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...
 A: 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
$S=2\pi\int_{0}^{\frac{\pi}{4}}2\cos\theta\cdot 2\cos\theta d\theta=8\pi\int_{0}^{\frac{\pi}{4}}\cos^{2}\theta d\theta=8\pi\int_{0}^{\frac{\pi}{4}}\frac{1}{2}(1+\cos 2\theta)d\theta$
$=4\pi\left[\theta+\frac{1}{2}\sin2\theta\right]_{0}^{\frac{\pi}{4}}=4\pi\left(\frac{\pi}{4}+\frac{1}{2}\right)=\pi^{2}+2\pi$.
A: Using the method of disks($V=π\int_{\alpha}^{\beta} f^2(x) dx$);
The function $y=\sqrt{\frac{2-x^2}{2}}$ has two distinct real roots $(\alpha,\beta)=(-\sqrt{2},\sqrt{2})$ and the solid of revolution of this half-elipsoid will have the same volume with the revolution of the whole ellipsoid as the whole elipsoid will just mark the solid of revolution's boundary one more time(symmetric), without doubling it's volume. So;
$π\int_{-\sqrt{2}}^{\sqrt{2}} (\sqrt{\frac{2-x^2}{2}})^2 dx = π\int_{-\sqrt{2}}^{\sqrt{2}} \frac{2-x^2}{2} dx= \frac {4π\sqrt{2}}{3}$
A: $$ 2\pi\ \int_0^{\sqrt{2}} \sqrt{4-x^{2}} $$ 
$$ \pi \left(x \sqrt{4-x^2} + 4 * \arcsin \frac {x}{2} \right)$$ 
from 0 to $\sqrt{2}$ which gives $\pi^{2}$ + 2$\pi$.
