Mathc initiation/005f
Apparence
Installer et compiler ces fichiers dans votre répertoire de travail.
c0b1.c |
|---|
/* ---------------------------------- */
/* save as c0b1.c */
/* ---------------------------------- */
#include "x_afile.h"
#include "fb.h"
/* ---------------------------------- */
int main(void)
{
double m = 0.;
v3d u = {1,0,1};
clrscrn();
printf(" Let S be the part of the graph of z = %s with z >= 0. \n\n", feq);
printf(" Let C be the trace of S on the x-y-plane. \n\n");
printf(" Verify Stokes's theorem for the vector field, \n\n");
printf(" F(x,y,z) = %si %sj %sk\n\n\n\n\n",Meq,Neq,Peq);
printf(" Stoke's theorem. \n\n"
" / // \n"
" | || \n"
" O F.T ds = || (curl F).dS \n"
" | || \n"
" / C // \n"
" S \n\n\n");
stop();
clrscrn();
printf(" // \n"
" || \n"
" || (curl F).dS = \n"
" || \n"
" // \n"
" S \n\n\n"
" with F = Mi + Nj + Pk \n\n"
" (curl F) = [(P_y-N_z)i + (M_z-P_x)j + (N_x-M_y)k] \n\n\n\n"
" dS = (-(u.i), -(u.j), 1) dA dA = dydx\n\n");
stop();
clrscrn();
printf(" With the Stokes's theorem you find :\n\n\n");
printf(" F : (x,y,z)-> %si %sj %sk \n\n",Meq,Neq,Peq);
printf(" f : (x,y)-> %s \n\n", feq);
printf(" y1 : (y)-> %s \n", y1eq);
printf(" y0 : (y)-> %s \n\n", y0eq);
printf(" x1 = %+.1f\n x0 = %+.1f \n\n",x1,x0);
m = stokes_dydx( M,N,P,
f,
y0,y1,LOOP,
x0,x1,LOOP,
u);
printf(" / x1 / y1(y)\n"
" | | \n"
" | | (curl F).(-(u.i), -(u.j), 1) dy dx = %.3f\n"
" | | \n"
" / x0 / y0(y)\n\n\n",m);
stop();
return 0;
}
/* ---------------------------------- */
/* ---------------------------------- */
Exemple de sortie écran :
Let S be the part of the graph of z = (1-x**2-y**2) with z >= 0.
Let C be the trace of S on the x-y-plane.
Verify Stokes's theorem for the vector field,
F(x,y,z) = + 2*yi + cos(z)*xj -(sin(x)*y)k
Stoke's theorem.
/ //
| ||
O F.T ds = || (curl F).dS
| ||
/ C //
S
Press return to continue.
//
||
|| (curl F).dS =
||
//
S
with F = Mi + Nj + Pk
(curl F) = [(P_y-N_z)i + (M_z-P_x)j + (N_x-M_y)k]
dS = (-(u.i), -(u.j), 1) dA dA = dydx
Press return to continue.
With the Stokes's theorem you find :
F : (x,y,z)-> + 2*yi + cos(z)*xj -(sin(x)*y)k
f : (x,y)-> (1-x**2-y**2)
y1 : (y)-> +(1)
y0 : (y)-> -(0)
x1 = +1.0
x0 = -0.0
/ x1 / y1(y)
| |
| | (curl F).(-(u.i), -(u.j), 1) dy dx = -0.755
| |
/ x0 / y0(y)
Press return to continue.