Computing an integral. How to compute $\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS$,
where $\Sigma=\Big\{(x,y,z)\Big| \cfrac{x^2}{2}+\cfrac{y^2}{2}+z^2=1,z\ge 0\Big\}$?
 A: The surface $\Sigma$ has equation $x^2+y^2=r(z)^2$ for $0\leqslant z\leqslant1$, with $r(z)=\sqrt{2}\sqrt{1-z^2}$. For each $z$, this is the equation of a circle with radius $r(z)$, whose length is $2\pi r(z)$, hence the integral is
$$
\int_0^1z\sqrt{1+z^2}\cdot2\pi r(z)\cdot\mathrm dz.
$$
The change of variable $z=\cos^2t$ yields the numerical value $\frac14\pi^2\sqrt2$ (to be checked).
A: Since in $\Sigma$, $z>0$ so our domain is $D:\sqrt{1-(1/2)x^2-(1/2)y^2}>0$ or $D: x^2+y^2<1$. Clearly $S: f=x^2+y^2+2z^2=2$ which is an ellipsoid which we regard just the upper part of it (top of the $x-y$ plane in $\mathbb R^3$). Now $$\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS=\displaystyle\iint\limits_{D}z\sqrt{1+z^2}d\sigma$$ where $d\sigma=\frac{||\nabla f||}{|\partial f/\partial z|}dxdy=\sqrt{1+z_x^2+z_y^2}dxdy$. I think you can evaluate the second double integral over $D$.
A: In Ellipsoidal coordinates, surfaces with constant $\lambda$ have:
$$\frac{x^{2}}{a^{2} + \lambda} +  \frac{y^{2}}{b^{2} + \lambda} + \frac{z^{2}}{c^{2} + \lambda} = 1$$
So that you can take for example: $a^2 = b^2 = 1, c^2 = 0$ and integrate over the other two coordinates (after applying the Jacobian of course).
