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

In summary, To compute the triple integral for f(x,y), where f(x,y,z) = 2x + 3y, on the tetrahedron T bounded by the coordinate planes and the first octant part of the plane 2x + 3y + z = 6, follow these steps: 1. Draw two diagrams, one of the solid and one of its projection on the x-y plane. 2. Solve the equation for z to get z = 2x + 3y - 6. 3. Determine the axis intersections to find the integral limits. 4. Set up the triple integral, with the innermost integral being with respect to z and limits from z =
  • #1
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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  • #2
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:
  • #3
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.
 

1. What is a triple integral?

A triple integral is a mathematical concept used in multivariable calculus to calculate the volume under a three-dimensional surface. It involves integrating a function over a three-dimensional region defined by three variables.

2. How is a triple integral different from a regular integral?

A triple integral is similar to a regular integral in that it involves finding the area under a curve. However, a triple integral involves integrating over a three-dimensional region, while a regular integral integrates over a one-dimensional interval.

3. What is the purpose of using a triple integral?

The main purpose of using a triple integral is to calculate the volume of a three-dimensional object or the volume under a three-dimensional surface. It is also used in physics and engineering to calculate the mass, center of gravity, and moment of inertia of three-dimensional objects.

4. What are the different types of triple integrals?

There are two types of triple integrals: single integrals and double integrals. A single integral involves integrating over a three-dimensional region defined by three variables, while a double integral involves integrating a function over a two-dimensional region defined by two variables.

5. How do you solve a triple integral?

To solve a triple integral, you first need to set up the limits of integration for each variable. Then, you can use the appropriate integration techniques, such as the Fubini's theorem or the substitution method, to evaluate the integral. Finally, you can solve the integral using algebraic and arithmetic operations.

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