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Why is the derivative of a circle's area its perimeter (and similarly for spheres)?

The derivative of the area $\pi r^2$ of a disk is the perimeter $2\pi r$ of the circumference. I know some explanations of this link.

1) Does there exist a explanation of this fact using Stokes's formula?

2) Does there exists a more general statement for some class of manifolds?

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marked as duplicate by joriki, Quixotic, Asaf Karagila, Zev Chonoles Dec 12 '11 at 9:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The content of your question does not seem to have anything to do with the title. The reason for this is very intuitive and does not require anything high-brow like Stokes's theorem or manifolds: it simply amounts to the fact that the boundary of the disc is a circle. –  Zhen Lin Dec 11 '11 at 9:31
    
Quite related... –  J. M. Dec 11 '11 at 9:31
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Wow -- question number 625 :-) On second thought, I was a bit rash to vote to close as duplicate, since this question asks about other aspects -- sorry about that; unfortunately close votes can't be retracted... –  joriki Dec 11 '11 at 9:39
    
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3 Answers

up vote 2 down vote accepted

For part 1 at least, let $$ F(x)=x $$ Then the Divergence Theorem, a case of Stoke's Theorem, says that when $\mathcal{D}$ is a ball of radius $r$ in $\mathbb{R}^n$, $$ \int_{\partial\mathcal{D}}F\cdot\vec{n}\;\mathrm{d}s=\int_{\mathcal{D}}\nabla\cdot F\;\mathrm{d}x $$ $$ \int_{\partial\mathcal{D}}r\;\mathrm{d}s=\int_{\mathcal{D}}n\;\mathrm{d}x $$ $$ r|\partial\mathcal{D}|=n|\mathcal{D}| $$ Thus, $|\partial\mathcal{D}|=\frac{n}{r}|\mathcal{D}|$. Since all spheres are geometrically similar, and the volume of similar objects in $\mathbb{R}^n$ grows as their size to the $n^{\text{th}}$ power, $|\mathcal{D}|=cr^n$. Therefore, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}r}|\mathcal{D}| &=cnr^{n-1}\\ &=\frac{n}{r}|\mathcal{D}|\\ &=|\partial\mathcal{D}| \end{align} $$

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Certainly nothing that has to do with Stokes theorem and with manifolds this has at first glimpse nothing to do as well. What we are dealing with here is firstly submanifolds that are embedded in some Euclidean space. Then, for the calculation, one has simply used a coordinate transformation. Please note that Stokes' theorem (one of the most beautiful theorems) expresses the duality between cohomology and homology (mostly in the de Rham sense). If you would like to establish some relationship between the volume of an n-disk and the volume of its boundary, then you will have to calculate both separately and afterwards compare the two results.

This may sound a bit harsh, but I really appreciate your questions because it is exactly what you have to think about when you work in the field of differential geometry and read more advanced books and research papers.

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The size of the boundary times the rate at which the boundary moves equals the rate at which the size of the bounded region changes.

There appears to be no conventional name for this fact. I've called it the boundary rule sometimes.

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