hint on stokes' theorem problem

Question

In my notes, I've just covered Stokes' theorem and so I expect the problem below to use this... but I'm just not seeing the connection. I would like a hint.

I want to compute $\iint_{S_1\cup S_2} \text{curl}F\cdot dS$ where $S_1=\{(x,y,z):x^2+y^2=1, 0\le z \le 1\}$ and $S_2=\{(x,y,z): x^2+y^2 +(z-1)^2 =1, z\ge 1\}$ and $F(x,y,z)=(zx+z^2y+x,z^2yx+y,z^4x^2)$.

I'm supposed to compute the union of these two surfaces and Stokes theorem (the "calc 3 version") is equipped to compute a surface when that surface is bounded by a loop. I guess I'm not seeing the loop.

Is this the approach I want to even use.... finding the boundary of a surface?

Answer 1

The boundary of $S_1\cup S_2$ is just $C = \{(x,y,z)\,|\,x^2+y^2=1,z=0\}$. Thus, if $D = \{(x,y,z)\,|\,x^2+y^2\leq1,z=0\}$, we have $$\iint_{S_1\cup S_2} \nabla \times F \,dS = \int_C F \,d\ell = \iint_D \nabla \times F \, dS$$ Given that the $D$ lies on the $xy$-plane, that last integral should be straightforward. This works because $C$ is the boundary of both $S_1 \cup S_2$ and $D$.

Answer 2

Notice that $S_1$ is a cylinder centered at the origin and $S_2$ is half of a sphere translated upward along the $z$ axis by one unit, you can see that the sphere forms a cap for the cylinder, and the loop is the bottom of the cylinder, or a circle parametrized by $(\cos t, \sin t, 0)$
Now your integral becomes

$$\int_0^{2\pi}(\cos t,\sin t,0)\cdot (-\sin t, \cos t, 0) \mathrm{d}t=0$$