Re: GPT4 + Mathematica

```On Tue, Apr 11, 2023 at 11:06 PM smitra <smi...@zonnet.nl> wrote:
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> *> So, it has noticed that it used the wrong formulas and it has found
> thecorrect ones. However, it was not able to do the calculation
> usingspherical symmetry. It only noted in the previous answer that the
> answeris spherically symmetric because it doesn't depend on the orientation
> ofthe axis relative to which the moment of inertia is computed.So, it tells
> you the steps to compute the three dimensional integral inspherical
> coordinates, it is unable to see that adding up the threemoments of
> inertias about the x, y and z axis yields a sphericallysymmetric expression
> that can be computed using a one dimensionalintegral over the radius.Being
> unable to solve a problem in a much simpler way than the standardway is a
> big red flag against the idea that GPT can do any math at all.*

I think the problem is that you and GPT have a difference of opinion about
which is simpler, Cartesian coordinates or spherical coordinates. And there
is no disputing matters of taste.

I just add this conversation with GPT

> ME: If adding up the three moments of inertias about the x, y and z axis
yields a spherically symmetric expression can the moment of inertia formula
be derived by simply using a one dimensional integral over the radius?

*GPT: "*

*Yes, that's correct. Once we have established that the moment of inertia
of a solid sphere is spherically symmetric, we can compute it using a
one-dimensional integral over the radius. Specifically, the moment of
inertia of a solid sphere of radius R and mass M about an axis passing
through its center is given by:I = (2/5) MR^2This formula can be derived by
integrating the spherically symmetric expression for the moment of inertia
over the volume of the sphere using spherical coordinates, which simplifies
the calculation to a one-dimensional integral over the radius. Here's how
the derivation goes:The moment of inertia of a solid sphere of radius R and
mass M about an axis passing through its center is given by:I = ∫∫ (r^2
sin^2θ) ρ dVwhere ρ is the density and dV is the volume element in
spherical coordinates. Since the sphere has uniform density, we can write ρ
= M / (4/3 * π * R^3).Using the fact that the integrand is spherically
symmetric, we can write:I = ∫∫∫ (r^2 sin^2θ) ρ dV = ρ ∫∫∫ (r^2 sin^2θ) r^2
sin θ dr dθ dφSubstituting in the expression for ρ and using the fact that
the integral is spherically symmetric, we can simplify the integral to:I =
(3/5) MR^2 ∫∫ sin^3θ dθ dφThe integral over θ and φ can be evaluated to
give:∫∫ sin^3θ dθ dφ = 4π/3Substituting this into the expression for I, we
get:I = (2/5) MR^2*
*Therefore, the moment of inertia of a solid sphere about an axis passing
through its center can be derived by simply using a one-dimensional
integral over the radius, and is given by (2/5) MR^2."*

John K Clark    See what's on my new list at  Extropolis
5vd

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