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 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
 August 16th, 2017, 12:58 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,971 Thanks: 807 $\displaystyle x^2+ y^2= z$ 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
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Quote:
 Originally Posted by Country Boy $\displaystyle x^2+ y^2= z$ 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.
But why from r to 4 and not from 0 to r^2?

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: 2,971 Thanks: 807 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. Thanks from sarajoveska Last edited by skipjack; August 17th, 2017 at 05:39 AM.

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