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Find the volume of the solid bounded by the cylinders $x^2 + z^2 = a^2$ and $y^2 + z^2 = a^2$

I couldn't figure out which region is it and I can't even "see" the cilinders. Can you give me a little help with the problem interpretation?

thanks

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  • $\begingroup$ Try viewing it from "the top". Looking down on the $xy$-plane what would you see? More generally, what is the intersection of your set with a plane $z=c$? $\endgroup$ – mrf Sep 24 '13 at 12:00
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    $\begingroup$ I've find this and it is what I was seeing: math.tamu.edu/~Tom.Kiffe/calc3/newcylinder/2cylinder.html but I can't see a good method to calculate the volume without double integral (I can't use it yet) $\endgroup$ – Giiovanna Sep 24 '13 at 12:07
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If you intersect your solid with the plane $z=\text{const}$ you will get the square $$|x| \le \sqrt{a^2-z^2},\quad |y| \le \sqrt{a^2-z^2}$$ with area $A(z)=4(a^2-z^2)$. Imagine the solid being made up of "infinitesimally thin" slices of height $dz$. The volume of such a slice will then be $$ dV = A(z)\,dz = 4(a^2-z^2)\,dz $$ and the total volume will be $$ V = \int_{-a}^a dV = \int_{-a}^a 4(a^2-z^2)\,dz $$ since we have to let $z$ vary from $-a$ to $a$ in order to cover the entire solid.

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