tutor01 said:

The question stipulated that you should use a triple integral and a change of variable - testing the use of Jacobians for triple integral transformations.

Find the vertices of the pyramid (by solving 3 at a time of the 4 simultaneous equations describing the planes), they are:

(x,y,z) = (-4,0,-6); (5,0,-6); (2,6,-6); (-4,0,12)

define a,b, and c such that:

(x,y,z) = (2,6,-6) + ((-4,0,6) - (2,6,-6))*x + ((5,0,-6) - (2,6,-6))*y + ((-4,0,12) - (2,6,-6))*z

(x,y,z) = (2,6,-6) + (-6,-6,12)*a + (3,-6,0)*b + (-6,-6,18)*c

(x,y,z) = A(a,b,c) + C

where A is the matrix [(-6,3,-6);(-6,-6,-6);(12,0,18)] (rows separated by semicolons), and C is the vector (2,6,-6)

The volume of the pyramid is:

V = 1/2*∫(0->1)(∫(0->1)(∫(0->1) 1 det(A) da)db)dc) (**)

= 1/2*det(A) ∫(0->1)(∫(0->1)(∫(0->1) 1 da) db) dc

= 1/2*det(A)

det(A) = -6*(-6*18) + 6*(3*18) + 12*(-18-36)

= 324

hence

V = 324/2

= 162

(**) - The reason for this is that the integral ∫(0->1)(∫(0->1)(∫(0->1) 1 det(A) da)db)dc) actually gives the volume of the object spanned by the three column vectors in the matrix A, which is actually a parallelepiped (a prism with a parallelogram cross section). The pyramid you're interested in is actually obtained by cutting this parallelepiped in half, hence the factor 1/2)