"Calculate the area for the rotation ellipsoid you get by rotating the ellipsoid $\frac{x^2}{2}+y^2 = 1$ around the x-axis."
I solved for x:
$$ y = \pm \sqrt{1-\frac{x^2}{2}} $$
Then did $ y = 0$ to get the integration limits, $\pm \sqrt(2)$.
So I've ended up with an integral I don't know if it's correct, and even if it is, I can't solve it.
$$ 4\pi \int_{\sqrt{2}}^{\sqrt{2}} \sqrt{1-\frac{x^2}{2}} \sqrt{1-\frac{x}{2\sqrt{1-\frac{x^2}{2}}}}dx$$
Yours is not correct. Since $\frac{dy}{dx}=-\frac{x}{2y}$ and $4y^2=2(2-x^2)$, we have
$$\begin{align}2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{1-\frac{x^2}{2}}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{1-\frac{x^2}{2}}\sqrt{1+\left(-\frac{x}{2y}\right)^2}\ dx\\&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{\frac{\color{red}{2-x^2}}{2}\cdot \frac{2(2-x^2)+x^2}{2\color{red}{(2-x^2)}}}\ dx\\&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{\frac{4-x^2}{4}}\ dx\end{align}$$
I hope this helps!
Besides to mathlove's Cartesian version, you can form an integral based on below new change of coordinates too:
$$x=\sqrt{2}\cos t,~~y=\sin t,~~~(0\le t\le 2\pi)$$ Now, you need to follow up:
$$A=\int_0^{2\pi}ydl,~~~dl=\sqrt{x'^2+y'^2}dt$$
