(%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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