Help Needed: Volume & Centroid of Region, Integral of Parabolas

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I have been looking at these 3 problems for 2 days now and have gotten practically nowhere. Help please!

1. Find the volume V of the solid bounded by the graph of x2 + y2 = 9 and y2 + z2 = 9.

I know that both equations are cylinders on different planes and that I need the intersection. I can not figure out what my bounds are or how to set up the problem. I'm stuck.


2. Find the volume and the centroid (center of mass) of the region that is bounded above by the sphere ρ = a and below by the cone φ = c with 0 < c < π/2. Here you assume constant density.


3. Use the change of variables x = u2 - v2, y = 2uv to evaluate the ∫∫R ydA , where R is the region bounded by the x - axis and the parabolas y2 = 4 − 4x and y2 = 4 + 4x.


I have no idea how to even begin the last two. I've looked through the book and through my notes and can not come up with anything.
 
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One at a time, ok? Fix a value of y. Then x^2=9-y^2 and z^2=9-y^2. So both x and z range from -sqrt(9-y^2) to +sqrt(9-y^2). I.e. for fixed y the x-z cross-section is a SQUARE.
 
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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