Triple Integral: Volume Between Y=1-X & Y=Z^2-1

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Homework Statement


Volume between Y=1-X and Y = Z^2 -1

The Attempt at a Solution



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Sorry, not a good drawer.

0 < y < 1
0< x < SqRoot: 1-y

-1 < y < 0
0 < Z < Sqroot: 1 + y

I'm not even sure if this is right, we just started the triple integrals. Please Help.
 
Last edited by a moderator:
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The problem is that the plane y= 1- x and the parbolic cylinder [math]y= z^2- 1[/math] are not boundaries for a bounded region.
 
So there is not enough information?
 

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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