Is the Triple Integral in Cylindrical Coordinates Correctly Solved?

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The discussion centers on solving a triple integral using cylindrical coordinates, with a focus on correctly defining the region of integration and the limits for r and z. The user questions whether the value of r should range from 2 to 4 or if it should be evaluated as the difference between two integrals. A correction is made regarding the boundary of integration, confirming it should be √(2z) instead of 2z. Participants agree on the correctness of the revised integral setup. The conversation concludes with gratitude for the assistance provided.
daphnelee-mh
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Homework Statement
(question is attached below)
Relevant Equations
x^2+y^2=r^2
dV=rdrdzdΘ
1593762965635.png


1593762987337.png


I am trying to solve it using cylindrical coordinates, but I am not sure whether the my description of region E is correct, whether is the value of r is 2 to 4, or have to evaluate the volume 2 times ( r from 0 to 4 minus r from 0 to 2), and whether is okay to take z from r^2/2 to 8
 
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\int_2^8 dz \int_0^{2z} 2\pi \rho^3 d\rho
 
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May I know which method did you use? I have no idea of it, thanks
 
Find attached my handwriting.
 

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anuttarasammyak said:
\int_2^8 dz \int_0^{2z} 2\pi \rho^3 d\rho
Shouldn't the boundary of integration be ##\sqrt{2z}## instead of ##2z##?
 
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Yes. Thanks for correction.
 
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1593775272308.png

Can you help to see whether this is correct ? Thank you
 
Yes it looks correct to me.
 
Thank you so much
 
  • #10
daphnelee-mh said:
View attachment 265726
Can you help to see whether this is correct ? Thank you
The digits of 1008 add to 9, so...
 
  • #11
haruspex said:
The digits of 1008 add to 9, so...
?
 
  • #12
daphnelee-mh said:
?
So 1008 is divisible by 3.
 
  • #13
haruspex said:
So 1008 is divisible by 3.
okay , thank you
 

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