Surface area of ellipsoid given by rotating $\frac{x^2}{2}+y^2=1$ around the x-axis

Question

Calculate the surface area of the ellipsoid that is given by rotating $\frac{x^2}{2}+y^2=1$ around the x-axis. My idea is that if $f(x)=\sqrt{1-\frac{x^2}{2}}$ rotates around the x-axis we will end up with the same figure. The formula for calculating the surface area is: $ 2\pi \int_{-\sqrt{2}}^{\sqrt{2}} f(x)\sqrt{1+f'(x)^2}dx $ and with $f'(x)^2 = \frac{x^2}{4-2x^2}$ the integral becomes $$ \int \sqrt{1-\frac{x^2}{2}}\sqrt{1+\frac{x^2}{4-2x^2}} \space dx$$ Now how do I integrate this?

Answer 1

Looks like one of those times you'll just have to tangle with the algebra monster. $$\begin{align} \sqrt{1-\frac{x^2}{2}}\sqrt{1+\frac{x^2}{4-2x^2}} = \sqrt{\frac{2-x^2}{2}}\sqrt{\frac{4-x^2}{4-2x^2}} \\ = \sqrt{\frac{2-x^2}{2}}\sqrt{\frac{4-x^2}{2(2-x^2)}} \\ = \frac{1}{2} \sqrt{\frac{(2-x^2)(4-x^2)}{2-x^2}}\end{align} \\ = \frac{1}{2}\sqrt{4-x^2} $$ Can you take it from here?