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2022-04-29: Fabricating Evidence to catch Carmen Sandiego

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2020-05-04: Archimedes and the sphere

2019-05-16: Glow worms return

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2016-09-25: Amath Timeline

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2016-02-20: Apple VS FBI

2016-02-19: More Zika may be better than less

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2015-03-24: Dawn of the CRISPR age

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2014-11-10: Vaccine mandates and bioethics

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2013-11-10: L'Hopital's Rule for Multidimensional Systems

2013-11-09: Using infinitessimals in vector calculus

2013-11-08: Functional Calculus

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2013-11-02: Proof of the circle area formula using elementary methods

Using infinitessimals in vector calculus

All right. So infinitessimals rock. One of the first examples of this is how you used them in elementary calculus to construct integration formula's like the disk method and the shell method for finding volumes of some objects. But some people don't like them, so they've been largely removed from most calculus curriculums. Never fear -- you can keep using them, even if your teacher won't.

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!

What are we missing?

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. \[ (dx + dy ) \wedge dz = dx \wedge dz + dy \wedge dz \] \[ dz \wedge (dx + dy ) = dz \wedge dx + dz \wedge dy \]

Fixing our simple linear change of variables

So, let's go back and rewrite our integral using the wedge product for infinitessimal vectors. \[ \iint_{\Omega} f(x+y,x-y) dy \wedge dx \] \[ \iint_{\Omega^*} f(u,v) \frac{1}{4} (du-dv) \wedge (du+dv) \] \[ \iint_{\Omega^*} f(u,v) \frac{1}{4} ( du \wedge du + du \wedge dv - dv \wedge du - dv \wedge dv ) \] Now, applying our new rules, \[ \iint_{\Omega^*} f(u,v) \frac{1}{4} 2 du \wedge dv \] \[ \iint_{\Omega^*} f(u,v) \frac{1}{2} du \wedge dv \]

Deriving the rule for a change of variables to polar coordinates

Starting with an integral in cartessian coordinates, \[ \iint_{\Omega} f(x,y) dx \wedge dy \] The transform to polar coordinates is \[ x = r \cos \theta, \; y = r \sin \theta \] so the infinitessimal transform is \[ dx = \cos \theta dr - r \sin \theta d\theta , \; dy = \sin \theta dr + r \cos \theta d\theta \] \begin{align} dx \wedge dy &= ( \cos \theta dr - r \sin \theta d\theta ) \wedge ( \sin \theta dr + r \cos \theta d\theta ) \\ &= \cos \theta \sin \theta dr \wedge dr + r \cos^2 \theta dr \wedge d\theta \\ & \quad\quad - r \sin^2 \theta d\theta \wedge dr - r^2 \sin \theta \cos \theta d\theta \wedge d\theta \\ &= r \cos^2 \theta dr \wedge d\theta + r \sin^2 \theta dr \wedge d\theta \\ &= r (dr \wedge d\theta) \end{align} Notice that in this calculation, we've used the rule for switching the order of infinitessimals and the nilpotence rule.

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.