Scaling factor required for change of coordinates for integration but not for integration of parametric forms of surfaces?

Question

When we use change of variables for integration, we are required to also multiply the integrand by a scaling factor:

For change of coordinates with double integrals, the scaling factor is found by $dxdy = \left| \det \left( \dfrac{\partial g(u,v)}{\partial(u,v)} \right) \right| dudv$.

For change of coordinates with triple integrals, the scaling factor is found by $dxdydz = \left| \det \left( \dfrac{\partial g(u,v,w)}{\partial(u,v,w)} \right) \right| dudvdw$.

For change of coordinates with polar coordinates, the scaling factor is found by $\left| \det \left( \dfrac{\partial g(\rho, \theta)}{\partial(\rho, \theta)} \right) \right| = \rho$.

For change of coordinates with cylindrical coordinates, the scaling factor is also $\rho$.

For change of coordinates with spherical coordinates, there scaling factor is found by $\left| \det \left( \dfrac{\partial g(r, \theta, \phi)}{\partial(r, \theta, \phi)} \right) \right| = r^2\sin(\phi)$.

These scaling factors are used when using change of coordinates to calculate the area or volume of an object.

However, when we parameterise surfaces, such as when finding the area of a surface $\left( \iint_S \left| \dfrac{\partial S}{\partial u} \times \dfrac{\partial S}{\partial v} \right| dudv \right)$ or calculating a surface integral $\left( \iint_S f(g(u, v)) \cdot \left| \dfrac{\partial S}{\partial u} \times \dfrac{\partial S}{\partial v} \right| dudv \right)$, we do not utilise a scaling factor; instead, we just parameterise and apply the relevant formulae. I can parameterise a surface using cylindrical coordinates or spherical coordinates, but unlike when using change of variables to integrate an object for volume, I do not need to multiply the integrand by a scaling factor.

One would assume that a scaling factor would be required in both cases, since they both involve converting from one coordinate system to another. Why is this not the case? The only reason I can think of is that the scaling factor is already implicitly imbedded in our formulae, since, similarly to the formulae used to find the scaling factors, we are taking the determinant of the partial derivatives with respect to each coordinate when we use the cross product $\left( \iint_S \left| \dfrac{\partial S}{\partial u} \times \dfrac{\partial S}{\partial v} \right| dudv, \iint_S f(g(u, v)) \cdot \left| \dfrac{\partial S}{\partial u} \times \dfrac{\partial S}{\partial v} \right| dudv \right)$. Is my hypothesis correct? Or is there another reason for this?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Using the divergence theorem, it is possible to link definition of the surface integral, in particular, the magnitude of the cross-product with the Jacobian determinant in the change of variables. I wrote about this here if you're interested. In that article, I take the divergence theorem as given and then try to deduce the scaling factor for change in variables. But this approach is shaky since the divergence theorem is a deeper result than change in variables. But at least, you can see that they are connected.

So to answer your question, the definition of the surface integral incorporates scaling factor for the change of variables. In fact, there's a more general concept of integrating differential forms which extends surface integrals and for which the scaling factor follows immediately as well. From this perspective, the Jacobian and the cross-product are sort of different facets of the same thing.

Answer 2

The geometric objects $O$ (curves, surfaces, solids) we are dealing with in geometry or physics are embedded in a euclidean space ${\mathbb R}^n$ (mostly $n=2$ or $n=3$) and inherit their metric characteristica (length, area, curvature, volume, etc.) from the a priori metric in ${\mathbb R}^n$.

The description of such an $O$ can be in terms of equations in the variables $x_1$, $\ldots$, $x_n$, in terms of inequalities, in terms of geometrical constructions, and similar. But for computations we mostly rely on a parametric representation. If $O$ has intrinsic dimension $d$ then we typically have a simple parameter domain $B$ (a segment, a rectangle, a circular disk, a $3$-simplex) in $d$-space ${\mathbb R}^d$ and a differentiable map $${\bf f}:\quad B\to{\mathbb R}^n,\qquad(u_1,\ldots, u_d)\mapsto{\bf x}={\bf f}(u_1,,\ldots, u_d)\ ,$$ which maps $B$ essentially in a $1:1$ way, i.e., bijectively, onto $O$.

The parameter variables can be "anonymous", like here, or be geometric quantities relevant in the situation, like geographical latitude and longitude, etc. But in any case $B\subset{\mathbb R}^d$, so that $B$ carries a natural $d$-dimensional measure ${\rm d}(u_1,\ldots,,u_d)$, or ${\rm d}(r,\phi,\theta)$.

Now comes the difficult part: The scaling factor. It is a fact of life that under a differentiable map a tiny $d$-dimensional piece of $B$ is mapped onto a tiny piece of the object $O$, and that the $d$-dimensional measures of these tiny pieces are related by a nonconstant scaling factor $J\geq0$ . This is expressed by equations of the form $${\rm d}\omega=J({\bf u})\cdot{\rm d}(u_1,\ldots,u_d)\ .\tag{1}$$ Here ${\rm d}\omega$ denotes the $d$-dimensional length-, area-, or volume element on $O$. Look at the formulas you quote: They are all of type $(1)$, with integral signs in front.

What I have written up here is more of a grand view nature. There can be no question of regurgitating the processes leading to the exact formulas you quote in your question.