My Math Forum Volume of Intersection of Two Cylinders

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 April 8th, 2017, 07:37 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Volume of Intersection of Two Cylinders Given the cylinders:z^2 + x^2 = 1 z^2 + y^2 = 1 Set up and evaluate the triple integral that gives the volume of their intersection. I cannot figure out what the bounds on my integrals should be and I'm not even sure which coordinate system to use to solve this. Can anybody help?
 April 9th, 2017, 12:29 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1041 If you look at this carefully you'll find that cross sections parallel to the xy plane have identical limits $x:(-\sqrt{1-z^2},\sqrt{1-z^2})$ $y:(-\sqrt{1-z^2},\sqrt{1-z^2})$ thus the volume of a cross section with infinitesimal thickness is $dV = (2\sqrt{1-z^2})(2\sqrt{1-z^2}) ~dz= 4(1-z^2)~dz$ $V = \displaystyle \int_{-1}^1 ~4 (1-z^2)~dz = \dfrac{16}{3}$ If you need to cast this as a triple integral just write $V = \displaystyle \int_{-1}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} ~dx~dy~dz$ Thanks from John Travolski

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