Mathc initiation/Fichiers c : c64cc
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c00c.c |
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/* --------------------------------- */
/* save as c00c.c */
/* --------------------------------- */
#include "x_afile.h"
#include "fc.h"
/* --------------------------------- */
int main(void)
{
double m = flux_dydzdx( M,N,P,
Y0, Y1,LOOP,
z0, z1,LOOP,
x0,x1,LOOP);
clrscrn();
printf(" Use the divergence theorem to find,\n\n");
printf(" the flux of F through S.\n\n");
printf(" // /// \n");
printf(" || ||| \n");
printf(" || F.n dS = ||| div F dV \n");
printf(" || ||| \n");
printf(" // /// \n");
printf(" S Q \n\n\n");
printf(" If F = Mi + Nj + Pk \n\n\n");
printf(" /// /// \n");
printf(" ||| ||| \n");
printf(" ||| div F dV = ||| M_x + N_y + P_z dV \n");
printf(" ||| ||| \n");
printf(" /// /// \n");
printf(" Q Q \n\n\n");
stop();
clrscrn();
printf(" / x1 / z1(x) / y1(x, z) \n");
printf(" | | | \n");
printf(" | | | M_x + N_y + P_z dydzdx = %.3f \n",m);
printf(" | | | \n");
printf(" / x0 / z0(x) / y0(x, z) \n\n\n");
printf(" With.\n\n\n");
printf(" F : (x,y,z)-> (%s)i + (%s)j + (%s)k \n\n",Meq,Neq,Peq);
printf(" y1 : (x,z)-> %s \n", y1eq);
printf(" y0 : (x,z)-> %s \n\n", y0eq);
printf(" z1 : (x)-> %s \n", z1eq);
printf(" z0 : (x)-> %s \n\n", z0eq);
printf(" x1 : -> %s \n", x1eq);
printf(" x0 : -> %s \n\n", x0eq);
stop();
clrscrn();
printf(" With.\n\n\n");
printf(" F : (x,y,z)-> (%s)i + (%s)j + (%s)k \n\n",Meq,Neq,Peq);
printf(" Compute the partial derivative M_x, N_y, P_z:\n\n"
" (x**3 + sin(z))_x = (3 x**2) \n"
" (x^2*y + cos(z))_y = (x**2) \n"
" (exp(x^2 + y^2))_z = (0) \n\n"
" ((3 x**2)+(x**2)+(0)) \n\n"
" Code Mathematica : S = %.3f \n\n"
" integral ((3 x**2)+(x**2)+(0)) dy dz dx"
" from (0) to (5-z) from -2 to 2 from 0 to (4-x**2)\n\n",m);
stop();
return 0;
}
/* --------------------------------- */
/* --------------------------------- */
Ce travail consiste à adapter l'intégrale triple au calcul du flux en 3d par le théorème de la divergence : (M_x + N_y + P_z)
Exemple de sortie écran :
Use the divergence theorem to find,
the flux of F through S.
// ///
|| |||
|| F.n dS = ||| div F dV
|| |||
// ///
S Q
If F = Mi + Nj + Pk
/// ///
||| |||
||| div F dV = ||| M_x + N_y + P_z dV
||| |||
/// ///
Q Q
Press return to continue.
/ x1 / z1(x) / y1(x, z)
| | |
| | | M_x + N_y + P_z dydzdx = 131.657
| | |
/ x0 / z0(x) / y0(x, z)
With.
F : (x,y,z)-> (x^3 + sin(z))i + (x^2*y + cos(z))j + (exp(x^2 + y^2))k
y1 : (x,z)-> 5-z
y0 : (x,z)-> +0
z1 : (x)-> 4-x^2
z0 : (x)-> +0
x1 : -> +2
x0 : -> -2
Press return to continue.
With.
F : (x,y,z)-> (x^3 + sin(z))i + (x^2*y + cos(z))j + (exp(x^2 + y^2))k
Compute the partial derivative M_x, N_y, P_z:
(x**3 + sin(z))_x = (3 x**2)
(x^2*y + cos(z))_y = (x**2)
(exp(x^2 + y^2))_z = (0)
((3 x**2)+(x**2)+(0))
Code Mathematica : S = 131.657
integral ((3 x**2)+(x**2)+(0)) dy dz dx from (0) to (5-z) from -2 to 2 from 0 to (4-x**2)
Press return to continue.