Let S be the union of two surfaces, $S_1$ and $S_2$, where $S_1$ is the set of $(x,y,z)$ with $x^2+y^2 =1$ ,$0 \le z \le 1$ and $S_2$ is the set of $(x,y,z)$ with $x^2+y^2+(z-1)^2 =1$, $z \ge 1$. Set $F(x,y,z)$=$(zx+z^2y+x)i + (z^3yx+y) +(z^4x^2)k$
Compute $\int\int_S \nabla \times F \cdot dS$
Solution says that the boundary $\Gamma$ is unit circle on the $xy$-plane.
I wonder that;
I learned $D$: $\mathbb{R}^2$ region;
$$T: \mathbb{R}^2 \to \mathbb{R}^3 ,T(D)=S$$
If $c(t)=(u(t),v(t))$; parameterization of $D$ in the positive direction.
then $\Gamma$ ;oriented simple closed curve
$$p: t \to T(u(t),v(t))$$
But I cannot see what is $D$ and $T$ is...
I don't understand why the boundary $\Gamma$ is a unit circle on the $xy$-plane.
Thinking about this geometrically helps. (The equations defining $S_1$ and $S_2$ should look familiar hopefully!)
Note that $S_2$ is the upper hemisphere of the unit sphere centered at $(x,y,z) = (0,0,1)$, while $S_1$ is the part of the cylinder of radius $1$ between $z=0$ and $z=1$. Hence $S_1$ has boundary the unit circle in the $xy$-plane, and the unit circle in the $z=1$ plane, while the boundary of $S_2$ is the unit circle in the $z=1$ plane.
So their union sticks them together along the shared boundary on the $z=1$ plane, and the boundary of their union is only the unit circle in the $xy$-plane.