The integral with the prescribed order of integration would be
[tex]\displaystyle \iiint_D f(x,y,z) \, dV = \int_{-\sqrt5}^{\sqrt5} \int_{x^2}^5 \int_{4-y}^{y-6} f(x,y,z) \, dz \, dy \, dx[/tex]
• z is bounded between two planes, z = 4 - y and z = y - 6. These planes meet in a line with coordinates y = 5 and z = -1. Deciding which plane lies above the other comes down to checking the value of z for some y-coordinate of any point that we know belongs to D. The parabolic face y = x² contains the origin (0, 0, 0), so we know there are points in D with y-coordinate = 0. When y = 0,
z = 4 - y = 4
z = y - 6 = -6
and this tells us that 4 - y is the "upper" plane.
• y is bounded between the parabolic cylinder y = x² and the plane y = 5, which we determined earlier when finding where the two planes intersect. This also tells us that 0 ≤ y ≤ 5 for any point in D.
• x is bounded between -√5 and +√5, since at most y = 5, so that
y = 5 = x² ⇒ x = ±√5
