DIFFERENTIATION
Given a scalar field f, grad f is the vector field (@f/@x, @f/@y, @f/@z).
Given a vector field F = (m,n,o), curl f is the vector field
(@o/@y - @n/@z, @m/@z - @o/@x, @n/@x - @m/@y).
Given a vector field F = (m,n,o), div f is the scalar field @m/@x+@n/@y+@o/@z.
Now, there is a basic formula for scalar fields which states that
df = @f/@x dx + @f/@y dy + @f/@z dz.
So df is like the gradient of f.
Now suppose F is the covector field m dx + n dy + o dz,
which is just another way to look at the vector field (m,n,o).
Then dF = d(m dx) + d(n dy) + d(o dz).
Following the product rule d(fg) = df g + f dg
(at least when f is a scalar field),
dF = dm dx + m ddx + dn dy + n ddy + do dz + o ddz.
But dd = 0, a basic and fundamental theorem that can be understood intutively
in many interesting ways, some of which I'll describe later.
Therefore, dF = dm dx + dn dy + do dz
= @m/@x dx dx + @m/@y dy dx + @m/@z dz dx
+ @n/@x dx dy + @n/@y dy dy + @n/@z dz dy
+ @o/@x dx dz + @o/@y dy dz + @o/@z dz dz.
Because multiplication of differentials is antisymmetric (df dg = - dg df),
dF = (@o/@y - @n/@z) dy dz + (@m/@z - @o/@x) dz dx + (@n/@x - @m/@y) dx dy.
This corresponds to the curl of (m,n,o).
Now suppose F is the 2form m dy dz + n dz dx + o dx dy.
Then dF = dm dy dz + dn dz dx + do dx dy
= @m/@x dx dy dz + @n/@y dy dz dx + @o/@z dz dx dy
= @m/@x dx dy dz + @n/@y dx dy dz + @o/@z dx dy dz
(all the other terms being 0 either because dd = 0 or because df df = 0).
So this is just like the divergence of (m,n,o).
INTEGRATION
0 dimensional integration is the evaluation of scalar fields at points.
For example, the integral of the scalar field f at the point p is f(p).
You might want to be funny and orient a point negatively;
then the value of the integral would be -f(p).
1 dimensional integration is the evaluation of covector fields along curves.
You can evaluate the vector field (m,n,o) along a curve using a line integral.
If F = (m,n,o), just integrate F . dr = m dx + n dy + o dz.
But this is just a covector field, so you can integrate it.
The direction along the curve matters;
changing this direction changes the sign of dr = (dx,dy,dz).
For example, the integral of x dx + z dy + y dz
along the line x = y in the xy plane from (0,0,0) to (1,1,0) is
int_0^1 x dx + int_0^1 z dy + int_0^0 y dz
= 1/2 + int_0^1 0 dy + 0 = 1/2.
2 dimensional integration is the evaluation of 2forms on surfaces.
You can evaluate the vector field (m,n,o) on a surface using a surface integral.
If F = (m,n,o), just integrate F . dS = m dy dz + n dz dx + o dx dy.
Here, dS = (dy dz, dz dx, dx dy) points in a direction normal to the surface.
The notation "dS" is historical; it's not the differential of any S.
In fact, dS is the cross product dr x dr / 2.
3 dimensional integration is the evaluation of 3forms in volumes.
You can evaluate the scalar field f in a volume using a volume integral.
dV = dr . dS / 3 = dr . dr x dr / 3!, and f dV is a 3form.
It makes sense to integrate all these forms over appropriate dimensions
even when working with manifolds of arbitrary dimension.
But you need a metric (dot product) to do line integrals,
and you need a metric in 3 dimensions for surface and volume integrals.
(Some of this can be generalized partially to other dimensions,
but you always need a metric to integrate vector fields.)
So the forms approach is better.
STOKES'S THEOREM
Suppose f is a scalar field.
The 2nd fundamental theorem of calculus for line integrals states that
the line integral of grad f along a curve from p to q is f(q) - f(p).
Let the boundary of a curve from p to q be q - p.
(That is, q is positively oriented and p is negatively oriented.)
Then the 2nd fundamental theorem is equivalent to saying
that the integral of df along a curve c is the integral of f at c's boundary.
Suppose F is a vector field.
The classical Stokes's theorem states that
the surface integral of curl F on a bounded surface S
is the line integral of F along the boundary of S.
Suppose F is now a covector field.
Stokes's theorem is equivalent to saying that
the integral of dF on S is the integral of F along the boundary of S.
Suppose F is a vector field.
Gauss's theorem states that the volume integral of div F in a bounded volume Q
is the surface integral of F on the boundary of Q.
Now suppose F is a 2form.
Gauss's theorem is equivalent to saying that
the integral of dF in Q is the integral of F on the boundary of Q.
In general, the generalized Stokes's theorem say that for any form F,
the integral of dF over an appropriate bounded region R
is the integral of F over the boundary of R.
This works in any dimension with any form.
WHY dd = 0
I never actually said what d means except when it acts on scalar fields.
One thing you can do is define d as an operator taking nforms to (n+1)forms
such that dd = 0, d satisfies the product rule,
and d takes a scalar field to its differential.
Then you can calculate d of any field the way I did in my post.
(Incidentally, the product rule is not that dFG = dF G + F dG.
Rather, if F is an nform, dFG = dF G + (-1)^n F dG.
So d acts like multiplication by a 1form for purposes of commutativity.
This makes sense, since d takes nforms to (n+1)forms,
just like multiplication by a 1form does.)
Alternatively, you can define d using Stokes's theorem.
Then you can prove that dd = 0, d satisfies the product rule,
and d takes a scalar field to its differential.
The key point here is that the boundary of a boundary is empty.
If R is some (n+2) dimensional region and @R is its boundary,
then @R is necessarily closed, so @@R is the empty set.
Imagine a ball, whose boundary is a sphere, whose boundary is empty.
Then just calculate: int_R ddF = int_@R dF = int_@@R F = 0.
Since int_R ddF = 0 for any R, it must be that ddF = 0.
But you should probably get a feeling for what d means in and of itself.
When it acts on a scalar field,
everyone is familiar with d in the sense of an infinitesimal change.
This is what it means when it acts on any form,
except that we're only interested in antisymmetric changes.
This is because the result has to be an antisymmetric form.