[ot][ot][ot][personal] whirligig small thoughts
Undescribed Horrific Abuse, One Victim & Survivor of Many
gmkarl at gmail.com
Tue Aug 1 06:02:42 PDT 2023
>>>> [if output is wrong, derivation of wedge integration expression is
>>>> pending check from start]
>>>>
>>>> wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx
>>>>
>>>> integration domain: x=(-r_sphere, +r_sphere)
>>>> wedge_volume = dtheta/2 [ integrate(r_sphere^2) - integrate(x^2) ]
>>>
>>> this looks reasonable r_sphere^2 >= x^2
>>>
>>>>
>>>> indefinite integral of r_sphere^2 dx = r_sphere^2 * x
good, 1 in denominator
>>>> indefinite integral of x^2 dx = 2x^3
wrong, used derivative rule instead of integration rule.
I(x^2,dx) = x^3/3 = (1/3) x^3 . can check rule by deriving area of triangle.
it really should be (1/3) x^3, see derivation of triangle area.
it is more right than 2x^3 for sure.
>>>
>>> noting that both of these have x raised to an odd power (a little hard
>>> to notice given the location of the ^2's). because x is raised to an
>>> odd power in both, they will both have the negative and positive parts
>>> of the integral in the same relation, and their inequality relation
>>> should be preserved. [at first i thought one had an even power, and
>>> then imagining plugging in made it not preserved]
>>>
>>>> evaulate fo rdomain
>>>> r_sphere^2 * r_sphere - r_sphere^2 * -r_sphere = 2rsphere^3
>>> this above line looks reasonably likely to be right
>>>> 2(r_sphere)^3 - 2(-r_sphere)^3 = 4 r_sphere^3
using (1/3)x^3,
(1/3)r_sphere^3 - (1/3)(-r_sphere)^3 = 2/3 r_sphere^3
:D
>>> here the errormistake is findable. the above two lines reveal it. i
>>> haven't found it yet
>>
>> here are parts: r^2 > x^2
>> we expect when integrating dx -r..+r for the relation to hold.
>> what changed?
>> integral r^2 dx = 2r^2 x
r^2 dx => r^2 x .
the rule and denominator comes from the variable of integration, not
the coefficient.
the rule is to divide, not multiply as in differentiation.
r^2 x is correct.
>> ok there's one mistake i had dropped that 2
this was not the mistake, which was a further mistake [thinking some
on dissociative delusion, seems there was one here
>> integral x^2 dx = 3x^2
> here it is, this is 3x^3 not 3x^2
this is (1/3)x^3 as described above :) see derivation of triangle area
to support rule
>> let's subtract before plugging
>> 2r^2 x - 3x^2 | x=-r..+r
> 2r^2 x - 3x^3 | x=-r..+r
r^2 x - (1/3)x^3 | x=-r..+r (note this is already done once above)
>> ohh then i can handle positive and negative separately
>> for +r: 2(r^2)r - 3(r^2) = 2r^3 - -- mistake indicated. [parts should
>> match in exponent
>
> +r: 2(r^2)r - 3(r^3) = 2r^3 - 3r^3 negative.
> is there a reason for this to be negative? it represents a partial area,
> right?
+r: r^2 r - (1/3)r^3 = r^3 - (1/3) r^3 = (2/3) r^3
-r: r^2 (-r) - (1/3)(-r)^3 = -r^3 + (1/3)r^3 = -(2/3) r^3
difference = 4/3 r^3
then finish other avenue where the parts were subtracted separate and
then summed
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