I am confused about the boundary required for Stokes' theorem to hold. Most of the time, examples I have encountered in textbooks and school courses show examples of the theorem holding for hemispheres or relatively simple surfaces which projections' are well contained within the bounds of the surface. I am curious about cases where the surface's projection "stretches" beyond the boundaries without having the actual boundary change.
For example, suppose you have an inflated balloon that is not tied up (the balloon is going to be considered the surface), then the boundary of the curve is the circle formed by the opening. From what I understand, Stokes' theorem does not hold in such a case, because of the width of the balloon is larger than the boundary curve (the shadow/projection of the balloon is not contained within the boundary curve). Is this correct?
Also, assuming that Stokes does not hold for that balloon case, is there another similar theorem that holds for such objects?
Thank you and I apologize if this is confusingly written.
A simple example: Consider the vector field $\mathbf F=\langle z,x,y\rangle$, which has curl $\langle1,1,1\rangle$. Suppose you want to find the circulation of $\mathbf F$ along $C$, the unit circle $x^2+y^2=1$ in the plane $z=0$.
Now consider two different surfaces that share the boundary $C$,
$$S_1=\{(x,y,z)\mid x^2+y^2+z^2=1,z\ge0\}$$
$$S_2=\{(x,y,z)\mid x^2+y^2+(z-\sqrt3)^2=2,z\ge0\}$$
$S_1$ is the top half of the unit sphere, and $S_2$ is the upper portion of a sphere with radius $2$ that has been translated up by $\sqrt3$ so that it intersects the plane $z=0$ at $C$. It fits the description you mention, since the sphere bows out and stretches beyond the "wall" of the cylinder $x^2+y^2=1$.
Stokes' theorem says
$$\int_C\mathbf F\cdot\mathrm d\mathbf r=\iint_S\operatorname{curl}\mathbf F\cdot\mathbf n\,\mathrm d\Sigma$$
regardless of the choice of $S$, so long as $S$ is any smooth surface with $C$ as its boundary. For the purposes of this example, take $\mathbf n$ to be the normal vector pointing away from the origin.
Indeed, we have
$$\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_0^{2\pi}\langle0,\cos t,\sin t\rangle\cdot\langle-\sin t,\cos t,0\rangle\,\mathrm dt=\pi$$
$$\iint_{S_1}\operatorname{curl}\mathbf F\cdot\mathbf n\,\mathrm d\Sigma=\int_0^{\pi/2}\int_0^{2\pi}\langle1,1,1\rangle\cdot\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle\,\mathrm du\,\mathrm dv=\pi$$
$$\iint_{S_2}\operatorname{curl}\mathbf F\cdot\mathbf n\,\mathrm d\Sigma=\int_0^{5\pi/6}\int_0^{2\pi}\langle1,1,1\rangle\cdot\langle4\cos u\sin^2v,4\sin u\sin^2v,4\cos v\sin v\rangle\,\mathrm du\,\mathrm dv=\pi$$
The only requirement for the theorem to hold is that $S$ is smooth. Compare the above to what happens if we consider the surface
$$S_3=\{(x,y,z)\mid(z-1)^2=x^2+y^2\}$$
which is a cone whose base is the disk $x^2+y^2\le1$ and whose apex (a "sharp" point) occurs at the point $(0,0,1)$.
We would get
$$\iint_{S_3}\operatorname{curl}\mathbf F\cdot\mathbf n\,\mathrm d\Sigma=\int_0^{2\pi}\int_{-1}^0\langle1,1,1\rangle\cdot\langle u\cos v,u\sin v,-u\rangle\,\mathrm du\,\mathrm dv=0$$