Volume of a tetrahedron of a function

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Homework Help Overview

The discussion revolves around calculating the volume integral of the function T = z^2 over a tetrahedron defined by specific corner points in three-dimensional space. The problem involves understanding the setup of the volume integral and determining the appropriate bounds for integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the volume integral and the necessary bounds for integration. There is an exploration of how to incorporate the function T = z^2 into the volume calculation. Some participants express uncertainty about their initial attempts and seek clarification on the bounds.

Discussion Status

The discussion is ongoing, with some participants providing insights into the bounds of integration. There is a recognition of the need for further exploration of the integration limits, particularly in relation to the tetrahedron's geometry. While some guidance has been offered, explicit consensus on the approach has not been reached.

Contextual Notes

Participants are working within the constraints of the first octant and the geometric boundaries defined by the tetrahedron. There is an acknowledgment of the need for confidence in the answers provided, indicating a learning process in progress.

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Homework Statement


Calculate the volume integral of the function T = z^2 over the tetrahedron with corners at
(0,0,0), (1,0,0) , (0,1,0), and (0,0,1)

The Attempt at a Solution



z to x (1,0,-1)
z to y (0,1,-1)

Then i crossed them to get (1,1,1)

Found the plane n dot (x-1, y , z) = x+y+z-1=0

Normally i'd have no problem from here, but I'm not sure how I'm supposed to incorporate the T=z^2
 
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So you're dealing with the volume integral:

\int \int \int z^2 dz dy dx

First thing is to find the bounds. You know that this is the volume contained in the first octant and bound by x+y+z-1=0. This can be rearranged to z=1-x-y, and the lower bound of z is zero (because it's in the first octant). So we can get our first bounds:

\int \int \int_0^{1-x-y} z^2 dz dy dx

From there can you see the other two sets of bounds?
 
okay, so when z is 0 y= 1-x and x goes from 0 to 1. If that is right i did this initially and i need to have more confidence in my answers...

thank you for the clarification
 
No worries. Have a great day!
 

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