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Often times in multi-variable calculus you would have expressions for the differentials of area and volume like this $dA =dxdy$ or $dV = dxdydz$ which we are supposed to just accept because it makes sense in that if you take a tiny piece of area/volume it looks like a square blah blah….But cannot get my head around it especially when doing integral substitutions. In single variable calculus it made sense, for example consider the integral

$$\int f'(g) \frac{dg}{dx} dx = \int f'(g)dg = f(x) + c$$

this makes sense as $f'(x) = g'(x)f'(g)$ by the chain rule but when preforming a substitution in a double integral

$$\int \int f(x,y) dx dy = \int \int f(u,v) \det(J)du dv$$ where $$J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u}

& \frac{\partial y}{\partial v}\end{pmatrix}$$

How on earth have they magically jumped and concluded that $dxdy$ (which itself is not explained) equal to $\det(J)dudv$ ? This has caused me so many problems conceptually and i was not able to find any clear explanation to why this is the case and even what is meant by the term $dxdy$

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$\DeclareMathOperator{\Area}{Area}$**Edit**: In the original answer, I was a bit careless with signed *versus* unsigned area. The original question implicitly asks about signed area (i.e., area where “handedness” matters; $dv\, du = -du\, dv$), while most accounts in multivariable calculus treat unsigned area (i.e., the “geometric notion of content”; $|dv\, du| = |du\, dv|$).

The argument below is tweaked to incorporate sign consistently. Particularly, the “$(u, v)$-plane” is oriented, and $\Area$ refers throughout to *signed* area. Algebraically, the arguments can be made “unsigned” by placing absolute value signs around determinants, deleting the adjectives “signed” and “oriented” where they appear, and interpreting $\Area$ as unsigned area.

To give a geometric interpretation: Suppose you apply a *linear* change of variables $(x, y) = T(u, v)$ to the plane:

$$

\begin{aligned}

x &= au + bv, \\

y &= cu + dv;

\end{aligned}

\quad\text{i.e.,}\qquad

\left[\begin{array}{c}

x \\

y \\

\end{array}\right]

= \left[\begin{array}{cc}

a & b \\

c & d \\

\end{array}\right]

\left[\begin{array}{c}

u \\

v \\

\end{array}\right].

$$

Since $T$ is linear, $T = dT(u_{0}, v_{0})$ for every point $(u_{0}, v_{0})$.

The oriented rectangle $[u_{0}, u_{0} + \Delta u] \times [v_{0}, v_{0} + \Delta v]$, which has signed area $\Delta u\, \Delta v$, maps to a parallelogram whose signed area is, from linear algebra,

$$

(ad – bc)\, \Delta u\, \Delta v = \det(dT(u_{0}, v_{0}))\, \Delta u\, \Delta v.

$$

If instead your change of variables $(x, y) = F(u, v)$ is continuously-differentiable, the preceding discussion still holds “approximately at small scales”: The oriented rectangle $[u_{0}, u_{0} + \Delta u] \times [v_{0}, v_{0} + \Delta v]$ at left maps to a near-parallelogram at right whose signed area is

$$

\Area(F(R)) = \det(dF(u_{0}, v_{0}))\, \Delta u\, \Delta v + \text{error},

$$

with error asymptotically small in absolute value compared to $\Delta u\, \Delta v$. Using infinitesimal notation, this state of affairs may be expressed by saying

$$

\Area\bigl(F([u_{0} + du] \times [v_{0} + dv])\bigr) = \det dF(u_{0}, v_{0})\, du\, dv.

$$

To connect this with integration, let $D$ denote the oriented rectangle on the left, think of a continuous, real-valued function $f$ defined over the region $F(D)$ on the right, and consider the problem of expressing the integral as an integral over $D$ itself. The change of variables formula says (assuming $F$ is one-to-one)

$$

\iint_{F(D)} f(x, y)\, dx\, dy = \iint_{D} f(F(u, v)) \det dF(u, v)\, du\, dv.

$$

This is the sum of infinitesimal contributions of the type

\begin{align*}

\iint_{F(R)} f(x_{0}, y_{0})\, dx\, dy

&= f(F(u_{0}, v_{0})) \Area(F(R)) \\

&= f(F(u_{0}, v_{0})) \det dF(u, v) \Area(R) \\

&= \iint_{R} f(F(u_{0}, v_{0})) \det dF(u, v)\, du\, dv.

\end{align*}

(If $R$ is sufficiently small, the continuous functions $f$ and $f \circ F$ are nearly constant.)

Analogous pictures hold in arbitrary (finite) dimension.

The explanation goes through the observation that the symbol $\mathrm d x\,\mathrm d y$ is actually the *exterior product* of differential forms $\;\mathrm d x\wedge\mathrm d y$.

