Area calculation of ellipse $x^2/2+y^2=1$ Calculate the area of the ellipse that you get when you rotate the ellipse $$\frac{x^2}{2}+y^2= 1$$ around the x-axis.
My approach has been to use the formula for rotation area from $-2$ to $2$. But this gives a complicated integral and I'm unsure about the limits $-2$ to $2$.
I would really appriciate a detailed solution to this

Surface
$\frac{x^2}{2}+y^2=1$
$S=2\pi\int_{-a}^{a}yds=4\pi\int_{0}^{a}yds$
$y'= -\frac{x}{2y}$
$ds=\sqrt{1+y'^2}dx=\sqrt{1+\frac{x^2}{4y^2}}dx=\frac{\sqrt{4y^2+x^2}}{2y}dx$ 
$S=4\pi\int_{0}^{\sqrt{2}}y\frac{\sqrt{4y^2+x^2}}{2y}dx=2\pi\int_{0}^{\sqrt{2}}\sqrt{4y^2+x^2}dx=2\pi\int_{0}^{\sqrt{2}}\sqrt{4-x^2}dx$
Substitution: $x=2sint, dx=2cost\cdot dt$
$S=2\pi\int_{0}^{\pi/4}\sqrt{4-4sin^2t}\cdot 2costdt=8\pi\int_{0}^{\pi/4}cos^2tdt=$
$=4\pi\int_{0}^{\pi/4}(1+cos2t)dt=4\pi[t+\frac{sin2t}{2}]_{0}^{\pi/4}dt=\pi(\pi+2)$
