I am interested in showing that the vector area $$\int_{\mathcal{S}}da$$ can be equivalently given by $$\int_{\mathcal{S}} da = \frac{1}{2}\oint(r \times dI).$$ I am mostly interested in getting a geometric idea of what this means with reference to the following which is stated in the literature I am working through: "Draw the cone subtended by the loop at the origin. Divide the conical surface up into infinitesimal triangular wedges, each with vertex at the origin and opposite $dI$, and exploit the geometrical interpretation of the cross product".
The idea I have is that it has something to do with $r \times dI$ being the area of a parallelogram, with direction perpendicular to the surface thereof. So then we have some kind of connection between the path integral and the surface integral. I'm specifically interested in the geometric interpretation of the right-hand side, I don't see the connection with a cone or conical section.
Thanks for any assistance.
Applying the proof from this post I can see that taking $\vec{v} = \vec{c}$, where $c$ is a constant vector , that we get my result above:
$$ \vec{c} \cdot\int d\vec{a} =\int \vec{c} \cdot d\vec{a}$$ then using Stokes Theorem as in this post we get $$\oint (\vec{c} \times \vec{r}) \cdot d\vec{I} = \int \nabla \times (\vec{c} \times \vec{r}) \cdot d \vec{a} = \int 2 \vec{c} \cdot d \vec{a}$$ therefore $$\int \vec{c} \cdot d \vec{a} = \frac{1}{2} \oint (\vec{c} \times \vec{r}) \cdot d\vec{I} = \frac{1}{2} \oint \vec{c} \cdot (\vec{r} \times d \vec{I}) = \vec{c} \cdot \frac{1}{2} \oint (\vec{r} \times d \vec{I}).$$ Since $\vec{c}$ is an arbitrary constant the result follows. But I am interested in the geometric interpretation mentioned above?