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Triple integrals

  1. Jun 29, 2003 #1


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    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, plz?
  2. jcsd
  3. Jun 30, 2003 #2
    I dont 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:
    Then determine the axis intersections in order to find the integral limits.
    Set up your triple integral, solve.
    Last edited: Jun 30, 2003
  4. Jun 30, 2003 #3


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    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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