with(linalg):
F:=vector([6*x-2*exp(2*x)*y^2,-2*y*exp(2*x),cos(z)]);
[ 2 ]
F := [6 x - 2 exp(2 x) y , -2 y exp(2 x), cos(z)]
curl(F,[x,y,z]);
[0, 0, 0]
Therefore the vector field is conservative.
x:=t;y:=(t-1)*(t-2);z:=Pi/2*t^3;
x := t
y := (t - 1) (t - 2)
3
z := 1/2 Pi t
R:=vector([x,y,z]);
[ 3]
R := [t, (t - 1) (t - 2), 1/2 Pi t ]
dR:=map(diff,R,t);
[ 2]
dR := [1, 2 t - 3, 3/2 Pi t ]
intgr1:=dotprod(F,dR);
4 3 2
intgr1 := 6 t - 2 exp(2 x) t + 12 exp(2 x) t - 26 exp(2 x) t
+ 24 exp(2 x) t - 8 exp(2 x)
2
- 2 (t - 1) (t - 2) exp(2 x) (2 t - 3) + 3/2 cos(z) Pi t
intgr1:=simplify(intgr1);
4 3 2
intgr1 := 6 t - 2 exp(2 t) t + 8 exp(2 t) t - 8 exp(2 t) t
3 2
- 2 exp(2 t) t + 4 exp(2 t) + 3/2 cos(1/2 Pi t ) Pi t
int(intgr1,t=0..1);
8
This confirms the answer obtained by using the potential.
Obviously the potential method is much easier than trying to
evaluate the above integral by hand!