Find centroid of region - triple integrals, please

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SUMMARY

The discussion focuses on finding the centroid (x, y, z) of the region R defined by the inequalities 0 ≤ z ≤ 5√(x² + y²) and the cylinder x² + y² = 2x. The user attempted to convert the problem into polar coordinates using the equations x = r cos(θ) and y = r sin(θ), and set up the integral ∫_{-π/2}^{π/2}∫_0^{2cos(θ)}∫_0^{5r} r dz dr dθ. After initial confusion regarding the bounds and the setup, the user confirmed that the method worked upon careful re-evaluation of the integral and limits.

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  • Knowledge of cylindrical coordinates and their applications
  • Basic concepts of centroids in three-dimensional geometry
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Students and educators in calculus, particularly those focusing on multivariable calculus and applications of integrals in finding centroids of regions.

thaer_dude
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Homework Statement



Find the centroid [STRIKE]x[/STRIKE],[STRIKE]y[/STRIKE],[STRIKE]z[/STRIKE] of the region R cut out of the region 0<=z<=5sqrt(x2+y2) by the cylinder x2+y2=2x.

Homework Equations



x2+y2 = r2
x= rcosθ
y= rsinθ

The Attempt at a Solution



Centroid [STRIKE]x[/STRIKE] being Mx/m I'm guessing

I've been working on this problem forever and I'm just not sure how to do it

I tried converting to polars and computing the following integral: <br /> \int_{-pi/2}^{pi/2}\int_0^{2cosθ}\int_0^{5r} r dz dr dθ to get the integral that will be in the denominator (btw if you guys see the upper bound of the 2nd integral as 2cos952 it is 2cosθ)

and then for [STRIKE]x[/STRIKE] I replaced r by r^2cosθ and for [STRIKE]y[/STRIKE] replaced r by r^2sinθ and for [STRIKE]z[/STRIKE] replaced r by z*r

I'm not getting it right, and I'm going at this the completely wrong way or are my bounds incorrect or what? Please help, thanks so much :)
 
Last edited:
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The integral and limits look ok to me. What are you getting for numbers?
 
Thanks for the method confirmation, I tried it again, very carefully, and it worked! :)
 

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