Total Vector Area of Closed Surface is Zero

Theorem

Let $S$ be a closed surface.

Let $\d \mathbf S$ be an infinitesimal vector area around some point $P$ of $S$.

Then the total surface area of $S$ is given by:

$\ds \iint_S \d \mathbf S = \mathbf 0$

Proof

This theorem requires a proof. In particular: The book waffles on about slicing $S$ into vanishingly small tetrahedra, but the argument is flimsy. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area