Volume of a solid bound by four surfaces

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Homework Statement


Compute the volume of the solid bounded by the four surfaces x+z=1,x+z=−1,z=1−y2,z=y2−1

Homework Equations


Fubini's theorem?

The Attempt at a Solution


I have tried to visualize this solid and define the limits; when I attempted to integrate by dxdzdy (in that order), I set the limits for the first integral as -z-1 to -z+1, the second integral as y^2-1 to 1-y^2, and third integral as -1 to 1, but the I'm not getting a value due to the second integral evaluating as zero.

I feel almost certain that the problem mainly has to do with defining the correct limits and order of integration, but I'm having a bit of trouble here.

Thanks.
 
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How can the second integral be zero? Some sign error?
The order of integrals is useful.
 
Because y2-1=-(1-y^2) and there are no y terms after the first integral, the integral equals 0.
 
If you integrate f(y)=1 from y=-5 to y=5, is the result 0 - or maybe 10?
 
Wow. I can't believe that I made such a dumb mistake. I stared at it for at least 10 min straight.

Thanks for putting up with such a trivial problem. X(
 
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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