maxima's definite integration is quite buggy, we see dozens bugs like this
a year, and report them upstream, with limited success...
On Saturday, April 9, 2016 at 7:17:09 PM UTC+1, Sergey V Kozlukov wrote:
>
> (%i1) f(r,phi) := signum(r^2 - 4);
> 2
> (%o1) f(r, phi) := signum(r - 4)
> (%i2) integrate(integrate(r*f(r,phi), r, 0, 3), phi, 0, 2*%pi);
> (%o2) - 9 %pi
>
> That's strange
>
>
> суббота, 9 апреля 2016 г., 19:14:49 UTC+3 пользователь Dima Pasechnik
> написал:
>>
>> Try these computations directly in Maxima, and see whether it's still a
>> discrepancy there.
>>
>> On Saturday, April 9, 2016 at 1:31:44 PM UTC+1, Sergey V Kozlukov wrote:
>>>
>>> x, y, r, phi = var('x y r phi')
>>> f(x, y) = sign(x^2 + y^2 - 4)
>>> T(r, phi) = [r*cos(phi), r*sin(phi)]
>>> J = diff(T).det().simplify_full()
>>> T_f = f.substitute(x=T[0], y=T[1])
>>> int_f = integral(integral(T_f*abs(J), r, 0, 3), phi, 0, 2*pi).
>>> simplify_full()
>>> show(r"$\iint\limits_\Omega%s = %s$"%(latex(f(x)), latex(int_f)))
>>> Returns correct answer: $\pi$, while
>>> x, y, r, phi = var('x y r phi')
>>> f(x, y) = sign(x^2 + y^2 - 4)
>>> T(r, phi) = [r*cos(phi), r*sin(phi)]
>>> J = diff(T).det() #.simplify_full()
>>> T_f = f.substitute(x=T[0], y=T[1])
>>> int_f = integral(integral(T_f*abs(J), r, 0, 3), phi, 0, 2*pi).
>>> simplify_full()
>>> show(r"$\iint\limits_\Omega%s = %s$"%(latex(f(x)), latex(int_f)))
>>> Yields $-9\pi$
>>> The only thing simplify_full() changes here is it applies identity
>>> $\sin^2 + \cos^2 = 1$
>>>
>>> Code was executed in SMC worksheet
>>>
>>
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