Surface area of an elliptic paraboloid The elliptic paraboloid has height h, and two semiaxes a, b. How to find its surface area? Does it possible to use a direct formula without integrals?
 A: The elliptic paraboloid is represented parametrically as follows:
$$x=a \sqrt{u} \cos{v}$$
$$y=b \sqrt{u} \sin{v}$$
$$z=u$$
The surface area of this object is given by
$$\int_0^h du \: \int_0^{2 \pi} dv \: \sqrt{E \,G - F^2}$$
where the 1st fundamental form is given by
$$E=1+\frac{a^2 \cos^2{v} + b^2 \sin^2{v}}{4 u}$$
$$F=\frac{1}{4} (b^2-a^2) \sin{2 v}$$
$$G = (a^2 \sin^2{v}+b^2 \cos^2{v}) u$$
The integral simplifies to
$$\int_0^h du \: \left [\sqrt{b^2 \left(a^2+4 u\right)} E\left(\frac{4 \left(b^2-a^2\right) u}{b^2 \left(a^2+4 u\right)}\right)+\sqrt{a^2 \left(b^2+4 u\right)} E\left(\frac{4
   (a-b) (a+b) u}{a^2 \left(b^2+4 u\right)}\right)\right ]$$
where $E$ is the elliptic integral defined as
$$E(m) = \int_0^{\pi/2} dx \sqrt{1-m \sin^2{x}}$$
This is about the best you'll do as far as I can see.
A: The function in question is $z=\frac{x^2}{a^2}+\frac{y^2}{b^2}$.
It is more convenient to work in polar coordinates, so that $x=a r \cos \theta,y=b r\sin \theta,z=r^2$, and the parameters vary like so: $0\leq \theta \leq 2\pi,0 \leq r \leq \sqrt{h}$.
You can find the surfrace area according the the formula $A=\int_0^{2 \pi} d \theta \int_0^\sqrt{h} dr \| {\bf x}_r \times {\bf x}_\theta \|$, where ${\bf x}=(x,y,z)$. I couldn't evaluate this integral analytically.
