If I understand correctly, then any time we specify a particular surface patch of some non-closed surface, we also automatically specify not only the curve which bounds the surface patch but also the set of all vector fields that are usable for calculating the surface area with a line intergral.
Stokes' Theorem states that:
$$\oint_{∂P} \vec{F} \cdot d{\vec{r}}=\iint_{P} \nabla \times \vec{F} \cdot d{\vec{S}}$$
Here 'P' is for (surface) patch
Let $\vec{u}=\nabla \times \vec{F}$.
Let $P$ be an arbitrary surface patch of our choosing.
If $\vec{F}$ is any vector field such that its curl always makes an angle $\cos^{-1}(1/|\vec{u}|)$ with the unit normal of the surface then it seems to me that $\oint_{∂P} \vec{F} \cdot d{\vec{r}}=\iint_{P} dS$ $=$Surface area of the surface patch.
My question is essentially about the entire set of all options that one has when choosing a vector field $\vec{F}$. In practice, whenever I apply Stokes' Theorem in order to calculate a surface area with a line integral, I always choose a vector field $\vec{F}$ with all components equal to zero except one (that is, $\vec{F}$ has only one component). This is in order to make the calculation as easy as possible. What I want to know about is what subset of all possible vector fields on $ℝ^3$ will work.
Is it correct to say that a subset of all vector fields that will work are those for which the curl of the vector field always makes an angle $\cos^{-1}(1/|\vec{u}|)$ with the unit normal of the surface?
Let $\vec{F}=\langle f(x,y,z),g(x,y,z),h(x,y,z)\rangle$
Let $\vec{u}=\nabla \times \vec{F}$.
Then $\vec{u}=\nabla \times \vec{F}=\langle\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z},\frac{\partial f}{\partial z}-\frac{\partial h}{\partial x},\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\rangle$
For this special case what we want is for the dot product of $\vec{u}$ with the surface's unit normal to be equal to 1. It seems to me that this can be achieved if and only if $\vec{F}$ is a vector field such that its curl always makes an angle $\cos^{-1}(1/|\vec{u}|)$ with the unit normal of the surface
Let's consider a right circular cone $z=\sqrt{x^2+y^2}\cos\phi$. Choosing the upward-pointing normal, the unit normal for the right circular cone is $\langle \frac{-x}{\sqrt{x^2+y^2}}\cos\phi,\frac{-y}{\sqrt{x^2+y^2}}\cos\phi,\sin\phi \rangle$
Let $\vec{F}=\langle f(x,y,z),g(x,y,z),h(x,y,z) \rangle$
Then we have the PDE
$$\langle h_y-g_z,f_z-h_x,g_x-f_y \rangle \cdot \langle \frac{-x}{\sqrt{x^2+y^2}}\cos\phi,\frac{-y}{\sqrt{x^2+y^2}}\cos\phi,\sin\phi \rangle=1$$
One particular solution $\vec{F}$ is $\vec{F}=\langle 0,x\csc(\phi),0\rangle$.
Another particular solution $\vec{F}$ is $\vec{F}=\langle-y\csc(\phi),0,0\rangle$
Thus, a third particular solution is $\vec{F}=\langle -ky\csc\phi,(1-k)x\csc\phi,0 \rangle$ where $k$ $\in$ $ℝ$