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August 16th, 2017, 12:07 PM   #1
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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?
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August 16th, 2017, 01:58 PM   #2
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$\displaystyle x^2+ y^2= z$ 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 06:39 AM.
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August 16th, 2017, 02:08 PM   #3
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Quote:
Originally Posted by Country Boy View Post
$\displaystyle x^2+ y^2= z$ 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 06:40 AM.
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August 17th, 2017, 05:31 AM   #4
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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 06:39 AM.
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