Question

Find the lower limit of x in the triple integral ${{{ \displaystyle \bold \red{ \int \int \int_D}}}}$f(x,y,z) dV if D is solid bounded z=4-y, z=y-6 and y=x², taking the order of integration as dzdydx.



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Answer 1

The integral with the prescribed order of integration would be

$\displaystyle \rm \iiint_D f(x,y,z) \, dV = \int_{-\sqrt5}^{\sqrt5} \int_{x^2}^5 \int_{4-y}^{y-6} f(x,y,z) \, dz \, dy \, dx$

• 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