I'll reply to this with what I can help you with since no one else has replied yet. I remember posting a very similar problem a while ago but I've forgotten all about it now.
The following is just what I can come up with right now so its probably best to wait for further replies. I would also suggest including the answer which you seem to have.
When doing these sorts of problems I think you need to view projections wrt a single plane - for example yhe yz plane or the xy plane. This is just arbitrary but I would 'project' everything onto the xz plane. So you can 'ignore' the equation x^2 + z^2 = 1 for now. You also have y^2 + z^2 = 1 where y is the 'vertical' part of the coordinate system since the projection is onto the xz plane. Solving for y yields y = +/-sqrt(1-z^2).
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V = \int\limits_{ - 1}^1 {\int\limits_{ - \sqrt {1 - x^2 } }^{\sqrt {1 - x^2 } } {\int\limits_{ - \sqrt {1 - z^2 } }^{\sqrt {1 - z^2 } } {dydzdx} } } <br />
Note: I checked on mathworldhttp://mathworld.wolfram.com/SteinmetzSolid.html) and the volume is given by the integral above. I checked it so that I could verify my answer. Although I have little idea as to how this why the limits are the way that they are. I mean, if I had to guess I would've gone with those limits but I just don't understand them. The limits seem like they would only work if the cylinders intersected in perfect circles but this doesn't appear to be the case.
mugizee - don't worry about the paragraph above this, those are just things that have been troubling me. Also, you see that I have used a triple integral but it really just reduces to a double integral if that's what you really need to use. This is because the integrand is equal to one and so after the first integration it's equivalent to starting off with a double integral. I used a triple integral because I prefer to use them for volumes.
So V = \int\limits_{ - 1}^1 {\int\limits_{ - \sqrt {1 - x^2 } }^{\sqrt {1 - x^2 } } {\int\limits_{ - \sqrt {1 - z^2 } }^{\sqrt {1 - z^2 } } {dydzdx} } }
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= \int\limits_0^{2\pi } {\int\limits_{ - \sqrt {1 - \sin ^2 \theta } }^{\sqrt {1 - \sin ^2 \theta } } {\int\limits_{ - \sqrt {1 - \cos ^2 \theta } }^{\sqrt {1 - \cos ^2 \theta } } {rdydrd\theta } } } <br />...where x = sin(theta) and z = cos(theta), assigned this way because the y-axis is the 'vertical' axis wrt to the chosen projection
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= \int\limits_0^{2\pi } {\int\limits_{ - \cos \theta }^{\cos \theta } {\int\limits_{ - \sin \theta }^{\sin \theta } {rdydrd\theta } } } <br />
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= 2\int\limits_0^{2\pi } {\int\limits_{ - \cos \theta }^{\cos \theta } r } \sin \theta drd\theta <br />
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= 2\int\limits_0^{2\pi } {\cos ^2 \theta \sin \theta d\theta } <br />
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= - \frac{2}{3}\left[ {\cos ^3 \theta } \right]_0^{2\pi } <br />
Ok there's an error somewhere since the 'answer' comes out to be zero. It should be 16/3. Perhaps someone else could give you some help with this because I can't get the answer.
