Centroid of a Solid (triple integral)

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SUMMARY

The discussion focuses on calculating the centroid of a tetrahedron in the first octant defined by the equation x + y + z = 1. The user initially attempts to compute the x-coordinate of the centroid using the formula xcenter = (∫∫∫_G x dV) / V but encounters an issue where the volume integral evaluates to zero, leading to an undefined expression. A fellow user suggests correcting the bounds of integration, specifically changing the limits for y to integrate from 0 to (1 - x), which resolves the issue and aligns with the expected centroid coordinates of (1/4, 1/4, 1/4).

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with the concept of centroids in geometry
  • Knowledge of integration limits in multiple dimensions
  • Experience with evaluating volume integrals
NEXT STEPS
  • Review the calculation of centroids for various geometric shapes
  • Practice solving triple integrals with different bounds
  • Learn about the application of centroids in physics and engineering
  • Explore advanced integration techniques such as change of variables in multiple integrals
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Students in calculus, geometry enthusiasts, and anyone studying applications of triple integrals in physics or engineering will benefit from this discussion.

Knissp
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Homework Statement


Find the centroid of the solid:
the tetrahedron in the first octant enclosed by the coordinate planes and the plane x+y+z=1.

Homework Equations



xcenter = \frac{\int\int\int_G x dV}{V}

ycenter = \frac{\int\int\int_G y dV}{V}

zcenter = \frac{\int\int\int_G z dV}{V}

The Attempt at a Solution



I have shown my attempt for xcenter, as the same problem arises for each one.

\frac{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} x dzdydx}{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV}

but \int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV is equal to zero,
so the above expression is undefined.

According to my text, the answer should be (1/4,1/4,1/4). Could someone point out what I did wrong? (Perhaps my bounds of integration?)
 
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Knissp said:
\frac{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} x dzdydx}{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV}

but \int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV is equal to zero,
so the above expression is undefined.

Hi Knissp! :smile:

Try \int_{x=0}^1\int_{y=0}^ {1-x}\int_{z=0}^{1-y-x} :wink:
 
Oh, I see now! Thank you!
 

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