Cylindrical Coordinates Domain

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Homework Help Overview

The discussion revolves around calculating the volume of a region defined by the paraboloids z = x² + y² and z = 8 - x² - y², using cylindrical coordinates.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the setup of a triple integral for volume, discussing the correct limits and integrand. There are attempts to clarify the integration order and the role of dz in the integral.

Discussion Status

Participants are actively questioning and refining their understanding of the integral setup. Some guidance has been provided regarding the correct form of the integrand and the limits of integration, but there is no explicit consensus on the final expression yet.

Contextual Notes

There are indications of confusion regarding the limits of integration and the proper formulation of the volume element in cylindrical coordinates. The discussion reflects a learning process with multiple interpretations being explored.

DrunkApple
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Homework Statement


Let W be the region between the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2


Homework Equations





The Attempt at a Solution


for each domain,
0 ≤ z ≤ 8
0 ≤ r ≤ 2
0 ≤ θ ≤ 2[itex]\pi[/itex]

So...

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ
 
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Assuming what you are calculating is the volume, yes.
 
wait wait shouldn't it be
[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ
 
DrunkApple said:
wait wait shouldn't it be
[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ

No. That is wrong on several levels. If you are doing a triple integral for volume the integrand is always 1dV = 1 rdzdrdθ.

1. You are missing the dz.
2. Your z limits are not correct; they would depend on r and maybe θ and you would want the dz on the inside so it is integrated first.
3. Your integrand was zupper-zlower in your double integral, which was correct, and that is what you should get when you integrate the 1dz in the triple integral.
 
oh so this is the final one then

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex](8-2[itex]r^{2}[/itex])r dzdrdθ

Yes I forgot dz accidently but I knew that :p
 
DrunkApple said:
oh so this is the final one then

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex](8-2[itex]r^{2}[/itex])r dzdrdθ

Yes I forgot dz accidently but I knew that :p

Still wrong. Your integrand should be 1dV.
 
How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ
 
DrunkApple said:
How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ

No, not correct.
 
but isn't dV = dzdrdθ?
 
  • #10
No, it is r dz dr dθ which you had correct. But now you messed up the limits again.
 
  • #11
ARRRGGGG this thing is driving me nuts
ok phew
one more time

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex]r dzdrdθ
 
  • #12
Finally. Notice that when you work the dz integral what you have left with is exactly the double integral which you had correct in your initial post.
 
  • #13
Yes i got it
 

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