Example Question:
Let $\vec{F} = <y, -z, e^{-y^2}+y^{1+x^2}+cos(z)>$
Let S the the portion of the surface which consists of two parts:
1) the portion of the paraboloid $y^2+x^2=4(z+1)$ satisfying $0\leq z \leq 3$ and
2) the portion of the sphere $x^2+y^2+z^2=4$ satisfying $z>0$ and is oriented upward.
Compute $\iint_s{curl \vec{F} d\vec{S}}$.
I know that $d\vec{S} = \vec{n}ds$, however I dont know how to find the actual $\vec{n}$ here. I also don't know how to do the set up of the integrals to find the $\iint_s$ using divergence theorem. How can divergence theorem be implemented into this question? Does divergence thoerem only apply on solids that are fully enclosed?
What I mean by 'setting up the equation', is that, in the solution for this question, these steps were taken:
Close the surface by adding (I assume a plane) z=3 and let $\partial E = S'\cup S''$
$\iint_{S'\cup S''} ... = \iiint_{E} div(curl(F)dV$ = ...
Which I don't understand, and need help understanding the conceptual side to this.
The divergence theorem with curl(F) as your vector field yields:
$\iint_M$curl(F)*ds=$\iiint_E$div(curl(F))$dV$
The surfaces $z=3$, $x^2+y^2=4(z+1)$, and $x^2+y^2+z^2=4$ form $\partial E$=M in the above equation. It is required that E be a closed volume to apply the divergence theorem.
But div(curl(F))=$0$, so $\iint_M$curl(F)*ds=$0$.
To answer the question, you could try computing $\iint_S$curl(F)*ds directly (Here S is the union of the surfaces $x^2+y^2=4(z+1)$ and $x^2+y^2+z^2=4$ constrained to the specified bounds on z).
For the sphere portion, n= ${\nabla F \over |\nabla F|}$, where $F$ = $x^2+y^2+z^2$. It is easy to check that this normal vector points outwards.
For the paraboloid portion, n= -${\nabla F \over |\nabla F|}$, where $F$ = $x^2+y^2-4(z+1)$. This works because $\nabla F$ is perpendicular to the level set $x^2+y^2-4(z+1)$=$0$, which is the surface in question. Note that this yields an inward normal vector. This orientation is consistent with the given orientation of the sphere (the normal vectors from each surface induce opposite orientations to the shared curve between them).
Then, you would just parametrize the surfaces that form S and compute the surface integral.
However, this is prohibitive. So, we utilize the Kelvin-Stokes Theorem which states:
$\iint_S$curl(F)* ds=$\int_C$(F)*ds, where C is the (closed) boundary curve of S.
The boundary curve is the curve obtained by the intersection of $z=3$ and $x^2+y^2=4(z+1)$. Therefore, the curve can be described by $x^2+y^2=16$, where $z=3$ (Why?).
Note that the induced orientation on this boundary curve from the inward-pointing normals on $x^2+y^2=4(z+1)$ is counterclockwise when looking down the + z-axis.
We parametrize C as:
X(t)=($4$cos(t),$4$sin(t),$3$), where t $\epsilon$ [$0$,$2$$\pi$] (Why?).
We also know that:
X'(t)=(-$4$sin(t),$4$cos(t),$0$).
Hence, we have that $\iint_S$curl(F)* ds=$\int_C$(F)*ds=$\int_C$(F(X(t))*X'(t)d$t$.
The computation from here on should be fairly straightforward.