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Today in lecture, we derived the formula for change of variables in multivariable integration, i.e. $dV' = \det(J) dV$. Here is our derivation:

Let $\vec{F}: U \subseteq \mathbb{R}^2 \to \mathbb{R}^2$ defined by $\vec{F}(x, y)$ be a bijection which represents a coordinate transformation $x = x(u, v)$ and $y = y(u, v)$.

Consider an area element $dV$ bounded by $(u_0, v_0), (u_0 + \Delta u, v_0), (u_0, v_0 + \Delta v)$ (and $(u_0 + \Delta u, v_0 + \Delta v)$). Under $\vec{F}$, the transformed area element $dV'$ will be bounded by $\vec{F}(u_0, v_0), \vec{F}(u_0 + \Delta u, v_0), \vec{F}(u_0, v_0 + \Delta v)$. As $\Delta u, \Delta v \to 0$, the transformed area element $dV'$ will approach a parallelogram, so we can approximate the area by

$$\begin{align*} \Delta a &\approx |\vec{a} \times \vec{b}| \\ &= |\color{purple}{(\vec{F}(u_0 + \Delta u, v_0) - \vec{F}(u_0, v_0)}) \times (\color{blue}{\vec{F}(u_0, v_0 + \Delta v) - \vec{F}(u_0, v_0)})| \\ &\approx |\color{purple}{(\vec{F}_u(u_0, v_0)\Delta u)}\times\color{purple}{(\vec{F}_v(u_0, v_0)\Delta v)}| \\ &= |\vec{F}_u\times\vec{F}_v|\Delta u\Delta v \end{align*}$$

Therefore, as $\Delta u, \Delta v \to 0$, $dA = \left|\begin{pmatrix}\nabla x \\ \nabla y\end{pmatrix}\right|dudv$.

And then of course this can be generalised to higher dimensions blablabla. However, I do not understand why approximating $\vec{F}(u_0 + \Delta u, v_0)$ by its linear approximation $\vec{F}(u_0 + \Delta u, v_0) \approx \vec{F}(u_0, v_0) + \Delta u\vec{F}_u(u_0, v_0)$ is sufficient. In particular, what happens if we instead approximate by say a quadratic approximation around $(u_0, v_0)$? And what happens to the $dV' = \cdots dV$? Does it change the scaling factor?

Thank you!

Gareth Ma
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  • Loosely, when you do the integration you take smaller and smaller boxes around each point and the only part that 'remains' is the linear part. – copper.hat Feb 28 '22 at 18:33
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    The change of variables theorem tells you that this is sufficient. A-priori, it is of course unjustified to drop all the other terms. What this "derivation" gives you is an inutition for why $|\det DF|$ appears in the change-of-variables formula. It is a completely separate matter to then provide an airtight rigorous proof. – peek-a-boo Feb 28 '22 at 18:36
  • @copper.hat Yeah that's the explanation my prof. gave me as well, but it just felt a bit unjustified I guess - like the whole $\Delta x \to dx$ limit is already weird and saying e.g. $(\Delta x)^2\to 0$ (?) while $\Delta x\to dx$... hope you understand haha – Gareth Ma Feb 28 '22 at 18:42
  • @peek-a-boo I see, so this derivation is not rigorous, or not rigorous enough to understand why the other terms would vanish? I would look into the theorem, though the content wikipedia already seems quite dense – Gareth Ma Feb 28 '22 at 18:43
  • I understand, but it is one of those things where you need to grind through the details to see why. But the observation that for linear (affine) functions that $\det J$ is what matters is the essential central theme. – copper.hat Feb 28 '22 at 18:44
  • @GarethMa it's just not rigorous :) but it's perfect for the intuition. I have written a complete proof here, but of course it has a lot of technical prerequisites. But for your question specifically, the point to keep in mind is the "technical lemma" in that post (and also the Radon-Nikodym theorem, but that's way out of the scope for now). Essentially, the limit of ratios of volumes is the Jacobian determinant: $\lim\limits_{r\to 0^+}\frac{\text{vol} F(B_r(x))}{\text{vol}(B_r(x))}=|\det DF_x|$. – peek-a-boo Feb 28 '22 at 18:45
  • It is just an analogy but is is similar to why we can write $f(x)-f(x_0) = \int_{x_0}^x f'(t)dt$, we only need the linear approximation $f'(t)$ to recover $f$. – copper.hat Feb 28 '22 at 18:46
  • @peek-a-boo I see, I will take a look... a few years later xD Thank you everyone for your help! – Gareth Ma Feb 28 '22 at 18:48

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