# What does it mean to multiply differentials?

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 $dxdy$

• I guess what you want is differential forms May 25, 2017 at 10:03
• They haven't done magic, they've applied the 2D version of the chain rule, which is well enough known that the author feels you can go find a proof instead of having the proof included right here. May 25, 2017 at 18:20

$\DeclareMathOperator{\Area}{Area}$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.

• I prefer this explanation compared with the more abstract one to start with. From the point of view of its utility in physics, the abstract one lacks of the direct interpretation that has the one you gave. May 25, 2017 at 12:22
• @RafaBudría: As a geometer (not an algebraist) I tend to agree with you. :) On the other hand, the multilinear algebra of Bernard's (+1) answer is merely the algebraic expression of the geometric ideas here. On a tangent, J. W. Gibbs' AAAS address On Multiple Algebra might be of interest. May 25, 2017 at 13:44
• +1 -- Explaining why the Jacobian appears is important here, not to be hidden in the discussion of the "multiplication of differentials" May 25, 2017 at 15:33

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}

• There's an abstract and quite general definition of the exterior product (a quotient of the tensor product by the relations which make the tensor product alternating (in the quotient). For the second question, intuitively, the exterior product is more or less the cross-product, so it is ‘orthogonal’ to the plane defined by $\mathrm d x$ and $\mathrm d y$ (the tangent plane). May 25, 2017 at 11:31
• I do not quite agree with this explanation. It is true as a computation in the exterior algebra, however this is not the way to prove the statement that the OP talks about. In any text on integration on manifolds, one defines the integral of a volume form locally, and then USES the result stated by the OP to show that this volume-form integral is well-defined. In other words, the result stated by the OP should be proven using analysis, not the algebra of differential forms. May 25, 2017 at 13:18
• In other words, you cannot prove the change of variable formula using your argument. You USE the change of variable formula to show that integral on manifolds are well-defined. The change of variable formula itself has to be proven through analytic means. May 25, 2017 at 13:20
• @user2520938: I didn't say I was proving it. I explained where it comes from, and why it is a natural formula. May 25, 2017 at 15:39
• @user2520938: Maybe, but, as often, later discoveries such as Grassman's and the theory of differential forms were illuminating for what's in action behind the change of variable formula. May 25, 2017 at 15:49

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.