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[Furtuka]

I need some help with a multivariable calculus problem.

Spoiler: The instructions (click to show/hide)

Spoiler: What I did (click to show/hide)

Spoiler: My teacher's incomprehensible response (click to show/hide)

Your objective is to find the volume V of that four-dimensional sphere (let's call it B4, the set of all points (x, y, z, t) with x²+y²+z²+t²<=rad²), by integrating 1 over its contents. To do this, you'll first want to find a nice parametrization of all the points in that sphere. For any valid parametrization p from any parameter space A to B4, the volume of the sphere is given by V = int_A{det(Dp(x₁, x₂, x₃, x₄))}. (*)

The first and most obvious parametrization you could use is the canonical parametrization
    p1: B4 -> B4; (x, y, z, t) |-> (x, y, z, t).
In this case, det(Dp1) = 1, but B4 isn't the most Fubini-friendly set you could integrate over. Let's try and make our parameter space more rectangular, shall we?

So let's try a different parametrization using double polar coordinates:
    p2: B2' x [0, 2*pi] x [0, 2*pi] -> B4; (r, u, s, v) |-> (r*cos(s), r*sin(s), u*cos(v), u*sin(v)).
Here B2' is the set of all points (r, u) with r>=0 and u>=0 and r²+u²=rad² (it's a quarter-disk). Here, det(Dp2) = r*u, but our parameter space is much more comfortable to work with. It's still stupid, because r and u are still not independent.

To fix this, let's try a third parametrization based on the previous one:
    p3: [0, rad]x[0, pi/2]x[0, 2*pi]x[0, 2*pi] -> B4; (p, q, s, v) |-> (p*cos(q)*cos(s), p*cos(q)*sin(s), p*sin(q)*cos(v), p*sin(q)*sin(v)).
Note that this is very similar to the previous parametrization, we just replaced r and u by p*cos(q) and p*sin(q) respectively. Now our parametrization space is rectangular! Now det(Dp3) = p³*sin(q)*cos(q) = p³*sin(2*q)/2, which is not bad as far as Jacobians go. We'll go with this one to integrate over the hypersphere.

So, let's insert p3 into the formula (*) from above:

V=int_A{det(Dp3(p, q, s, v))}
=int(p³*sin(2*q)/2, p=0..rad, q=0..pi/2, s=0..2*pi, v=0..2*pi)
=pi²*rad⁴/2.

Hey, uh sorry it's taken me a while to respond... I've been trying to get it but I don't really understand the notation you're using >_<

[Pnx]

I have good news for you: Buoyancy is independent of fluid pressure (as long as the fluid is incompressible), and the buoyant force is exactly the negative of the gravitational force that would appear if the paddle were made of the fluid instead. So all you need to do is find out the volume of the paddle, multiply it by the fluid's density, then multiply it by -g.

I consulted my teacher about it, and actually I guess the term I should have been using was hydrostatic pressure, and also it turns out I was doing the problem wrong, I was supposed to find how high the water level was in the tank before figuring this out. Problem's all solved now.

[da_nang]

-snip-

Hey, uh sorry it's taken me a while to respond... I've been trying to get it but I don't really understand the notation you're using >_<

Spoiler: Does this help? (click to show/hide)

[Furtuka]

Yes! That was exactly what I needed. Thanks!

[Pnx]

There's a couple of integrals that are things like (9-x²)^(1/2) that I'm having trouble with, I mean I know I can make the substitution x=3sinθ which gets me 9cos²θ. But I'm not sure where to go from there?

Can anyone help?

[Karlito]

Know your trig identities. The power reducing formula is the one you want in this case.

[Skyrunner]

Here's another problem of the day!

Define F_n(x) = Integrate for dx: (x³ⁿ-1)/(x²+x+1)

F_n(1) = 1/(3n-1) - 1/(3n-2) + 1/(3n-4) - 1/(3n-5) + ... + 1/2 - 1

Represent F_n(0) in terms of n.

[da_nang]

Here's another problem of the day!

Define F_n(x) = Integrate for dx: (x³ⁿ-1)/(x²+x+1)

F_n(1) = 1/(3n-1) - 1/(3n-2) + 1/(3n-4) - 1/(3n-5) + ... + 1/2 - 1

Represent F_n(0) in terms of n.

Spoiler (click to show/hide)

[Lightningfalcon]

How would you find the coefficients for (x+y)^n?  I know you can use pascal's triangle, but that takes too long to construct if I just want to know the first few terms of, say, (x+y)^49.   

[Vector]

Binomial theorem.

[Lightningfalcon]

Binomial theorem.

Thank you.
I really need to start learning the proper names for things.    I kept getting stuff about how to multiply polynomials together when trying to google earlier.   

[Skyrunner]

Use C (combinations)!

For (x+y)^n, the coefficients are

0Cn, 1Cn, ... (n-1)Cn, nCn.



Edit: ninja'd by vector

Here's another problem of the day!

Define F_n(x) = Integrate for dx: (x³ⁿ-1)/(x²+x+1)

F_n(1) = 1/(3n-1) - 1/(3n-2) + 1/(3n-4) - 1/(3n-5) + ... + 1/2 - 1

Represent F_n(0) in terms of n.

Spoiler (click to show/hide)

How do you intuit the first step? Dividing the x^3n formula by (x-1) is so weird.

Also: how long did it take?

[Vector]

Binomial theorem.

Thank you.
I really need to start learning the proper names for things.    I kept getting stuff about how to multiply polynomials together when trying to google earlier.

You're welcome!  Happy multiplying : D

[da_nang]

Here's another problem of the day!

Define F_n(x) = Integrate for dx: (x³ⁿ-1)/(x²+x+1)

F_n(1) = 1/(3n-1) - 1/(3n-2) + 1/(3n-4) - 1/(3n-5) + ... + 1/2 - 1

Represent F_n(0) in terms of n.

Spoiler (click to show/hide)

How do you intuit the first step? Dividing the x^3n formula by (x-1) is so weird.

Also: how long did it take?

It's an identity.
aⁿ - bⁿ = (a-b)(a^n-1 + a^n-2b + a^n-3b² + ... + ab^n-2 + b^n-1)

Although I admit, I didn't have it remembered. See, I've got this "Engineering Mathematics Formula Collection" where I happened to stumble upon it after flailing around with the problem.
Like, "Hey, maybe it can be written as a double integral or perhaps Leibnitz's rule might make it less horrible or reduce the power of the numerator by using an integral identity" (which that book didn't have). Then I remembered to use partial fractions and with the numerator being of higher degree than the denominator, it's time for long division.

But long division with a polynomial of arbitrary degree is messy (but still possible if you know what you're doing), at which I point I vaguely remembered there being a identity for factoring that polynomial.

All in all, an hour spent banging my head.

[scrdest]

Halp. It shouldn't be a really that hard equation, but my brain won't cooperate now.

x/(c-x)(b-x)=a, abc are known constants, find x.

Or, just to be sure it's clear, could be formulated as:

x = (c-x)(b-x)a

E: Wolfram Alpha handled it for me. And good thing it did, the result is terrifying.