# 6

> restart : with(linalg): i:= [1,0,0] : j := [0,1,0] : k := [0,0,1]:

Warning, new definition for norm

Warning, new definition for trace

> G_1 := (1+x)* exp(x+y);

> G_2 := x* exp(x+y);

> G_3 := 5*y ;

> G := [G_1, G_2, G_3 ] ;

> curl_G := curl( G, [x,y,z] ) ;

> curl_G := [ dotprod(curl_G,i) , dotprod(curl_G,j) , simplify(dotprod(curl_G,k)) ];

> F_1 := G_1 ;

> F_2 := G_2 ;

> F_3 := 0;

> F := [F_1, F_2, F_3 ];

> curl_F := curl( F, [x,y,z] ) ;

> curl_F := [ dotprod(curl_F,i) , dotprod(curl_F,j) , simplify(dotprod(curl_F,k)) ];

So int_c G dot ds = [ int_c (G - F) dot ds ] + [ int_c F dot ds ]

>

For the [ int_c (G - F) dot ds] part:

>

> G_minus_F := [ G_1 - F_1 ,G_2 - F_2, G_3 - F_3 ];

> c_1 := (2-t)*exp(t);

> c_2 := t ;

> c_3 := 2*t ;

> c := [c_1, c_2, c_3 ];

> G_minus_F_on_c := subs( {x=c_1, y=c_2, z=c_3 }, G_minus_F );

> c_prime := [diff(c_1, t), diff(c_2, t), diff(c_3, t) ];

> integrand := dotprod(G_minus_F_on_c, c_prime);

> integral_1 := int(integrand, t = 0 .. 2) ;

>

For the [ int_c F dot ds ] part:

since F is conservative, the integral is independent of path ...

so we can use any path from c(0) to c(2) ...

so let's use the straight line segment ....

go back to informal summary - equation of a line - if you need:

> c_start := simplify( subs( {t = 0} , c ) ) ;

> c_end := subs( {t=2} , c ) ;

> v := c_end - c_start;

> p := c_start + t * v ;

> p := matadd( c_start , scalarmul( v , t) ) ;

> p_1 := dotprod(p, [1,0,0] );

> p_2 := dotprod(p, [0,1,0] );

> p_3 := dotprod(p, [0,0,1] );

> F_on_p := subs( {x=p_1, y=p_2, z=p_3} , F );

> p_prime := [diff(p_1, t), diff(p_2, t), diff(p_3, t) ];

> integrand := dotprod( F_on_p, p_prime) ;

> integrand := simplify( dotprod( F_on_p, p_prime) ) ;

> integral_2 := int(integrand, t = 0 .. 1) ;

>

adding we get:

> integral := integral_1 + integral_2 ;

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