Finding the z-coordinate of the center of gravity for a region

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(Moderator's note: thread moved from Calculus and Analysis)

I just need to know if my answer is correct for the following:

Find the z-coordinate of the center of gravity of the region:

(x/(z^3-1))^2 + (y/(z^3+1))^2<=1, 0<=z<=1.

I'm getting 7/16; am I in the ballpark? Thanks for any help.
 
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I got that value as well.
 
Thanks; glad to know I'm at least somewhat following this material.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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