(873, #13) Evaluate the triple integral

,

where E is bounded by the planes x = 0, y = 0, z = 0 y+z = 1, and x + z = 1.

Solution: The inequalities that define E are

For fixed z the set of (x, y) that satisfy the first two inequalities form a rectangle with corners at (0,0,z), (1 - z, 0 ,z), (0, 1 - z, z) and (1 - z, 1 - z, z). The rectangle corresponding to z = 0.4 is displayed at the left below. As z varies from 0 to 1we get the pyramid outlined at the right.

Consequently,

The integral is equal to 1/12:

lprint(int( z,x));
lprint(int( z,x = 0..1-z));
z*x
z-z^2

lprint(int(int( z,x = 0..1-z),x));
lprint(int(int( z,x = 0..1-z),x = 0 .. 1 -z));
(z-z^2)*x
z-2*z^2+z^3

lprint(int(int(int( z,x = 0..1-z),x = 0 .. 1 -z),z));
lprint(int(int(int( z,x = 0..1-z),x = 0 .. 1 -z),z = 0..1));
1/2*z^2-2/3*z^3+1/4*z^4
1/12

Maple Code

by: gm