The area of a circle is $\pi r^2$, and the circumference is the derivative of this: $2\pi r$.
The same holds in one higher dimension: the volume of a sphere is $\frac{4}{3} \pi r ^3$ and the derivative of this is the surface area: $4\pi r^2$.
These make sense to me geometrically – if you increase the volume by a tiny bit $dr$, the increase in volume will be in proportion to the surface area.
The generalized stokes theorem says that for a $(k-1)$-form $\omega$ on a $k$-dimensional manifold $M$ with boundary:
$$\int_M d\omega = \int_{\partial M} \omega$$
Here, the derivative is the volume form — which is the opposite of what I observed from the geometric formulas above. Is there a relationship between these formulas which are derivatives of each other and the generalized stokes theorem?
Best Answer
It's not the opposite.
Consider the disk $D_r$ with radius $r>0$ with area form $\operatorname{d}\!\omega = \operatorname{d}\!x \wedge \operatorname{d}\!y$. The boundary of the disk is $\partial D_r$ which is the circle $S^1_r$. Moreover, notice that $\operatorname{d}(x \operatorname{d}\!y) = \operatorname{d}\!x \wedge \operatorname{d}\!y$ meaning that $\omega = x\operatorname{d}\!y$. It follows that:
$$\int_{D_r} \operatorname{d}\!x \wedge \operatorname{d}\!y = \int_{S^1_r} x \operatorname{d}\!y$$
Let's evaluate the right hand side. We can parametrise the circle by $x(\theta) = r\cos\theta$ and $y(\theta) = r\sin\theta$ where $0 \le \theta < 2\pi$. Making a simple substitution gives:
$$\int_{S^1_r} x \operatorname{d}\!y = \int_0^{2\pi} r^2\cos^2\theta \, \operatorname{d}\!\theta = \frac{r^2}{2}\left[ \theta + \sin(2\theta)\right]_0^{2\pi} = \pi r^2 $$