I am trying to calculate the surface area of a prolate ellipsoid: $$a^2x^2+a^2y^2+c^2z^2=1 \ \ \ \ \ \ c>a>0$$ My idea was to represent the ellipsoid by a rotating ellipse around the $z$ - axis, so I parametrised it via spherical coordinates and ended up with the representation: $$\phi(z,\theta) = (r(z)\cos\theta,r(z)\sin\theta,z)$$ with $z \in (-\frac1c,\frac1c)$ and with a radius function $r(z) = \frac{\sqrt{1-z^2c^2}}{a}$. So far, everything seems to be correct. However, using the area formula for a surface of revolution I end up with:
$$\frac{2\pi}{a}\int_{-\frac1c}^{\frac1c} \sqrt{1+\frac{z^2c^2(a+c)(a-c)}{a^2}}dz$$ This is where I get stuck. Is there a rather elementary way of solving this integral? I know what the result should be but I cannot seem to solve this integral correctly. Any help would be appreciated!
Well, we have:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right):=\int_{-\text{n}}^\text{n}\sqrt{1+\text{m}\cdot x^2}\space\text{d}x\tag1$$
Substitute $\text{u}:=\arctan\left(x\cdot\sqrt{\text{m}}\right)$:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right)=\frac{1}{\sqrt{\text{m}}}\cdot\int_{\arctan\left(-\text{n}\cdot\sqrt{\text{m}}\right)}^{\arctan\left(\text{n}\cdot\sqrt{\text{m}}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}=$$ $$\frac{1}{\sqrt{\text{m}}}\cdot\int_{-\arctan\left(\text{n}\cdot\sqrt{\text{m}}\right)}^{\arctan\left(\text{n}\cdot\sqrt{\text{m}}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}\tag2$$
Using:
$$\int\sec^3\left(\text{u}\right)\space\text{d}\text{u}=\frac{\sec\left(\text{u}\right)\cdot\tan\left(\text{u}\right)+\ln\left|\sec\left(\text{u}\right)+\tan\left(\text{u}\right)\right|}{2}\tag3$$
So, we get:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right)=\frac{1}{2\cdot\sqrt{\text{m}}}\cdot\mathcal{Z}\tag4$$
Where:
$$\mathcal{Z}=\left[\sec\left(\text{u}\right)\cdot\tan\left(\text{u}\right)+\ln\left|\sec\left(\text{u}\right)+\tan\left(\text{u}\right)\right|\right]_{-\arctan\left(\text{n}\cdot\sqrt{\text{m}}\right)}^{\arctan\left(\text{n}\cdot\sqrt{\text{m}}\right)}=$$ $$2\cdot\text{n}\cdot\sqrt{\text{m}}\cdot\sqrt{1+\text{n}^2\cdot\text{m}}+\ln\left|\frac{\text{n}\cdot\sqrt{\text{m}}+\sqrt{1+\text{n}^2\cdot\text{m}}}{\text{n}\cdot\sqrt{\text{m}}-\sqrt{1+\text{n}^2\cdot\text{m}}}\right|\tag5$$