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.
Best Answer
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.