# problem from a uni student (1 Viewer)

#### tutor01

##### Member
Find the volume of the pyramid with base in the plane z=-6 and sides formed by the three planes y=0 and y-x=4 and 2x+y+z=4.

The question comes from Multivariable Calculus by McCallum et al (1997 version) page 239 Q12. It is at the end of a section on triple integrals and use of substitutions in triple integrals.

Anyone who has done Multivariable Calculus at uni and would like to try this one for me is most welcome!

I

#### icycloud

##### Guest
Hey there. I'm not in uni, but I'll attempt the question.

I think it is important to draw a diagram for questions like this.
First you want to find the base of this pyramid.

Plane z=-6
We have y=0, y-x=4 and 2x+y=10
Solve simultaneously: y-x=4 and 2x+y=10 to find the apex of the triangular base (2,6)

now, x=y-4 and x=(10-y)/2

Thus we have 0<=y<=6, and y-4<=x<=(10-y)/2
And with z: -6<=z<=4-2x-y

So we form the triple integral:
∫(0-->6) ∫(y-4-->(10-y)/2)∫(-6-->4-2x-y) dzdxdy
=∫(0-->6) ∫(y-4-->(10-y)/2) 10-2x-y dxdy
=∫(0-->6) -27y + 81 + 9/4y^2 dy {after simplifying}
= [-27/2 y^2 + 81y + 3/4 y^3] (0-->6)
= 162
#

Not sure if this is the right answer though... Check using the V=1/3 Ah formula.

#### tutor01

##### Member
The question stipulated that you should use a triple integral and a change of variable - testing the use of Jacobians for triple integral transformations. I agree that V=1/3Ah works for tetrahedra and that 162units^3 is the final result. Also agree with (and am heaps impressed by) the solution of icycloud.

#### wogboy

##### Terminator
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)