]p[/math] are the two canonical coordinates on the symplectic manifold [math]M[/math] whose lagrangian manifold is the phase space [math]D[/math]. Symplectic manifolds are [math]2n[/math] dimensional where[math]n \in \mathbb{N}[/math].
If you need me to BTFO you again just ring me up.

Take your pedophile cartoons back to .

the proof is trivial

Verbose.

What a load of bs

pi(((A+B)/2)^2)/2 - pi((B)^2)/2 + pi((A)^2)/2

Using only chords, divide a circle into equal area pieces with no two pieces congruent. Pic related almost does it except for that small intersection on the left

>Pic related almost does
How the hell do you figure

What do you mean? The structure is fairly simple to understand once you realize each piece has definite area since they divide the circle equally. So each chord is defined by how it divides the circle, and the separation of intersecting chords is defined by the area of its intersections.

I'm in college and I don't even know how to start this problem and how you got to that answer.

Could anyone provide a good explanation or resource? I'd appreciate it.

you have the big circle, diameter is a+b (radius is diameter halved)
on the left you have a smaller circle diameter of a
you have on the right half circle with a diameter of b

area of circle is r^2*pi

do you get it now?

Yeah but what specifically did you do to ensure the pieces were near equal? Just guess and check with different random chords?

I guess what I'm wondering is whether this is a problem with analytic solutions or an exercise in numerical optimization.

Simply plug it into your favorite symbolic geometry software package.

He just means x and p are vectors.

Good question. If we start with the property that the division must be equal then we can see that the number of possibilities for some number of chords is finite. There are only so many ways the chords can intersect. However the problem does not appear to be analytically solvable as the equations that govern the position of a chord based on the areas it divides the circle into can only be solved numerically.

Alright smartasses try this one on for size.

Find the volume of the intersection of eight spheres of radius [math]r[/math] centered at the corners of a cube of side length [math]r[/math].

Yeah but Poincare Bendixon only works in two dimensions. More than that and things get more complicated.

that's what makes the problem so difficult

Darn. I guess I'd need to read up on the differential geometry of symplectic manifolds then. Or just hamiltonian vector fields.

You can't solve it without more information. That's the point of the "you should be able to solve this" meme.

>Look at it as 4 half circles and its easy.
You don't know that that's what they are though.

not defined for all a,b; consider x =1 if [a,b]=[-1,1].

Don't fret. A lot of people know the answer or find it easy, so the answers without explanation make sense to them. Here's how it works (you can only solve this is you assume the line shown is dividing the circle):

The top half of the circle has a diameter of a+b. The area of a circle is pi * r^2. This is a half circle (so, half the area of a full circle. The radius is half the diameter, or 1/2 * (a+b) in this case. Therefore, the area is

pi/2 * ((1/2)(a+b))^2 = pi/2 * (1/2)^2 * (a+b)^2 = pi/2 * 1/4 * (a+b)^2 = pi/8 * (a+b)^2.

Now, we only want the shaded area. All we need to do is subtract the non-shaded region, which is simply half a circle with diameter b. So, let's apply a similar process to the top portion.

Area of non-shaded region of top part (another half circle or semicircle) is pi/2 * (b/2)^2 = pi/2 * b^2 * 1/4 = pi/8 * b^2

Okay. Now we need to look at the bottom half. We want the shaded region of the bottom half, which is simple half of the circle defined with diameter a.

Area of shaded part of bottom half (also a semicircle) = pi/2 * (a/2)^2 = pi/8 * a^2

- We know the whole area of the top half (both shaded and non-shaded) is:

pi/8 * (a+b)^2

- We know the area of the top half that is non-shaded (and thus needs to be subtracted from the area of the top half to leave only the shaded area):

pi/8 * (a+b)^2 - pi/8 * b^2

- We know the area of the bottom half that IS shaded and thus needs to be added:

[pi/8 * (a+b)^2] - [pi/8 * b^2] + [pi/8 * a^2]

The above is the final formula.

---

How do I use TeX here?

Don't you ever grow tired of this? You must be in every second thread on Veeky Forums

>Shaded area= [(piAB^2)/2]-(Bpi^2)+[(Api^2)/2]

Does this look correct?

Underrated meme

this is the hardest one.
bet you cant do it

why do you enable /a/utism on Veeky Forumsence ?

kek

So much irony

back to

u fukin pedophile stop stealing my meme fucking cancer go back

Not an argument

s-sorry senpai

impossible with calculus I

fuck, I just realized I'm a cargo culter
lmao, area=integration in my brain

do you know if it's possible?
how about we proof that first

Okay looks like no one is attempting this might as well post the answer because I thought it looked cool.

