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Cylindrical Coordinates Domain

  • Thread starter DrunkApple
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  • #1
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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
 

Answers and Replies

  • #2
LCKurtz
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Assuming what you are calculating is the volume, yes.
 
  • #3
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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
 
  • #4
LCKurtz
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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.
 
  • #5
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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
 
  • #6
LCKurtz
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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.
 
  • #7
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How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ
 
  • #8
LCKurtz
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How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ
No, not correct.
 
  • #9
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but isn't dV = dzdrdθ?
 
  • #10
LCKurtz
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No, it is r dz dr dθ which you had correct. But now you messed up the limits again.
 
  • #11
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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
LCKurtz
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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
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Yes i got it
 

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