# Computing an integral.

How to compute $\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS$, where $\Sigma=\Big\{(x,y,z)\Big| \cfrac{x^2}{2}+\cfrac{y^2}{2}+z^2=1,z\ge 0\Big\}$?

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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 $D$.

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I don't think this is a good method. too much computing work to do. –  Leitingok Aug 24 '12 at 14:30
@Leitingok: I wrote here what I had learnt about the surface integral. Your function in integrand is notable and it takes time to be done properly. :) –  Babak S. Aug 24 '12 at 14:50
you are right. I worked out –  Leitingok Aug 26 '12 at 11:00
@Leitingok: Just out of curiosity, I wonder whether you worked out a numerical value for your integral starting from the indications in Babak's post, and, if you did, what is this numerical value. –  Did Aug 26 '12 at 11:38
@did: Thanks. As I have seen through questions, your answers have been of the neat and perfect ones. I love this way of doing Maths. I am just at the beginning of this way here. :) –  Babak S. Aug 28 '12 at 13:55

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).

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+1 Numerical value of $\frac{\pi^2}{2\sqrt{2}}$ confirmed. –  Sasha Aug 25 '12 at 0:15
@Sasha Thanks for the appreciation and for the confirmation. –  Did Aug 25 '12 at 8:15
@did, by Babak Sorouh's way, I get diffenrent answer form yours. –  Leitingok Aug 26 '12 at 4:54
@Leitingok: Well, too bad. What do you suggest I should do about it? –  Did Aug 26 '12 at 7:52

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).

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