...

>Subtle homework help post

the anti derivative of 1/x is ln(x) therfore the limit does not exist.

The area is "infinite".

This is easy.
Let A be the area of the entire circle, so A = pi * (a+b)^2.
Call the areas of the grey and white regions G and W. Then
G = 0.5 * pi * b^2 + (0.5 * A - 0.5 * pi * a^2)
and
W = 0.5 * pi * a^2 + (0.5 * A - 0.5 * pi * b^2).
More simply,
G = pi * (b^2 + a * b)
and
W = pi * (a^2 + a * b).
As expected, G + W = A.

for given a, b:
grey semicircle area = pi/2 * a^2
white semicircle area = pi/2 * b^2
large semicircle area = pi/2 * (a + b)^2 = pi/2 * (a^2 + 2ab + b^2)

grey area = grey semicircle + (large semicircle - white semicircle)
= pi/2 * (a^2 + (a^2 + 2ab + b^2 - b^2))
= pi/2 * (2a^2 + 2ab)
= 2pi * a * (a + b)

Fuck me, I'm retarded. Replace a and b with a/2 and b/2 obviously.

Result is pi/2 * a * (a + b)

And I'm even more retarded because (assuming a, b are the radii of the smaller semicircles):
pi/2 * (2a^2 + 2ab) = pi * (a^2 + 2ab) = pi * a * (a + b)

Since a, b should be the diameters rather than radii, replace a

Should be easy

seems so, only tricky part is the integration necessary for the upper right portion of the shaded region, unless there's an easier method I am missing.

I'm more bored of your repetitive posts than manga scribbles.

[eqn]90-25 \pi +\frac{25}{2}\ \sin ^{-1}\left(\frac{3}{5}\right)[/eqn]

I'm sure there is. This is a sixth grade problem for Chinese kids, so no calc probably

Find the area of the region in terms of R or r given that: the radius of the big circle is R, the radius of the small circle is r, and the center of the small circle intersects the big circle.

Really?

I don't know what I'm missing.

[eqn]
\begin{cases}
\pi R^2 & R\leq \frac{r}{2} \\
\frac{1}{2} \left(-r \sqrt{4 R^2-r^2}-2 R^2 \tan ^{-1}\left(\frac{r \sqrt{4 R^2-r^2}}{r^2-2 R^2}\right)+2 r^2 \cot ^{-1}\left(\frac{r}{\sqrt{4 R^2-r^2}}\right)+2 \pi R^2\right) & \frac{r}{2}

Why are people saying this is impossible without calculus?

It's literally just difference and sums of areas of semicircles.

[math] \frac{1}{2} \pi ( \frac{a+b}{2} )^2 - \frac{1}{2} \pi ( \frac{b}{2} )^2 + \frac{1}{2} \pi ( \frac{a}{2} )^2 [/math]

One can use the formula for the area of a circle cut by a chord.

mathworld.wolfram.com/CircularSegment.html

The Chinese kids probably learned that.

Oh boy, hope you guys are ready for this.
As someone pointed out, only the upper right shaded area takes work. I won't simply calculations with the hope that the work is clearer.
Call the area of the right circle above the cord A. Noting that the line forms an angle of a = arctan(1/2), we see that
A = ((2*(pi/2 - a))/(2*pi))*(pi*5^2) - 2*(0.5*5*cos(a)*5*sin(a))
= 17.67871794...
Let S = 25*(1 - pi/4), the area of the lower right shaded region. Let T be the area of the shaded upper right region.
Then
T = A + 2*S - 25
= 3.40880977.

Assuming no circle completely overlaps the other:
Find the two points of intersection, draw a line between them (the chord), and just add the two areas "above" the cord.

The area of the entire rectangle minus the area of the two circles is ~42.

You'd expect the area of the shaded region to be a little less than half of that, no?

Because Veeky Forums is infested with unenlightened engineer half-wits.

Using my answer, we get that the shaded area is about 19: the "easy" shaded regions are about 5 and the smaller one is about 3.
Maybe you misread my post.

Yeah I did my bad.

...

Not exaggerating when I say you should be able to do this in your head.
I didn't write anything down I just used my browser calc to get an approximate answer.
it's 18.7776607

protip: you can divide the shaded area into 3 and a half equal portions. You can calculate a single portion by subtracting 1/4 of the area of a circle from a square.