Triple Integral in Cylindrical Coordinates

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Homework Help Overview

The problem involves finding the mass and center of mass of a solid bounded by a paraboloid and a plane, using cylindrical coordinates. The solid has a constant density.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of the triple integral for mass and center of mass, questioning the volume element used in cylindrical coordinates.
  • Some participants express uncertainty about the correctness of their integrals and the method for calculating the center of mass components.
  • There is a suggestion to convert to Cartesian coordinates for certain calculations.

Discussion Status

Participants are exploring different approaches to set up the integrals correctly. Some have provided hints and guidance regarding the volume element and the necessity of coordinate conversion for specific calculations. There is an ongoing examination of assumptions related to the center of mass in cylindrical coordinates.

Contextual Notes

There is a mention of the condition that the density is constant and that the variable 'a' is greater than zero. Participants also note the implications of having r equal to zero in relation to the theta component.

daveyman
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Homework Statement


Find the mass and center of mass of the solid S bounded by the paraboloid [tex]z=4x^2+4y^2[/tex] and the plane [tex]z=a\;\;(a>0)[/tex] if S has constant density K.


Homework Equations


In cylindrical coordinates, [tex]x^2+y^2=r^2[/tex].


The Attempt at a Solution


In order to find the mass, I tried
[tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K\;drdzd\theta[/tex]
This didn't seem to work, though.
 
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daveyman said:
… I tried
[tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K\;drdzd\theta[/tex]
This didn't seem to work, though.

Hi daveyman! :smile:

Hint: the volume element isn't drdzdθ. :wink:
 
Okay, so this is the mass:

[tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}\text{Kr}\;\;drdzd\theta = \frac{1}{8} a^2 K \pi[/tex]
 
Last edited:
Now that I have the mass (m), I want to find the r component of the center of mass. I tried

[tex]\frac{1}{m}\int _0^{2 \pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K r^2drdzd\theta[/tex]

This is incorrect. What am I doing wrong?
 
Last edited:
daveyman said:
Okay, so now that I have the mass (m), I want to find the r component of the center of mass. I tried

[tex]\frac{1}{m}\int _0^{2 \pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K r^2drdzd\theta[/tex]

This is incorrect. What am I doing wrong?

That should come out as 0, which is the r component of the centre of mass. :wink:
 
Mathematica is telling me that this integral is not equal to zero. Have I made a mistake?
I've included the output as an attachment.
 

Attachments

daveyman said:
Mathematica is telling me that this integral is not equal to zero. Have I made a mistake?
I've included the output as an attachment.

ah … just realized … you can't do centre of mass that way in cylindrical coordinates … you'd have to convert to x and y coordinates first.

(but the z coordinate will be ok :smile:)
 
tiny-tim said:
ah … just realized … you can't do centre of mass that way in cylindrical coordinates … you'd have to convert to x and y coordinates first.

(but the z coordinate will be ok :smile:)

Right, that makes sense. So how can I solve for the r and theta components in cylindrical coordinates? I know qualitatively that they should be zero, but how do I show this mathematically?
 
daveyman said:
Right, that makes sense. So how can I solve for the r and theta components in cylindrical coordinates? I know qualitatively that they should be zero, but how do I show this mathematically?

I think you're perfectly entitled to say that it's obvious. :smile:

(btw, if r is zero, then θ is meaningless :wink:)
 
  • #10
tiny-tim said:
(btw, if r is zero, then θ is meaningless :wink:)

Very true. I think that's it. Thank you!
 

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