Moment of Inertia around z-axis: 1-x^2, 4x+3y+2z+12

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Homework Statement


Find the moment of inertia around the z-axis of the solid bounded by x=0, y=0,z=0,y=1-x^2 and 4x+3y+2z+12 assume density=1.


Homework Equations


please refer to my attachment.


The Attempt at a Solution


Attached
 

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You have written 4x+3y+2z+12, which I'm interpreting as 4x+3y+2z=12 => x/3+y/4+z/6=1. (All right, now I got the correct eqn from the pdf file.)

How can x go from 0 to 3? In the 1st octant, x is from 0 to 1. The rest looks OK.
 
Yes your assumption above was correct, and thank you for your assistance!
 

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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