Area and volume integral of vector field

[Jhenrique]

In 2 dimensions

given a scalar field f(x,y)

is possible to compute the line integral $\int f ds$ and area integral $\iint f d^2A$.

In 3D, given a scalar field f(x,y,z)

is possible to compute the surface integral $\iint f d^2S$ and the volume integral too $\iiint f d^3V$.

So, given a vector field in f in 2 and 3 dimensions, is possible to compute the line integral and the surface integral, respectively, but is possible to compute the area integral and the volume integral those vector fields? This make sense?

 

[Matterwave]

Yes, in regular (flat) space you can simply treat the integrals as 2 or 3 scalar integrals for each of the components: E.g. if I want to take the volume integral of $\vec{F}=x\hat{i}+y\hat{j}+z\hat{k}$ over the unit box with corner at the origin, I would have:

$$\int_0^1\int_0^1\int_0^1( x\hat{i}+y\hat{j}+z\hat{k}) dxdydz = \frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}$$

We just don't use these very much. There is one "fundamental theorem of calculus"-like formula associated with these:

$$\iint_{\partial V} \left(\hat{n}\times\vec{A}\right) dS = \iiint_V \left(\vec{\nabla}\times\vec{A}\right) dV$$

The first integral is a closed one over the boundary. I can't get oiint to work here.

 

[Jhenrique]

Yes, in regular (flat) space you can simply treat the integrals as 2 or 3 scalar integrals for each of the components: E.g. if I want to take the volume integral of $\vec{F}=x\hat{i}+y\hat{j}+z\hat{k}$ over the unit box with corner at the origin, I would have:

$$\int_0^1\int_0^1\int_0^1( x\hat{i}+y\hat{j}+z\hat{k}) dxdydz = \frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}$$

We just don't use these very much. There is one "fundamental theorem of calculus"-like formula associated with these:

$$\iint_{\partial V} \left(\hat{n}\times\vec{A}\right) dS = \iiint_V \left(\vec{\nabla}\times\vec{A}\right) dV$$

The first integral is a closed one over the boundary. I can't get oiint to work here.

Interetering! But in the eletromagnetism, we have the vectorial current density, that is the quantity of current by unit of volume/area. But this is the unique example that I know...

 

[Jhenrique]

Hey man, you realize that we define eight integral operations, but I can't think in eight differential operations that are inverse to those. Given a function f(x) and f(x,y) too, we have 2 operations for each case: df/dx, ∫fdx, ∫∫fdxdy and d²f/dxdy. So, can you think in all inverse operations for each case in my first post?

 

[Matterwave]

Usually one expects that given an n-dimensional integral of a derivative of some kind, the derivative might "negate" in a VERY rough sense (I want to STRESS STRONGLY to not take this "negation" too seriously) one of the integrals so that you can obtain a n-1 dimensional integral along the boundary of the original volume.

These laws are therefore ANALOGOUS to the fundamental theorem of calculus (telling us differentiation is inverse to integration). They basically all arise, in one form or the other, from Stoke's theorem, which tells us that integration over a volume is kind of an inverse to the exterior derivative:

$$\int_\Omega d\omega=\oint_{\partial\Omega}\omega$$

For vector calculus, this results in many different identities, with many different forms (this is because the differential operators one sees in vector calculus are kind of like avatars of the exterior derivative, but they are sometimes non-trival constructions of it). You can find most of them in this article:

http://en.wikipedia.org/wiki/Vector_calculus_identities#Summary_of_important_identities

Scroll down to the "integration" section.