This is a well-known identity:
$$ \left( \int_{\Delta x}g(x)dx \right)\left( \int_{\Delta y} f(y) dy \right)=\int_{\Delta x}\int_{\Delta y}g(x)f(y)dxdy \tag{1} $$
But I am wondering if we may be missing out on something important by doing so.
Consider two parametrized line elements :
$$ (ds_\lambda)^2 = \eta_{\mu\nu} \frac{\partial g(\lambda)^\mu}{\partial \lambda} \frac{\partial g(\lambda)^\nu}{\partial \lambda} \\ (ds_\beta)^2 = \eta_{\mu\nu} \frac{\partial f(\beta)^\mu}{\partial \beta} \frac{\partial f(\beta)^\nu}{\partial \beta} \\ $$
Now applying Identity (1) to the line elements, I get
$$ \left( \int_{\Delta \lambda} \sqrt{\eta_{\mu\nu} \frac{\partial g(\lambda)^\mu}{\partial \lambda} \frac{\partial g(\lambda)^\nu}{\partial \lambda}} d\lambda \right)\left( \int_{\Delta \beta} \sqrt{\eta_{\mu\nu} \frac{\partial f(\beta)^\mu}{\partial \beta} \frac{\partial f(\beta)^\nu}{\partial \beta}} d\beta \right) \\ = \int_{\Delta \lambda} \int_{\Delta \beta} \sqrt{\eta_{\mu\nu} \frac{\partial g(\lambda)^\mu}{\partial \lambda} \frac{\partial g(\lambda)^\nu}{\partial \lambda} \eta_{\mu\nu} \frac{\partial f(\beta)^\mu}{\partial \beta} \frac{\partial f(\beta)^\nu}{\partial \beta}} d\lambda d\beta $$
This seems close to the definition of the area integral involving the determinant, but not quite.
However, if I multiply the integrals using the wedge product:
$$ \left( \int_{\Delta x}g(x)dx \right) \wedge \left( \int_{\Delta y} f(y) dy \right) $$
Can I recover the correct area integration formula:
$$ \iint \sqrt{\det g_{\mu\nu}}dxdy $$
I expect that it is possible?