# How to show algebraically that the determinant of the Jacobian the scaling factor for change in variables?

I am currently trying to learn about Jacobians (self study). In particular struggling to understand the geometry of change in variables and why the determinant of the Jacobian is the scaling factor when going from $$dudv \to dxdy$$.

My current understanding is that the Jacobian represents the best approximation of a linear transformation of the change in a function given a tiny change in its inputs at a given point. At least in my head this makes sense as:

Let: $$\begin{matrix} x = f_1(u,v) \\ y = f_2(u,v) \end{matrix}$$

Then based on the definition of the Jacobian:

$$\left(\begin{matrix} dx \\ dy \end{matrix}\right) = J \cdot \left(\begin{matrix} du \\ dv \end{matrix}\right)$$

Where I get lost is how to relate this back to the transformation of $$dxdy$$. I tried to calculate it as follows though it doesn't work so am assuming I've made (several) invalid steps along the way...

$$\left(\begin{matrix} dx \\ dy \end{matrix}\right) = J \cdot \left(\begin{matrix} du \\ dv \end{matrix}\right) = \left(\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix}\right) \cdot \left(\begin{matrix} du \\ dv \end{matrix}\right) = \left(\begin{matrix} \frac{\partial x}{\partial u} \cdot du\ + \frac{\partial x}{\partial v} \cdot dv \\ \frac{\partial y}{\partial u} \cdot du\ + \frac{\partial y}{\partial v} \cdot dv \end{matrix}\right)$$

Thus:

\begin{align} dxdy &= det\left[ \left(\begin{matrix} dx & 0 \\ 0 & dy \end{matrix}\right) \right]\\ \\&= det\left[ \left(\begin{matrix} \frac{\partial x}{\partial u} \cdot du\ + \frac{\partial x}{\partial v} \cdot dv & 0 \\ 0 & \frac{\partial y}{\partial u} \cdot du\ + \frac{\partial y}{\partial v} \cdot dv \end{matrix}\right) \right] \\ \\&= \left(\frac{\partial x}{\partial u} \cdot du\ + \frac{\partial x}{\partial v} \cdot dv \right) \cdot \left( \frac{\partial y}{\partial u} \cdot du\ + \frac{\partial y}{\partial v} \cdot dv \right) \end{align} \\ \\ \\ = \text{...?}

I know that this should resolve to $$det(J)dudv$$ but I just don't know how to equate that. Any hints / advice on where I might have gone conceptually wrong would be appreciated.

I have already read the answers to similar questions but they haven't helped me clarify where I've gone wrong :(

• This works out if you are thinking of $\mathrm{d}x$ and $\mathrm{d}y$ as differential forms. However, that's probably not the answer you are looking for? Aug 11, 2023 at 13:41

For the $$2$$-D Jacobian, consider a $$1$$-$$1$$ continuous transformation from $$(x, y)$$ to $$(u, v)$$. Then you would write $$x=x(u, v)$$ and $$y=y(u, v)$$, furthermore

\begin{align} \int \int_R f(x, y)\mathcal{d}x\mathcal{d}y = \int \int_{R'}f(x(u, v), y(u, v)) \left| \frac{\partial(x, y)}{\partial(u,v)}\right|\mathcal{d}u\mathcal{d}v \end{align}

Where the Region R in the $$xy$$-plane maps to the region R' in the $$uv$$-plane. Setting \begin{align} \left| \frac{\partial(x, y)}{\partial(u,v)}\right| = x_uy_v-x_vy_u \end{align} Now the transformation from R to R' distorts the cartesian 'flat' $$xy$$-plane to a distorted version in $$u$$ and $$v$$, which for small enough $$(du, dv)$$ rectangles map to parallelograms, $$\hat{du}=(du, 0)$$ and $$\hat{dv}=(dv, 0)$$ with

\begin{align} \hat{du} = \begin{pmatrix} x_u & x_v\\ y_u & y_v\\ \end{pmatrix} \begin{pmatrix} du\\ 0 \\ \end{pmatrix} \end{align} and similarly for $$\hat{dv}$$.

The Area isnthen found by $$dA = |\hat{du} \times \hat{dv}|= \begin{pmatrix} x_u du\\ y_u du\\ \end{pmatrix} \times \begin{pmatrix} x_v dv\\ y_v dv\\ \end{pmatrix} = (x_u y_v - x_v y_u )du dv$$ I wonder if this helps?

Ok I think I have come up with an explanation for why this approach doesn't work that at least makes sense to me (e.g. I'm not sure it's completely technically accurate but it at least makes intuitive sense :) ). I just hope this is useful to someone else in the future !

So to understand what's going on I think it's good to be clear about what we are actually doing. Lets start by defining the initial integral:

$$\int\int_R f(x,y) \ dx \ dy$$

This essentially represents the area calculated by performing the infinite sum of cuboids in the x/y plane who have a base area of dx/dy and a height of $$f(x,y)$$.

By changing the basis of the integral to $$du$$, $$dv$$ we are saying that we instead want to define our cuboids in the u/v plane and that the base of these cuboids should be a square of $$du \cdot dv$$. But what is the "height" of these cuboids ? We don't actually know this, but we would know the "height" if we were map them from the u/v plane into the x/y plane; that is to say that the area of our cuboid in the u/v plane = "height in the x/y plane" x base area in the x/y plane.

Thankfully the Jacobian allows us to map between the 2 planes and is defined as: $$\left(\begin{matrix} dx \\ dy \end{matrix}\right) = J \left(\begin{matrix} du \\ dv \end{matrix}\right)$$

Or more graphically:

The key bit to realise is that the area of this shape in the x/y plane (this "shape" being our square in u/v plane after its been mapped to the x/y plane) is equal to the determinant of the matrix of the transformed basis vectors e.g.

$$\left| J\left(\begin{matrix} du & 0 \\ 0 & dv \end{matrix}\right)\right| = \left| J\right| \cdot \left|\left(\begin{matrix} du & 0 \\ 0 & dv \end{matrix}\right)\right| = \left| J\right| \cdot du \cdot dv$$

Thus we can now say that the volume of our cuboid in the u/v plane equals:

$$\underbrace{f\Big(f_1(u,v), f_2(u,v)\Big)}_\text{height in the x/y plane} \cdot \underbrace{\left| J\right| \cdot du \cdot dv}_{ \substack{ \text{area of the base square} \\ \text{in the x/y plane} }}$$

So how does all this relate / answer the original question?

Well my question was why can't we just do:

$$det \left| \left(\begin{matrix} dx & 0 \\ 0 & dy \end{matrix}\right) \right | = det \left| \left(\begin{matrix} \frac{\partial x}{\partial u} \ du + \frac{\partial x}{\partial v} \ dv& 0 \\ 0 & \frac{\partial y}{\partial u} \ du + \frac{\partial y}{\partial v} \ dv \end{matrix}\right) \right |$$

Well the issue is that by doing this we are actually finding the volume of the square that is created by just taking the x/y coordinates of our original u/v square after its been mapped to the x/y plane. This however is not equal to the volume of our u/v square in the x/y plane which is what we are actually after in order to adjust the integral. This is perhaps better shown as a diagram:

That is, in order to correctly adjust our integral for defining the base of our cuboids as a du/dv square we need to find the area of the green shape. However, by using direct substitution (as I was doing in the opening question) what I'm actually doing is finding the area marked by the blue shape.