Surface area of the ellipsoid $\frac{x^2}{16}+\frac{y^2}{8}+z^2=1$ My professor gave us this question on a calculus II quiz. One of my calculus III pals suggested I use surface integrals, but that tool is not available to us (I don't know how to use it yet, nor do my classmates). We only know how to use the surface area formula for a curve rotated about a line or axis, which is
S= $\int_a^b 2π*f(x)*\sqrt{1+f'(x)^2} dx$
Surface area of the ellipsoid $\frac{x^2}{16}+\frac{y^2}{8}+z^2=1$
We have tried to find an equation for the curve of the ellipse and rotate it about the x-axis, removing the variable $z^2$ and solving $\frac{x^2}{16}+\frac{y^2}{8}=1$ for y, which yielded $y=\sqrt{8-\frac{x^2}{2}}$ 
We then solved to find where the curve crossed the x-axis, at -4 and 4. Then, we took the derivative of that and plugged all components into the formula above. Somehow, we are not getting a correct answer. If anyone else knows of a different approach we could try, please let us know!
Additionally, if you can modify the equation so that there is a coefficient of $z^2$ that makes the problem solvable, I may be able to try it from there.
Thank you!
 A: let $x=4\sin{u}\cos{v},y=2\sqrt{2}\sin{u}\sin{v},z=\cos{u}$.
then
$$E=x''_{u}+y''_{u}+z''_{u}=16\cos^2{u}\cos^2{v}+8\cos^2{u}\sin^2{v}+\sin^2{u}$$
$$F=x''_{v}+y''_{v}+z''_{v}=16\sin^2{u}\sin^2{v}+8\sin^2{u}\cos^2{v}$$
$$G=x'_{u}x'_{v}+y'_{u}y'_{v}+z'_{u}z'_{v}=-16\sin{u}\cos{u}\sin{v}\cos{v}+8\sin{u}\cos{u}\sin{v}\cos{v}$$
so
\begin{align*}EG-F^2&=(128\cos^2{u}+16\sin^2{u}\sin^2{v}+8\sin^2{u}\cos^2{v})\sin^2{u}\\
&=8\sin^2{u}+120\sin^2{u}\cos^2{u}+8\sin^4{u}\cdot \sin^2{v}
\end{align*}
so Surface area is
$$I=\int_{0}^{\pi}\int_{0}^{2\pi}\sqrt{8\sin^2{u}+120\sin^2{u}\cos^2{u}+8\sin^4{u}\cdot \sin^2{v}}dvdu$$
then use Elliptic integral
A: The normal to the surface is
$$
\left(\frac x8,\frac y4,2z\right)
$$
The ratio of the normal component to the $z$-component is
$$
\frac{\sqrt{\frac{x^2}{64}+\frac{y^2}{16}+4z^2}}{2z}
$$
Thus, using $z^2=1-\frac{x^2}{16}-\frac{y^2}8$, the surface area is
$$
\begin{align}
&2\iint_{\frac{x^2}{16}+\frac{y^2}{8}\lt1}\frac{\sqrt{4-\frac{15x^2}{64}-\frac{7y^2}{16}}}{2\sqrt{1-\frac{x^2}{16}-\frac{y^2}8}}\,\mathrm{d}y\,\mathrm{d}x\\
&=2\int_0^4\int_0^{\sqrt{8-\frac{x^2}2}}\sqrt{\frac{256-15x^2-28y^2}{16-x^2-2y^2}}\,\mathrm{d}y\,\mathrm{d}x\\
&=\sqrt2\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{\frac{256-15x^2-14y^2}{16-x^2-y^2}}\,\mathrm{d}y\,\mathrm{d}x
\end{align}
$$
In the last step, we substituted $y\mapsto y/\sqrt2$.
Numerically, this is $83.974845470544$, but I don't have a closed form for the integral.
Unless I am missing some key simplification, this seems a bit advanced for Calc II.
