<s>[INST] Calculate the area of the ellipsoid obtained from ellipse $\frac {x^{2}}{2}+y^{2} = 1$ rotated around the $x$-axis So we are about to calculate the area of the ellipsoid around the $x$-axis. 
$$ \frac {x^{2}}{2}+y^{2} = 1 \implies x=\sqrt{2-y^{2}}$$ 
We are squaring it so the sign shouldn't matter.
I was thinking of it may be easier to solve for $x$ and then set up a integration, with the integration from $-1$ to $1$. And use : 
$$ 2\pi\int_{-1}^{1} f(y) \sqrt{1+f'(y)^{2}} dy  $$
I also tried with solving for $y$, but got a very complicated integral. Or maybe I have done errors down the way?
Now I have solved for y instead and got the integral:
$$\pi\ \int_{-\sqrt{2}}^{\sqrt{2}} \sqrt{4-x^{2}} dx = 2\pi\ \int_0^{\sqrt{2}} \sqrt{4-x^{2}} dx $$
I have no idea how to solve this or how I should proceed, variable change? Partial integration? Fill me in...
 [/INST] I am assuming that you want to find the area of the surface generated by revolving the ellipse $\frac{x^2}{2}+y^2=1$ about the x-axis.  In that case,
Then $y^2=1-\frac{x^2}{2}=\frac{2-x^2}{2}$, so $y=\frac{1}{\sqrt{2}}\sqrt{2-x^2}$ on the top half of the ellipse; and
$S=\int_{-\sqrt{2}}^{\sqrt{2}}2\pi y\sqrt{1+(\frac{dy}{dx})^2}\;dx=\int_{-\sqrt{2}}^{\sqrt{2}}2\pi\left(\frac{1}{\sqrt{2}}\sqrt{2-x^2}\right)\sqrt{1+\frac{x^2}{2(2-x^2)}}\;dx$
$\;\;\;=2\int_{0}^{\sqrt{2}}2\pi\frac{\sqrt{2-x^2}}{\sqrt{2}}\frac{\sqrt{4-x^2}}{\sqrt{2}\sqrt{2-x^2}}dx = 2\pi\int_{0}^{\sqrt{2}}\sqrt{4-x^2}dx$.  
Now let $x=2\sin\theta$ and $dx=2\cos\theta d\theta$ to get
$S=2\pi\int_{0}^{\frac{\pi}{4}}2\cos\theta\cdot 2\cos\theta d\theta=8\pi\int_{0}^{\frac{\pi}{4}}\cos^{2}\theta d\theta=8\pi\int_{0}^{\frac{\pi}{4}}\frac{1}{2}(1+\cos 2\theta)d\theta$
$=4\pi\left[\theta+\frac{1}{2}\sin2\theta\right]_{0}^{\frac{\pi}{4}}=4\pi\left(\frac{\pi}{4}+\frac{1}{2}\right)=\pi^{2}+2\pi$.
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