Mathc initiation/0038
Installer et compiler ces fichiers dans votre répertoire de travail.
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c00f.c |
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/* --------------------------------- */
/* save as c00f.c */
/* --------------------------------- */
#include "x_afile.h"
#include "ff.h"
/* --------------------------------- */
int main(void)
{
double m = flux_dxdzdy( M,N,P,
x0, x1,LOOP,
z0, z1,LOOP,
Y0, Y1,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(" / y1 / z1(y) / x1(y, z) \n");
printf(" | | | \n");
printf(" | | | M_x + N_y + P_z dxdzdy = %.3f \n",m);
printf(" | | | \n");
printf(" / y0 / z0(y) / x0(y, 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(" x1 : (x,y)-> %s \n", x1eq);
printf(" x0 : (x,y)-> %s \n\n", x0eq);
printf(" z1 : (y)-> %s \n", z1eq);
printf(" z0 : (y)-> %s \n\n", z0eq);
printf(" y1 : (y)-> %s \n", y1eq);
printf(" y0 : (y)-> %s \n\n", y0eq);
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"
" (2*x*z)_x = (2*z) \n"
" (x*y*z)_y = (x*z) \n"
" (y*z)_z = (y) \n\n"
" ((2*z)+(x*z)+(y)) \n\n"
" Code Mathematica : S = %.3f \n\n"
" integral ((2*z)+(x*z)+(y)) dx dz dy"
" from (0) to 2 from (0) to (4) from 0 to ((4-y)/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.
/ y1 / z1(y) / x1(y, z)
| | |
| | | M_x + N_y + P_z dxdzdy = 77.333
| | |
/ y0 / z0(y) / x0(y, z)
With.
F : (x,y,z)-> (2*x*z)i + (x*y*z)j + (y*z)k
x1 : (x,y)-> (4-y)/2
x0 : (x,y)-> 0
z1 : (y)-> 4
z0 : (y)-> 0
y1 : (y)-> 2
y0 : (y)-> 0
Press return to continue.
With.
F : (x,y,z)-> (2*x*z)i + (x*y*z)j + (y*z)k
Compute the partial derivative M_x, N_y, P_z:
(2*x*z)_x = (2*z)
(x*y*z)_y = (x*z)
(y*z)_z = (y)
((2*z)+(x*z)+(y))
Code Mathematica : S = 77.333
integral ((2*z)+(x*z)+(y)) dx dz dy from (0) to 2 from (0) to (4) from 0 to ((4-y)/2)
Press return to continue.