Example 25.1.1
Evaluate the line integral , where C is the boundary of the region in the first quadrant fenced by ,  and   oriented in counterclockwise direction.  First evaluate the integral directly, then use Green's theorem.

To answer part a, we code the vector field, parametrize the boundaries and visualize the region.

>	F:=map(unapply, [x^2+y^2+1, 2*x+y], (x, y)); r1:=[t, t^2]; r2:=[t, 4]; r3:=[0, t];

>	plot([[op(r1), t=0..2], [op(r2), t=2..0], [op(r3), t=4..0]], color=[red, blue, green], axes=frame);

The integral is now readily evaluated.

>	e1:=Int(dotprod(F(r1[1], r1[2]), map(diff, r1, t), orthogonal), t=0..2)+Int(dotprod(F(r2[1], r2[2]), map(diff, r2, t), orthogonal), t=2..0)+Int(dotprod(F(r3[1], r3[2]), map(diff, r3, t), orthogonal), t=4..0);

>	answer:=value(e1);

As solution to part b, we apply Green's theorem.

>	a1:=Doubleint(diff(F(x, y)[2], x)-diff(F(x, y)[1], y), y=x^2..4, x=0..2);

>	check:=value(a1);

>

Example 25.1.2
If ,  compute  a decimal approximation for the integral  along the positively oriented curve .

Code the vector field, the integration curve and make a picture.

>	F:=[x^3*y, x^4+x*y^2]; c:=x^4+y^4=1;

>	p1:=fieldplot(F, x=-1..1, y=-1..1, arrows=THICK, color=blue):

>	p2:=implicitplot(c, x=-1..1, y=-1..1, thickness=3):

>	display([p1, p2], scaling=constrained, axes=frame);

Because it is non-trivial to parametrize this curve, Green's theorem provides a welcome alternative solution.

>	bnd:=solve(c, y);

>	e1:=Doubleint(diff(F[2], x)-diff(F[1], y), y=bnd[3]..bnd[1], x=-1..1);

>	answer:=evalf(e1);

>

Example 25.1.3
Compute the area of the region bounded by the closed curve , 0 <= t <= 1.

>	r:=[cos(t)^3, sin(t)^5];

>	plot([op(r), t=0..2*Pi]);

We will use Green's theorem in reverse, with  =  =  to obtain that the area of this region equals  =  .  Where the latter denotes the line integral over the positively oriented boundary of the region.

>	Q:=(x, y)->x;

>	e1:=Int(Q(r[1], r[2])*diff(r[2], t), t=0..2*Pi);

>	answer:=value(e1);

>	evalf(answer);

>