I'll present a calculation of the surface area of a sphere in $N$-dimensions. This calculation is performed in cartesian coordinates. I haven't seen the computation done this way before (though I haven't done an exhaustive search of the literature).
Then I'll attempt to generalize the method to compute the surface area of the ellipsoid. Perhaps unsurprisingly, it fails. I'm unsure why and lack an interpretation for what has actually been computed in this calculation.
An $N-1$ sphere (in $N$ dimensions) of radius $\chi$ has surface area given in terms of the Dirac $\delta$ function by \begin{align} S_N &= \int_{-\infty}^\infty du_1\, \int_{-\infty}^\infty du_2\, \cdots \int_{-\infty}^\infty du_N\ \delta\left(\chi-\sqrt{\sum_{i=1}^N u_i^2}\right)\\ &= 2\chi\int_{-\infty}^\infty du_1\, \int_{-\infty}^\infty du_2\, \cdots \int_{-\infty}^\infty du_N\ \delta\left(\chi^2-{\sum_{i=1}^N u_i^2}\right)\\ &=2\chi \int_{-\infty}^\infty du_1\, \int_{-\infty}^\infty du_2\, \cdots \int_{-\infty}^\infty du_N\, \int_{-\infty}^\infty \frac{d\xi}{2\pi} \, e^{i(\chi^2 - u_1^2 - u_2^2 -\cdots- u_N^2)\xi}. \\ &= (-i\pi)^{N/2} \frac{\chi}{\pi} \int_{-\infty}^\infty d\xi \, \frac{e^{i\chi^2\xi}}{(\xi-i\epsilon)^{N/2}}. \end{align} The identity \begin{align} \int_{-\infty}^\infty dx\, e^{-ix^2(\xi-i\epsilon)} &= \sqrt{\frac{-i\pi}{\xi-i\epsilon}} \end{align} has been used, where $\epsilon>0$. We can evaluate the remaining integral as: \begin{align} I_N &= \int_{-\infty}^\infty d\xi \, \frac{e^{i\chi^2\xi}}{(\xi-i\epsilon)^{N/2}}, \\ &= \frac{2e^{iN\pi/4}\pi\chi^{N-2}}{\Gamma(\tfrac{N}{2})}. \end{align} Substitution of this integral result and using $(-i)^{N/2}=e^{-iN\pi/4}$ gives: \begin{align} S_N &= \frac{2\pi^{N/2}}{\Gamma(\tfrac{N}{2})}\chi^{N-1}. \end{align} Substituting $N=2,3$ into the above gives the expected results for the circumference of the circle and surface area of the 2-sphere; indeed it holds for arbitrart $n\in\mathbb{Z}>0$. (Note: $\Gamma({3}/{2}) = {\sqrt\pi}/{2}$.)
Before moving on to the ellipsoid, perhaps it's a good idea to comment on the use of the $\delta$ function. First, the $\delta$ is a well defined distribution -- that is, a limit of a sequence of functions. For example, if we define \begin{align} \delta_{\tilde\epsilon}(x) &= \frac{1}{\sqrt{\tilde\epsilon\pi}} e^{-x^2/\tilde\epsilon^2}, \end{align} and perform the above computations for $\tilde\epsilon^2>0$ substituting $\delta(\cdots)\to\delta_{\tilde\epsilon}(\cdots)$, taking $\tilde\epsilon\to 0$ at the end of the calculation, we recover the stated results for both the sphere and, below, for the ellipsoid.
Now, we attempt to generalize this result to that of an ellipsoid. \begin{align} E_N &= \int_{-\infty}^\infty du_1\, \int_{-\infty}^\infty du_2\, \cdots \int_{-\infty}^\infty du_N\ \delta\left(\chi-\sqrt{\sum_{i=1}^N\frac{u_i^2}{\tilde a_i^2}}\right)\\ &= \prod_{j=1}^N {\tilde a}_j \int_{-\infty}^\infty du_1\, \int_{-\infty}^\infty du_2\, \cdots \int_{-\infty}^\infty du_N\ \delta\left(\chi-\sqrt{\sum_{i=1}^N u_i^2}\right)\\ &= S_N \prod_{j=1}^N {\tilde a}_j. \end{align} Numerical comparison via Mathematica (http://mathworld.wolfram.com/Ellipsoid.html) shows disagreement. So two questions arise. 1) What error has been made in the attempt to identify $E_N$ as the surface area of the ellipsoid of semi-principal axes $(\tilde a\chi,\tilde b\chi,\tilde c\chi)$. 2) If $E_N$ is not the surface area of the ellipsoid, what has been calculated?
The problem with the above approach is that, in a heuristic sense, the integral over the $\delta$ function has contributions irrespective of the direction that one approaches the "surface" defined by the zero of the argument. The correct way to generalize the $\delta$ function to a "surface $\delta$ function" is to define it as (minus) the normal derivative of the $\theta$ function. Here is the solution.
\begin{align} E_N &= -\int_{\mathbb{R}^n} d^nx\, n\cdot\nabla\theta(-f), \\ n &= \frac{\nabla f}{|\nabla f|}, \\ f &= \sum_i \frac{x_i^2}{a_i^2}-1, \\ E_N &= \int_{\mathbb{R}^n} d^nx\, \delta(f) |\nabla f|,\\ &= \int_{\mathbb{R}^n} d^nx\, \delta(f) \sqrt{\sum_i \frac{x_i^2}{a_i^4}}. \end{align}