As $\;\mathrm d x=\dfrac{\partial x}{\partial u}\mathrm d u+\dfrac{\partial x}{\partial v}\mathrm d v $, and similarly for $\;\mathrm d y$, we obtain, following the computation rules of exterior algebra:

\begin{align}

\mathrm d x\wedge\mathrm d y&=\biggl(\dfrac{\partial x}{\partial u}\mathrm d u+\dfrac{\partial x}{\partial v}\mathrm d v\biggr)\wedge\biggl(\dfrac{\partial y}{\partial u}\mathrm d u+\dfrac{\partial y}{\partial v}\mathrm d v\biggr)\\&=\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial u}\;\mathrm d u\wedge\mathrm d u +\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v}\;\mathrm d u\wedge\mathrm d v+\dfrac{\partial x}{\partial v}\dfrac{\partial y}{\partial u}\;\mathrm d v\wedge\mathrm d u +\dfrac{\partial x}{\partial v}\dfrac{\partial y}{\partial v}\;\mathrm d v\wedge\mathrm d v\\

&=\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v}\;\mathrm d u\wedge\mathrm d v-\dfrac{\partial x}{\partial v}\dfrac{\partial y}{\partial u}\;\mathrm d u\wedge\mathrm d v =\dfrac{\partial(x, y)}{\partial(u,v)}\;\mathrm d u\wedge\mathrm d v.

\end{align}

Firstly, the role of change of variables in integrals such as the general double integral you wrote above introduces *new* variables in terms of *old* variables.

Explicitly this means that We are transforming from $(x,y)$ to $(u,v)$ coordinates, and the transformation is given as ‘old in terms of new’ variables.

That is:

$$x = x(u, v),\, \quad y = y(u, v)$$

Thus, if one has an initial $f=f(x, y)$ then we now seek $F(u, v)=f(u(x, y),v(x, y))$

Using the Chain Rule, one obtains

\begin{align}

\frac{\partial f}{\partial u} &= \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u} \tag{1}\\

\frac{\partial f}{\partial v} &= \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} \tag{2}

\end{align}

however, this is all well and good, but what if the transformation to *new* variables is **not** invertible? Well, we could form the following

\begin{align}

\frac{\partial f}{\partial x} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \tag{3}\\

\frac{\partial f}{\partial y} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} \tag{4}

\end{align}

Treating $(3),(4)$ as simultaneous equations in unknowns $\partial f/\partial u$ and $\partial f /\partial v$ and rearranging, one obtains

\begin{align}

\frac{\partial f}{\partial u} &= \left(\frac{\partial f}{\partial x}\frac{\partial v}{\partial y} – \frac{\partial f}{\partial y}\frac{\partial x}{\partial v}\right)/\left(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y} – \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}\right) \tag{5}\\

\frac{\partial f}{\partial v} &= \left(\frac{\partial f}{\partial y}\frac{\partial u}{\partial x} – \frac{\partial f}{\partial x}\frac{\partial u}{\partial y}\right)/\left(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y} – \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}\right) \tag{6}

\end{align}

The denominator is of very special significance. It is the Jacobian. Usually denoted

\begin{align}

J(u, v) &= \left(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y} – \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}\right) \\

&=

\begin{vmatrix}

\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\

\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\

\end{vmatrix} \\

&= \frac{\partial (u, v)}{\partial (x, y)}

\end{align}

Importantly, if $(u, v)$ and $(x, y)$ are functionally independent then $J \neq 0$.

So this is all about differentiation, but the role of the Jacobian in multiple integrals arises very much in the same way. The transformation from one region in the $xy$-plane to another in the $uv$-plane requires us to define what we mean by a small infinitesimal area slice in the *new* variables that arise from the *old*.

Symbolically one would need to redefine a new rectangular (in the limit, for simplicity) are element, so that we can evaluate a multiple integral over this new are in new coordinates.

Symbolically this is

$$dA = |d \mathbf{u} \times d \mathbf{v} |$$

Using $\hat{\mathbf{i}},\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ as unit vectors in the $x,y$ and $z$ directions one obtains

\begin{align}

d \mathbf{u} &= \frac{\partial x}{\partial u}du \hat{\mathbf{i}}+ \frac{\partial y}{\partial u}du \hat{\mathbf{j}} \\

d \mathbf{v} &= \frac{\partial x}{\partial v}dv \hat{\mathbf{i}}+ \frac{\partial y}{\partial v}dv \hat{\mathbf{j}}

\end{align}

Taking the vector cross product of these yields,

$$d \mathbf{u} \times d \mathbf{v} = \hat{\mathbf{k}}\left(\frac{\partial x}{\partial u}du \frac{\partial y}{\partial v}dv-\frac{\partial x}{\partial v}du \frac{\partial y}{\partial u}dv \right)$$

Since $\hat{\mathbf{k}}$ is a unit vector, taking the modulus of the above yields,

$$d A = \Big |\frac{\partial (x,y)}{\partial (u, v)}\Big |dudv$$

This is why, in both differentiation and multiple integration, when transformation of coordinate systems is considered, the Jacobian arises in both cases.

My best effort came when using differentials in a straight forward use of small linr-segments or ssmall cubes (rectangular boxes) and explained the the Theorems of Calculus (Namly the Fundamental Theorem of Calculus to *”talk away”*how the limit process being the formal result og performing an infinite number of operations in one-fell-swoop.

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