August 16th, 2017, 11:07 AM  #1 
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  Triple integral
x^2+y^2=z 0<z<4 if I substitute z in the first equation I get x^2+y^2=4. So my bounds would be: 0<phi<2pi 0<r<2 and now I'm not sure whether z will be from 0 to 4 or 0 to r^2? 
August 16th, 2017, 12:58 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
$\displaystyle x^2+ y^2= z$ is the surface of a cone. Yes, for each z, the cross section is the circle $\displaystyle x^2+ y^2= z$ which has center at (0, 0, z) and radius $\displaystyle \sqrt{z}$. Using cylindrical coordinates, the most obvious way to do it is to take $\displaystyle \theta$ running from 0 to $\displaystyle 2\pi$, z from 0 to 4, and, for each z, r from 0 to $\displaystyle \sqrt{z}$. If you really want to integrate with respect to z first and then r, take r from 0 to 2, and, for each r, z from r to 4. Last edited by skipjack; August 17th, 2017 at 05:39 AM. 
August 16th, 2017, 01:08 PM  #3  
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  Quote:
Last edited by skipjack; August 17th, 2017 at 05:40 AM.  
August 17th, 2017, 04:31 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
I misread the problem  this is a paraboloid, not a cone. The integral should be from $\displaystyle r^2$ to 4, not r to 4. But 0 to $\displaystyle r^2$ would be below, so outside the cone.
Last edited by skipjack; August 17th, 2017 at 05:39 AM. 

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