Finding the surface area of the spheroid $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$ I'm asked to evaluate this:

What is the surface area of the surface defined by $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$?

I first parameterized it with spherical coordinates and then I took the cross product and then the magnitude of the cross product of $r(\phi) \times r(\theta)$.
\begin{equation}
\left.
\begin{aligned}
x &= \sqrt{3} \cos\theta \sin\phi, \\
y &= \sqrt{3} \sin\theta \sin\phi, \\
z &= 2\cos\phi;
\end{aligned}\right\}\qquad
0 < \theta < 2\pi,\quad
0 < \phi < \pi.
\end{equation}
I took the partials with respect to $\phi$ and $\theta$ and arrived at
\begin{align*}
r(\theta) &= (-\sqrt{3} \sin\theta \sin\phi, \sqrt{3} \cos\theta \sin\phi, 0), \\
r(\phi) &= (\sqrt{3} \cos\theta \cos\phi, \sqrt{3} \sin\theta \cos\phi, -2 \sin\phi).
\end{align*}
Once I found the partials, I took the cross product between them and then took its magnitude to get
$$
\int \sin\phi \sqrt{12\sin^2\phi + 9\cos^2\phi}\, d\phi.
$$
Now, I've tried plenty of trig things but I just can't solve this integral. Can someone help me?

Note that two of the axes of the ellipsoid are the same, and the third is longer.  Thus it is actually a prolate spheroid.  The formula for the surface area of the prolate spheroid $\frac{x^2 + y^2}{a^2} + \frac{z^2}{c^2}=1$ is:
$$ S=2\pi a^2 + 2\pi\frac{ac}{e}\sin^{-1}e $$
where $e$ is the ellipticity $\sqrt{1-\frac{a^2}{c^2}}$.
This formula is derived by considering the spheroid as a surface of revolution about the $z$-axis and doing the usual integration to find its area:
$$S = 2\pi \int_{-c}^{c}{r(z) \sqrt{1+(r'(z))^2}\,dz}$$
where $r(z) = a\sqrt{1-\frac{z^2}{c^2}}$ is the radius of the circular cross section at height $z$.
Source:


*

*Wolfram MathWorld - Prolate Spheroid
