where E = { (x,y,z) : 0 <= z <= 1, 0<= y <=2z, 0 <= x <= z+2 } and then evaluate the integral.
Solution: Perhaps the easiest way to see the domain of integration is to first consider z fixed. Then x varies from 0 to z + 2 and y varies from 0 to 2z. This defines a rectangle whose vertices are (0,0,z), (z + 2, 0, z), (0, 2z, z) and (z+2, 2z,z). This plane is located z units above the x-y plane. A typical one would look like this:
Next, as z increases, the rectangles become larger. If we stack them, one above the other, we get the solid domain of integration:
As for evaluation of the integral, we have
Geometrically, when we hold z fixed and integrate over x and y then we are integrating over the typical plane pictured above. Continuing, the last triple integral is equal to
