Now, we've moved on to vector calculus, and multidimensional integrals, and you want to keep using them. Unfortunately, things aren't working out so well. You're given a problem of finding a volume under a surface that looks like \[ \iint_{\Omega} f(x+y,x-y) dy dx \] and you want to do a change of variables $u = x+y, \; v=x-y$ to simplify the calculation. Looks simple. Pull out our infinitessimals, and find $2 dx = du + dv, \; 2 dy = du - dv$. \[ \iint_{\Omega^*} f(u,v) \frac{1}{4} (du-dv) (du+dv) \] \[ \iint_{\Omega^*} f(u,v) \frac{1}{4} (dudu-dvdu + dudv -dvdv) \] But now what? Well, we expect squared infinitessimals to vanish because they don't actually have any area measure. And $du dv = dv du$, so it looks like all our infinitessimal's really cancel out!

The key idea at hand is that we're now talking about geometry problems
in multiple dimensions. The infinitessimal $dydx$ does not represent
regular multiplication -- really, it represents an infinitessimal square in the coordinate plane of $x$ and $y$. $dx$ and $dy$ are infinitessimal vectors,
and we know vector multiplication is a more complicated thing than regular multiplication. So, we define a new product called the "wedge-product" to construct an infinitessimal square area $dA$ out of the sides of the square $dx$ and $dy$.
\[ dA := dy \wedge dx \]
But the rules a wedge product that we use to create these squares are a little different than regular multiplication.
The two most important ones are as follows.
The first rule is called **nilpotence**:
\[ dx \wedge dx = dy \wedge dy = 0 \]
Nilpotence means that you cann't use the same side twice to create an infinitessimal square.
The second less obvious rule is called **anti-commutativity**:
\[ dy \wedge dx = - dx \wedge dy \]
Anti-commutativity happens because our infinitessimal areas have not just magnitude, but also orientation, and assembling the edges in the opposite order flips that orientation.

The one other rule of importance for the moment is scalar multiplication of infinitiessimals. If we multiply one or the other of these infinitessimal's by a scalar $\alpha$, it changes the area by the same amount, and does not change any orientations, so \[ (\alpha dy ) \wedge dx = \alpha (dy \wedge dx) = dy \wedge (\alpha dx) \] And we have standard distributive laws. \[ (dx + dy ) \wedge dz = dx \wedge dz + dy \wedge dz \] \[ dz \wedge (dx + dy ) = dz \wedge dx + dz \wedge dy \]

Substituting, we obtain our standard formula \begin{align} \iint_{\Omega} f(x,y) dx dy &= \iint_{\Omega} f(x,y) dx \wedge dy \\ &= \int_{\Omega^*} f(r,\theta) r dr \wedge d\theta \\ &= \int_{\Omega^*} f(r,\theta) r dr d\theta \end{align} Now, this is the exact same thing you get if you use the standard formula involving determinants of the Jacobian matrix, but it's a much more direct derivation that you can do easily on a test if you forgot the formula.

Rules for spherical coordinates and radial coordinate transforms in 3 dimensions can be derived in the same way.