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April 8th, 2017, 07:37 PM   #1
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Question 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?
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April 9th, 2017, 12:29 PM   #2
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If you look at this carefully you'll find that cross sections parallel to the xy plane have identical limits



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

$V = \displaystyle \int_{-1}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} ~dx~dy~dz$
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cylinders, intersection, volume

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