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Isn’t your place in the politics thread?

Could you please provide problems that don't involve advanced mathematics? Have mercy on a humanities student.

Isn't your place the thread about politics?

We have two threads. It's not that many compared to the whole forum... And even then, they try to spoil it. :mad::mad::mad:

Could you please not post math problems? Have mercy on us humanities folks.

There are problems for humanities folks here too. The last one, by the way, isn't math, but physics... :)

Why are you simply arithmetically adding the forces ρg(H-h)*dS(h) with which the water presses on infinitesimally small strips?

Ah, yes, the projection onto the vertical is needed... The number of errors is overwhelming me.

mg=integrate (h from 0 to H) ρg(H-h)*(-sin(α))*dS(h) ----------- (1) It seems like there should be a "-" sign here. Because the direction is upwards, not downwards, and because I need to adjust the answer. ))
integrate (h from 0 to H) ρgHsin(α)*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgH*2π(Rcos(α))sin(α)*Rdα = πρgHR^2 integrate (α from 0 to arcsin(H/R)) sin(2α)dα = -0.5 πρgHR^2 cos(2α) dα (α from 0 to arcsin(H/R)) = -0.5 πρgHR^2 (cos(2arcsin(H/R))-1) = -0.5 πρgHR^2 (1-2(H/R)^2-1) = πρgH^3 -------------- (2)
integrate (h from 0 to H) ρghsin(α)*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgRsin(α)sin(α)*2π(Rcos(α))*Rdα = integrate (α from 0 to arcsin(H/R)) ρgsin(α)sin(2α)*πR^3 dα = ...
Let's recall the formula: sinx siny = 1/2 (-cos(x+y)+cos(x-y))
and continue:
... = integrate (α from 0 to arcsin(H/R)) ρg((cos(α)-cos(3α))/2)*πR^3 dα = 0.5 πρgR^3 (sin(α) - (1/3)sin(3α)) (α from 0 to arcsin(H/R)) = 0.5 πρgR^3 (H/R - (1/3)sin(3arcsin(H/R))) = ...
Sine of the triple angle sin (3x) = 3 sin x - 4 sin^3 x
... = 0.5 πρgR^3 *4 (H/R)^3 = 2 πρgH^3 ----------------- (3)
(1),(2),(3):
m = πρH^3

H=(m/πρ)^(1/3)

Now does the answer look more plausible? )

Almost. Mine is slightly different.

Added 1 minute ago

mg=integrate (h from 0 to H) ρg(H-h)*(-sin(α))*dS(h) ----------- (1) It seems like there should be a minus sign here.

Why the minus sign? Which direction are you pointing the axis onto which you are projecting the forces acting on the sphere: up or down?

Why the negative feedback?

Mainly because I got a negative answer and I couldn't be bothered to check it for the millionth time. Argh. Okay, I'll check it again anyway.

If you direct the axis upwards, then the equilibrium condition in its projection will be -mg + [your integral] = 0, and if downwards, then mg - [your integral] = 0.

Yes. That's why it's unclear why the answer is negative. In general, it's obvious that H >= h, so the integral should be positive...

There you can degrade yourselves, and leave this place for us...:mad::mad::mad:

And you go to hell

And you go to the воопу.

Is the letter "t" missing here? But there are dragons!

Added 4 minutes later

Yes. That's why it's unclear why the answer is negative. In general, it's obvious that H>=h, so the integral should be positive...

Just write it with a negative answer. There should definitely be a plus sign with the sine.

By the way, I'm a fan of "What? Where? When?" trivia games, and I can post some of the most interesting questions here...

Tox, to ask a question, you need to answer the previous one... :D

dfweafgrweguiafnwnefuweanfawiuo!!! >_____<

Trivia questions are relevant here!

At least write something with a negative answer. There should definitely be a positive sine.

H=-(m/πρ)^(1/3)
Nothing else changes.
Haven't found any errors yet.