Solving triple integrals in a tetrahedron: How do I get started?

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SUMMARY

The discussion focuses on computing the triple integral of the function f(x,y,z) = 2x + 3y over the tetrahedron defined by the coordinate planes and the plane equation 2x + 3y + z = 6. The solution process involves drawing diagrams, solving for z, determining integral limits, and setting up the triple integral. The final integral is expressed as ∫(x=0 to 3) ∫(y=0 to 2-(2/3)x) ∫(z=0 to 6-2x-3y)(2x+3y) dz dy dx.

PREREQUISITES
  • Understanding of triple integrals in multivariable calculus
  • Familiarity with the concept of tetrahedrons and their geometric properties
  • Ability to solve equations for variables in three-dimensional space
  • Knowledge of setting up and evaluating multiple integrals
NEXT STEPS
  • Study the method of setting up triple integrals in different coordinate systems
  • Learn about the geometric interpretation of triple integrals
  • Explore the use of Jacobians in changing variables for multiple integrals
  • Practice solving triple integrals involving different functions and boundaries
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Students and professionals in mathematics, particularly those studying calculus, as well as educators teaching multivariable calculus concepts.

Dx
Compute triple integral f(x,y)dV given f(x,y,z)=2x+3y. T is the tetrahedron bounded by the coordinate planes and first octant part of the plane with equation 2x + 3y + z = 6.

how do i solve for this, can someone get me started halfway, please?
Dx
 
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I don't know if you have solved this problem already, but here goes:

The first step you will want to do is draw two diagrams, one of the solid, and one projection of it on the x-y plane.

Then solve the equation for z (or x or y whichever is going to produce the easiest integral outcome).

In this case I would go for the z solution:
z=2x+3y-6
Then determine the axis intersections in order to find the integral limits.
Set up your triple integral, solve.
 
Last edited:
The "innermost" integral will be with respect to z (dz) and will have limits z= 0 to z= 6- 2x- 3y.

"projecting" down to the x,y plane (z= 0) gives the line where the plane 2x+ 3y+ z= 6 crosses the x,y plane: 2x+ 3y= 6. We need to integrate over the triangle with edges x=0, y=0, 2x+ 3y= 6.

Solving for y, y= 2- (2/3)x. The second integral will be with respect to y (dy) and will have limits y= 0 and y= 2-(2/3)x.

Finally, project to the x-axis itself. The line 2x+3y= 6 crosses the x-axis when y= 0: 2x= 6 or x= 3. The final integral will be with respect to x (dx) and will have limits x= 0, x= 3.

The triple integral you want is

integral (x=0 to 3) integral (y=0 t0 2-(2/3)x) integral (z= 0 to 6- 2x+3y)(2x+3y)dzdydx.
 

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