Question

Let T be a tetrahedron with vertices

(0,0,0),(1,0,0),(0,2,0),(0,0,3) and f(x,y,z)=x+y. Calculate the
integral

[the integration limits for z should be [0; 3-3x-3/2y], for y
[0; 2-2x] and for x [0; 1],

Answer 1

The integral of f(x, y, z) = x + y over the tetrahedron T with the given limits is ∫[0 to 1] ∫[0 to 2-2x] ∫[0 to 3-3x-3/2y] (x + y) dz dy dx = 1/4.

Step-by-step explanation:

In this triple integral, we are evaluating the function f(x, y, z) = x + y over the specified tetrahedron T with the given limits for x, y, and z. The integral is set up in a way that the innermost integral is with respect to z, the middle integral with respect to y, and the outermost integral with respect to x.

Starting with the innermost integral, the limits for z are [0 to 3-3x-3/2y]. Moving to the middle integral, the limits for y are [0 to 2-2x]. Finally, the outermost integral has limits for x from [0 to 1].

The integral is solved step by step, following the order of integration. The resulting value, 1/4, represents the volume-weighted average of the function x + y over the specified tetrahedron T.

This calculation involves iterated integration, where each integration accounts for one dimension of the three-dimensional space defined by the tetrahedron. The given limits ensure that the integration is performed over the region defined by the vertices of the tetrahedron. The final answer, 1/4, provides the numerical result of the specified triple integral.