Is this a troll problem?

Is this a troll problem?

"Calculate the blue area"

Calculate with what? The 0 values actually given?

Brainlets, when will they learn?

are you retarded?

That hehe bit in the diagram has got me stumped

yea

It's not a troll problem by any means, you have more than enough information to solve it as given.

The hehe bit just means you have to do a little bit of calculus to subtract from the otherwise straightforward geometric value: half the rectangle area minus one circle minus the hehe bit.

Are you? Go ahead and "calculate" the area of the blue. I want the answer in units.

r^2*pi and you will know that in this case the bottom right blue piece would be around 0,05379 and the piece your looking for is exactly 1 fourth so 0,013475

Double integral who?

Use the formula for circular segment area.
You can find the point where the diagonal intersects the circle analytically.

If this was without your cheeky little [hehe] region, you would just have to do d^2-π(d/2)^2. If I have to explain why, this probably isn't a thread for you.

Now, to subtract [hehe] you would need to do some integral magic for witch I don't have the time right now.

>calculus

scratch that. you can pretty easily determine the arclength of that section and thereby the area of the accompanying wedge to the center of the circle. from there it is simple to figure out the rest.

Come to think of it I only need the point of intersection. Then just integrate from there to 0, and add the triangle formed by the straight line.

>any integral being necessary at all
it's literally just trig

>straight line

DELET

What the fuck?

>it's a youtube channel advertisement episode

d(2d-pi)

1/2 * (d * 2d - 2 * (pi * ( d / 2 ) ^ 2))

It clearly means in ters of d. Don't get hung up on the poor wording.

>not (9/10-pi/2-asin(3/5)/8)d^2
brainlets

This Is pretty straightforward. Write down the coordinates for the top right corner, the center of the right circle, and the point at which the left and right circle touch. Use the point slope formula to get the equation of the line connecting the tangent point to the corner point. Now writedown the equation of the right circle, and take the linear equation away from it, use the quadratic formula. Now that you know these three points, create two vectors centered at the center of the right circle, then take the dot product. Divide by the magnitudes, and arc is, you will get angle. Use this angle and the radius to compute the area of the white sector. Calculating the ehe area is trivial from here.

Arccos*

Easy
d^2 - 0.25*pi*d^2

The "hehe" area is tiny compared to the rest, so you can just neglect it.

You need calculus for the top right portion. Everything else you can get just from geometry.

>I'm autistic and I get hung up over the misuse of the words "compute" and "derive"

WRONG

There's a line that runs through the diagonal of the rectangle of lengths 2d and d that passes through 2 circles with diameters of d. The total area of the circles that are intersected equal to one circle. (If you flip one half of the divided rectangle over the dividing line they're the same, thus the sections of the circles add to one.) This means that all you have to do is subtract the area of one circle from half the area of the rectangle and you get the answer. I don't know why the fuck people are bringing trigonometry into this when it's extremely basic geometry. so it's (2d * d)/2 - pi*(d/2)^2, plain and simple. Maybe someone just psyched you out before starting this problem somehow and made you overthink it?

>wrote all that shit
>doesn't read the thread
>doesn't see the problem
>wonders why people are talking about trig
a losar is u

>Go ahead and "calculate" the third homology group of this manifold. I want the answer in units

You're wrong. Look at the picture again.

Fuck me, I just woke up. When I realized the twist I face-palmed. I'm going to go do other stuff and think about my life.

Oh well, it's something pretty damn close to
((2d * d)/2 - (pi * (d/2)^2)) - ((2d/5) * (d/5))
Good enough for government work.

And by that I meant
- ((2d/5) * d(/10))
I fucked up by using the radius for the vertical for some reason.

d^2 - (pi*d^2)/4
d^2(1-pi/4)
Fuck your hehe part

>negative number

[math]\left(\frac{9}{10} + \frac{\tan^{-1} (2)}{4} - \frac{5\pi}{16}\right) d^2[/math]

It is completly trivia if you know how to evaluate an Integral.

I displayed this on a photo, two blue spheres in the same Nebula.

Found the engineer

in terms of d, you cocksucking engineer

Due to the symmetry about the line, we only have to find the area of the (hehe). This should be achievable through polar coordinates.

We have a circle with centre [math](\frac{d}{2},\frac{d}{2})[/math] and radius [math]\frac{d}{2}[/math]. This gives us a centre in polar coordinates of [math](\frac{d}{\sqrt{2}},\frac{\pi}{2})[/math].

We can take this information to give the polar equation of the circle by plugging it into the general formula (easily derivable from the definition of a circle):
[eqn]r^2-2rr_c\cos(\theta-\phi)+r_c^2=R^2[/eqn]
Where [math]r_c[/math] is the the distance to the centre from the origin, [math]\phi[/math] is the angle, and [math]R[/math] the radius of the circle. Giving us the polar equation
[eqn]r^2-2r\frac{d}{\sqrt{2}}\cos(\theta-\arctan(\frac{1}{2}))+\frac{d^2}{2}=\frac{d^2}{4}[/eqn]

Plugging this into the quadratic formula gives us
[eqn]r=\frac{d}{\sqrt{2}}\cos(\theta-\frac{\pi}{4})+\sqrt{\frac{d^2}{4}-\frac{d^2}{2}\sin^2(\theta-\frac{\pi}{4})}[/eqn]
Which we can use to calculate the area of (haha) via
[eqn]\int_0^{\arctan(\frac{1}{2})}\frac{d}{\sqrt{2}}\cos(\theta-\frac{\pi}{4})+\sqrt{\frac{d^2}{4}-\frac{d^2}{2}\sin^2(\theta-\frac{\pi}{4})}\ \mathrm{d}\theta[/eqn]
Though, fuck knows what that is.

I lol'd

the 2d is unnecessary

>inb4 someone superimposes a graph on top of it and counts the squares

Youre the one talking about the d, faggot.

Apparently it's
[eqn]
\frac{1}{40} \left(-10 \pi d^2+36 d^2-5 d^2 \sin ^{-1}\left(\frac{3}{5}\right)+10 d^2 \tan ^{-1}\left(\frac{3}{4}\right)\right)
[/eqn]