Finding the Centroid of a Half-Cone: Tips and Troubleshooting

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SUMMARY

The discussion focuses on finding the centroid of a half-cone defined by the inequalities sqrt(x^2 + y^2) ≤ z ≤ 1 and x ≥ 0. The user, Wakko, attempted to use cylindrical coordinates with limits of integration set as 0 ≤ r ≤ 1, -π/2 ≤ θ ≤ π/2, and 0 ≤ z ≤ 1. The mass was calculated using the triple integral of [r dr dθ dz]. The user suspects that the error lies in the limits of integration or the choice of coordinate system, prompting a request for advice from the community.

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  • Cylindrical coordinates in multivariable calculus
  • Triple integrals for volume and mass calculations
  • Understanding of centroids in geometric shapes
  • Basic integration techniques
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wakko101
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The question is this: find the centroid of the half-cone
sqrt(x^2 + y^2) <(oet) z <(oet) 1
and x >(oet) 0
(oet being or equal to, I apologize for the lack of sophistocated symbols). I thought I was doing it correctly, but my answers do not match up with those in the book.

I assumed that it would be wise to use cylindrical coordinates which would make r <(oet) 1 and z <(oet) 1. Now, for the limits of integration, I used

0 < r < 1,
-pi/2 < theta < pi/2 and
0 < z < 1.

The mass is the triple integral of [r dr dtheta dz]. And to find the centroid coordinates I take the triple integral of my particular coordinate multiplied by r and over the mass. I'm pretty sure I know how to integrate, so I assume that my mistake was in my limits or perhaps my choice of coordinate system.

Advice please and thank you...

Cheers,
Wakko. =)
 
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It doesn't sound like you are doing anything obviously wrong. Maybe you could post your result and let people check it.
 

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