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that’s called the jacobian, it accounts for the change between coordinate systems (cartesian -> polar in this case). you’ll learn about it in calc 3

@@kathode1 I learned about it within the next few days after that 😁 Then there was even some guy named Jordan who apparently had a lemma

This is merely the preface for multiple integral of calc 3. The whole calculus is a baby stuff

@@spiderjerusalem4009 🤓

@@danialjeelani but those are vector based? Is there supposed to be vectors in the Gaussian integral

​@@BlueMushroomSmurfCatkinda, you're transforming the integral from R->R to R²->R. I wouldn't call them vectors (they are still outputting a scalar number), but there is a vectorial explanation of what is going on there

from what I understand, because you're transforming space from the xy basis to the rθ basis, and because you can interpret differential forms as covector fields, they transform with the Jacobian, which is the determinant of that matrix

​@@BlueMushroomSmurfCat 𝐫:uv-plane → xy-plane 𝐫(u,v)=(x,y,0) 𝝏ᵤ𝐫  Δᵤ ≈ 𝐫(u+Δᵤ,v)-𝐫(u,v) 𝝏ᵥ𝐫  Δᵥ ≈ 𝐫(u,v+Δᵥ)-𝐫(u,v) (Δᵤ,Δᵥ)→(0,0) 𝐫ᵤ = 𝝏ᵤ𝐫 = (𝝏ᵤx, 𝝏ᵤy, 0) 𝐫ᵥ = 𝝏ᵥ𝐫 = (𝝏ᵥx, 𝝏ᵥy, 0) Although cross products do only work in three dimension, we can extend those elements to have them lie on the plane z=0 Area = Sum of norm of cross products of 2 infinitesmall vectors, one of which goes in u-direction, whilst the other in v-direction = 𝜮 ||[𝐫(u+Δᵤ,v)-𝐫(u,v)]×[𝐫(u,v+Δᵥ)-𝐫(u,v)]|| (Δᵤ,Δᵥ)→(0,0) = 𝜮 ||(𝝏ᵤ𝐫  Δᵤ)×(𝝏ᵥ𝐫  Δᵥ)|| (Δᵤ,Δᵥ)→(0,0) = 𝜮 ||𝝏ᵤ𝐫 ×𝝏ᵥ𝐫|| |ΔᵤΔᵥ| (Δᵤ,Δᵥ)→(0,0) = 𝜮 |Δx Δy|, (Δx,Δy)→(0,0) whence Area = ∫ dA = ∫∫ dxdy = ∫∫  ||𝐫ᵤ×𝐫ᵥ|| dudv hence the imposition that dxdy = ||𝐫ᵤ×𝐫ᵥ|| dudv = |det((xᵤ,xᵥ),(yᵤ,yᵥ))| du dv

The Jacobian is just the first-order derivative for functions f : R^n -> R^m

Yep

This is actually the easiest. Nothing else works. Not u sub, trig sub, integration with parts. You could try this using Laplace transform but then again, why keep a table of Laplace transformations at all times

@@TheHellBoy05 yes, using polar coordinates sub for double integrals

​@@silveykmbro, that is what the jacobian does

​@@TheHellBoy05Feyman technique could do. Laplace solved it too (reference to the person, not his transformation. Bprp covered it)

Thanks! Didn't know that was possible. "pip uninstall manim" and "pip install manim==0.11.0" worked well for me

the gaussian integral, its just the normal distribution mean 0 sd 1 being integrated to infinity, comes out as sqrt(pi)

@@_mekk_ ah right thx mate

@the brain yea np, there's a nice video that explains it rather simply using polar coordinates, the premise is that the definite integral from 0-♾️ of e^-x^2 is = to 0-♾️ of e^-y^2 so u can then change it into double integral of e^-(y^2+x^2) and then u can draw it in 3d and find it using tubes and simple geometry and a simple improper integral

Just opening to calc 3 regarding multiple integrals. Nothing hard, notably if you're already into gamma/beta function

@spiderjerusalem4009 yeah im not american so i dont know what calc 3 is, but actually in my own time i found a way of doing this with volumes of revolutions rather that whatever the fuck this is and actually it turned out pretty nice. I also have no idea what the gamma or beta functions are, do you cover that at university?

Gross physicist notation, it hurts the eyes

because he deserves hell

Leibniz thats why

Weird preference

Don’t worry, you only have to integrate 1 from -inf to inf and multiply it by “e^-x^2”!!!

Yes, it is manim