The region common to the interiors of the cylinders x^{2}+y^{2}=1 and x^{2}+z^{2}=1 and the first octant.

bsmart36

bsmart36

Open question

2022-08-13

Finding the volume of this intersection
How to find the volume of this:
The region common to the interiors of the cylinders x 2 + y 2 = 1 and x 2 + z 2 = 1 and the first octant.
I tried finding the volume via double integration and the integration gets too complicated. I want to know how to solve this via triple integration. Usually there are x, y, z′s in both equations and that I can solve with triple but I dont know what to do here. I started of doing x 2 + z 2 = x 2 + y 2 so z = y but dont know what to do with this it doesnt look like the line of intersection is z = y in the diagram given.

Answer & Explanation

nedervdq3

nedervdq3

Beginner2022-08-14Added 13 answers

Step 1
Draw out the surfaces and see that x varies from 0 to 1, y varies from 0 to 1 x 2 , and z varies from 0 to 1 x 2 as well. There are 5 other ways to order the integration but aren't as useful as this ordering (check why) and this ordering makes use of squaring the square root V = 0 1 0 1 x 2 0 1 x 2 d z d y d x.
Step 2
This is the same as V = 0 1 0 1 x 2 1 x 2 d y d x.
Which is the area integral of the function f ( x , y ) = 1 z 2 , the surface of the "top" cylinder. The other cylinder, x 2 + y 2 = 1 serves as the boundary of the integration region. The triple and double integrals are equivalent.
When using the triple integral to find volume, the integrand is simply 1 and integrate wrt all variables whereas with using the double integral to find volume, the integrand is the function f(x,y) and then integrate wrt x and y
As for z = y, it appears on the y z plane as the cylinders make contact at (0, 1, 1)
makeupwn

makeupwn

Beginner2022-08-15Added 2 answers

Explanation:
Let A be the intersection of the two solid cylinders, considered only in the first octant x , y , z 0.
Use Fubini for the computation of the volume. It is clear that x [ 0 , 1 ]. Now for a fixed x we are searching for all (y,z) with the property 0 y , z 1 x 2 . This is a square with side 1 x 2 , its area is ( 1 x 2 ) so the needed volume is Volume ( A ) = 0 1 ( 1 x 2 ) d x = 1 1 3 = 2 3   .

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