How do I calculate surface of ellipse rotating around x axel? How to calculate surface of ellipse rotating around x axel when ellipse equation is :
$y^{2}+4x^{2} = 36$
I started with this formula for calculating surface of rotating object around x axel 
$ \int_{a}^{b}f(x) \sqrt{1 + f'(x)^{2}} dx$ 
but i always end up with complicated solution, I believe there is easier way of solving this but I am missing it. Any hint and help is welcome. 

The surface area is given by
\begin{equation}
A=2\pi\int_{a}^{b}y\sqrt{1+(y^\prime)^2}\,dx
\end{equation}
with $y=2\sqrt{9-x^2}$, $a=-3,\,b=3$. Because of symmetry we can double the integral from $0$ to $3$.
Since $\sqrt{1+(y^\prime)^2}=\sqrt{\dfrac{9-3x^3}{9-x^2}}$ we have
\begin{eqnarray}
A&=&4\pi\int_{0}^{3}2\sqrt{9-x^2}\sqrt{\dfrac{9+3x^2}{9-x^2}}\,dx\\
&=&8\sqrt{3}\pi\int_{0}^{3}\sqrt{3+x^2}\,dx
\end{eqnarray}
which can be done by trigonometric substitution or by using the formula
\begin{equation}
\int\sqrt{a^2+x^2}\,dx=\dfrac{1}{2}\left[x\sqrt{a^2+x^2}+a^2\ln\vert x+\sqrt{a^2+x^2}\vert\right]+c
\end{equation}
If I substituted correctly, this gives $A=\left[72+12\sqrt{3}\ln\left(1+\frac{2\sqrt{3}}{3}\right)\right]\pi$
